{"id":13703,"date":"2018-08-24T17:15:28","date_gmt":"2018-08-24T17:15:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13703"},"modified":"2020-11-22T19:58:52","modified_gmt":"2020-11-22T19:58:52","slug":"logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/chapter\/logarithmic-properties\/","title":{"raw":"Walkthrough of Unit 9: Logarithmic Functions","rendered":"Walkthrough of Unit 9: Logarithmic Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Convert between logarithmic to exponential form.<\/li>\r\n \t<li>Evaluate logarithms.<\/li>\r\n \t<li>Use common logarithms to model real-world problems.<\/li>\r\n \t<li>Use natural logarithms to model real-world problems.<\/li>\r\n \t<li>Identify the domain of a logarithmic function.<\/li>\r\n \t<li>Graph logarithmic functions.<\/li>\r\n \t<li>Use the product rule for logarithms.<\/li>\r\n \t<li>Use the quotient rule for logarithms.<\/li>\r\n \t<li>Use the power rule for logarithms.<\/li>\r\n \t<li>Expand logarithmic expressions.<\/li>\r\n \t<li>Condense logarithmic expressions.<\/li>\r\n \t<li>Use the change-of-base formula for logarithms.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<figure id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/> <strong>Figure 1.\u00a0<\/strong>Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137557013\">In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote] One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0like those shown in the picture above. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0whereas the Japanese earthquake registered a 9.0.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013.[\/footnote]<\/p>\r\n<p id=\"fs-id1165137760714\">The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\r\n\r\n<h2>Convert from logarithmic to exponential form<\/h2>\r\n<section id=\"fs-id1165137644550\">\r\n<p id=\"fs-id1165135192781\">In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex], where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?<\/p>\r\nWe have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3, since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/> <b>Figure 2<\/b>[\/caption]\r\n<p id=\"fs-id1165137662989\">Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\r\n<p id=\"fs-id1165137404844\">We read a logarithmic expression as, \"The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,\" or, simplified, \"log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.\" We can also say, \"<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,\" because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as \"log base 2 of 32 is 5.\"<\/p>\r\n<p id=\"fs-id1165137597501\">We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\r\n\r\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]<\/div>\r\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive.<span id=\"fs-id1165137696233\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137400957\">Because logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex], using parentheses to denote function evaluation, just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<span id=\"fs-id1165137771679\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137575165\">Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.<\/p>\r\n\r\n<div id=\"fs-id1165137472937\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Definition of the Logarithmic Function<\/h3>\r\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165137584967\">For [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137433829\" class=\"equation\" style=\"text-align: center\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137893373\">where,<\/p>\r\n\r\n<ul id=\"fs-id1165135530561\">\r\n \t<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\" or the \"log base <em>b<\/em>\u00a0of <em>x<\/em>.\"<\/li>\r\n \t<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137547773\">Also, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\r\n\r\n<ul id=\"fs-id1165137643167\">\r\n \t<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137677696\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1549475\"><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\r\n<p id=\"fs-id1165137653864\"><em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137874700\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137806301\">How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form.<\/h3>\r\n<ol id=\"fs-id1165137641669\">\r\n \t<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\r\n \t<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_01\" class=\"example\">\r\n<div id=\"fs-id1165135570363\" class=\"exercise\">\r\n<div id=\"fs-id1165137557855\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Converting from Logarithmic Form to Exponential Form<\/h3>\r\n<p id=\"fs-id1165137580570\">Write the following logarithmic equations in exponential form.<\/p>\r\n\r\n<ol id=\"fs-id1165137705346\">\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"799633\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"799633\"]\r\n<p id=\"fs-id1165137408172\">First, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165137705659\">\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]\r\n<p id=\"fs-id1165137602796\">Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\r\n<p id=\"fs-id1165137698078\">Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equivalent to [latex]{3}^{2}=9[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137418681\">Write the following logarithmic equations in exponential form.<\/p>\r\n<p style=\"padding-left: 60px\">a. [latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">b. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/p>\r\n[reveal-answer q=\"234346\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"234346\"]\r\n\r\na.\u00a0[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equivalent to [latex]{10}^{6}=1,000,000[\/latex]\r\nb. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equivalent to [latex]{5}^{2}=25[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]151465[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">\u00a0Convert from exponential to logarithmic form<\/span>\r\n\r\n<\/section>\r\n<p id=\"fs-id1165137933968\">To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n\r\n<div id=\"Example_04_03_02\" class=\"example\">\r\n<div id=\"fs-id1165135168111\" class=\"exercise\">\r\n<div id=\"fs-id1165137727912\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Converting from Exponential Form to Logarithmic Form<\/h3>\r\n<p id=\"fs-id1165137804412\">Write the following exponential equations in logarithmic form.<\/p>\r\n\r\n<ol id=\"fs-id1165135192287\">\r\n \t<li>[latex]{2}^{3}=8[\/latex]<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"693170\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"693170\"]\r\n<p id=\"fs-id1165137474116\">First, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165137573458\">\r\n \t<li>[latex]{2}^{3}=8[\/latex]\r\n<p id=\"fs-id1165137466396\">Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex]\r\n<p id=\"fs-id1165135193035\">Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\r\n<p id=\"fs-id1165135187822\">Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equivalent to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137566762\">Write the following exponential equations in logarithmic form.<\/p>\r\n<p style=\"padding-left: 60px\">a. [latex]{3}^{2}=9[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">b. [latex]{5}^{3}=125[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">c. [latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/p>\r\n[reveal-answer q=\"932188\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932188\"]\r\n\r\na.\u00a0[latex]{3}^{2}=9[\/latex] is equivalent to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\r\nb. [latex]{5}^{3}=125[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]\r\nc. [latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equivalent to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]14387[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Evaluate logarithms<\/h2>\r\n<section id=\"fs-id1165137530906\">\r\n<p id=\"fs-id1165137422589\">Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, \"To what exponent must 2\u00a0be raised in order to get 8?\" Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].<\/p>\r\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137937690\">\r\n \t<li>We ask, \"To what exponent must 7 be raised in order to get 49?\" We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex]<\/li>\r\n \t<li>We ask, \"To what exponent must 3 be raised in order to get 27?\" We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137456358\">Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137584208\">\r\n \t<li>We ask, \"To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? \" We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137455840\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137453770\">How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally.<\/h3>\r\n<ol id=\"fs-id1165134079724\">\r\n \t<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\r\n \t<li>Use previous knowledge of powers of <em>b<\/em>\u00a0identify <em>y<\/em>\u00a0by asking, \"To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?\"<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_03\" class=\"example\">\r\n<div id=\"fs-id1165137732842\" class=\"exercise\">\r\n<div id=\"fs-id1165135296345\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Solving Logarithms Mentally<\/h3>\r\n<p id=\"fs-id1165135393440\">Solve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"960893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960893\"]\r\n<p id=\"fs-id1165137611276\">First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, \"To what exponent must 4 be raised in order to get 64?\"<\/p>\r\n<p id=\"fs-id1165137661814\">We know<\/p>\r\n<p style=\"text-align: center\">[latex]{4}^{3}=64[\/latex]<\/p>\r\n<p id=\"fs-id1165137619013\">Therefore,<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137745041\">Solve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"650736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"650736\"]\r\n\r\n[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recalling that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex])\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_03_04\" class=\"example\">\r\n<div id=\"fs-id1165137663658\" class=\"exercise\">\r\n<div id=\"fs-id1165137680390\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Evaluating the Logarithm of a Reciprocal<\/h3>\r\n<p id=\"fs-id1165137938805\">Evaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"177970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177970\"]\r\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, \"To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]\"?<\/p>\r\n<p id=\"fs-id1165137552085\">We know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{3}^{-3}&amp;=\\frac{1}{{3}^{3}} \\\\ &amp;=\\frac{1}{27} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135437134\">Evaluate [latex]y={\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)[\/latex] without using a calculator.<\/p>\r\n[reveal-answer q=\"708400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"708400\"]\r\n\r\n[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]35042[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">\u00a0Use common logarithms<\/span>\r\n\r\n<\/section><section id=\"fs-id1165137405741\">\r\n<p id=\"fs-id1165137661970\">The most frequently used base for logarithms is <em>e<\/em>. Base <em>e<\/em>\u00a0logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base <em>e<\/em>\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137473872\">Most values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137452317\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Definition of the Natural Logarithm<\/h3>\r\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165135613642\">For [latex]x&gt;0[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137580230\" class=\"equation\" style=\"text-align: center\">[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equivalent to }{e}^{y}=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>\" or \"the natural logarithm of <em>x<\/em>.\"<\/p>\r\n<p id=\"fs-id1165137566720\">The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137705251\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for <em>x\u00a0<\/em>&gt; 0.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137409558\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137832169\">How To: Given a natural logarithm with the form [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex], evaluate it using a calculator.<\/h3>\r\n<ol id=\"fs-id1165135407195\">\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter the value given for <em>x<\/em>, followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_08\" class=\"example\">\r\n<div id=\"fs-id1165137731536\" class=\"exercise\">\r\n<div id=\"fs-id1165137434974\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Evaluating a Natural Logarithm Using a Calculator<\/h3>\r\n<p id=\"fs-id1165137573341\">Evaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.<\/p>\r\n[reveal-answer q=\"847079\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847079\"]\r\n<ul id=\"fs-id1165137563770\">\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter 500, followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137645024\">Rounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137435623\">Evaluate [latex]\\mathrm{ln}\\left(-500\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"393229\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"393229\"]\r\n\r\nIt is not possible to take the logarithm of a negative number in the set of real numbers.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]35022[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135194555\">In <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction-to-graphs-of-exponential-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of Exponential Functions<\/a>, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the <em>cause<\/em> for an <em>effect<\/em>.<\/p>\r\n<p id=\"fs-id1165137603580\">To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year <em>t<\/em>\u00a0can be found with the equation [latex]A=2500{e}^{0.05t}[\/latex].<\/p>\r\nBut what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1\u00a0shows this point on the logarithmic graph.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_0012.jpg\" alt=\"A graph titled, \" width=\"900\" height=\"459\" \/> <b>Figure 1<\/b>[\/caption]\r\n<p id=\"fs-id1165135161452\">In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions.<\/p>\r\n\r\n<h2>Identify the domain of a logarithmic function<\/h2>\r\n<p id=\"fs-id1165137748716\">Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.<\/p>\r\n<p id=\"fs-id1165137758495\">Recall that the exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number <em>x<\/em>\u00a0and constant [latex]b&gt;0[\/latex], [latex]b\\ne 1[\/latex], where<\/p>\r\n\r\n<ul id=\"fs-id1165137736024\">\r\n \t<li>The domain of <em>y<\/em>\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of <em>y<\/em>\u00a0is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135641666\">In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:<\/p>\r\n\r\n<ul id=\"fs-id1165137656096\">\r\n \t<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]:[latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135245571\">Transformations of the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations\u2014shifts, stretches, compressions, and reflections\u2014to the parent function without loss of shape.<\/p>\r\n<p id=\"fs-id1165137653624\">In <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction-to-graphs-of-exponential-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of Exponential Functions<\/a> we saw that certain transformations can change the <em>range<\/em> of [latex]y={b}^{x}[\/latex]. Similarly, applying transformations to the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can change the <em>domain<\/em>. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists <em>only of positive real numbers<\/em>. That is, the argument of the logarithmic function must be greater than zero.<\/p>\r\n<p id=\"fs-id1165137851584\">For example, consider [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex]. This function is defined for any values of <em>x<\/em>\u00a0such that the argument, in this case [latex]2x - 3[\/latex], is greater than zero. To find the domain, we set up an inequality and solve for\u00a0<em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-318\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&amp;2x - 3&gt;0 &amp;&amp; \\text{Show the argument greater than zero}. \\\\ &amp;2x&gt;3 &amp;&amp; \\text{Add 3}. \\\\ &amp;x&gt;1.5 &amp;&amp; \\text{Divide by 2}. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137645047\">In interval notation, the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex] is [latex]\\left(1.5,\\infty \\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137423048\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135173951\">How To: Given a logarithmic function, identify the domain.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165137823224\">\r\n \t<li>Set up an inequality showing the argument greater than zero.<\/li>\r\n \t<li>Solve for <em>x<\/em>.<\/li>\r\n \t<li>Write the domain in interval notation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_01\" class=\"example\">\r\n<div id=\"fs-id1165137846475\" class=\"exercise\">\r\n<div id=\"fs-id1165137460694\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Identifying the Domain of a Logarithmic Shift<\/h3>\r\n<p id=\"fs-id1165135209576\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?<\/p>\r\n[reveal-answer q=\"174870\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174870\"]\r\n<p id=\"fs-id1165137693442\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]x+3&gt;0[\/latex]. Solving this inequality,<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;x+3&gt;0 &amp;&amp; \\text{The input must be positive}. \\\\ &amp;x&gt;-3 &amp;&amp; \\text{Subtract 3}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137638183\">The domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137645484\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x - 2\\right)+1[\/latex]?<\/p>\r\n[reveal-answer q=\"405290\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"405290\"]\r\n\r\n[latex]\\left(2,\\infty \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_04_02\" class=\"example\">\r\n<div id=\"fs-id1165137894615\" class=\"exercise\">\r\n<div id=\"fs-id1165134108527\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Identifying the Domain of a Logarithmic Shift and Reflection<\/h3>\r\n<p id=\"fs-id1165135499558\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?<\/p>\r\n[reveal-answer q=\"675604\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"675604\"]\r\n<p id=\"fs-id1165137780875\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]5 - 2x&gt;0[\/latex]. Solving this inequality,<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;5 - 2x&gt;0 &amp;&amp; \\text{The input must be positive}. \\\\ &amp;-2x&gt;-5 &amp;&amp; \\text{Subtract }5. \\\\ &amp;x&lt;\\frac{5}{2} &amp;&amp; \\text{Divide by }-2\\text{ and switch the inequality}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137656879\">The domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137453336\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(x - 5\\right)+2[\/latex]?<\/p>\r\n[reveal-answer q=\"87516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"87516\"]\r\n\r\n[latex]\\left(5,\\infty \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174284[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Graph logarithmic functions<\/h2>\r\n<p id=\"fs-id1165134104063\">Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] along with all its transformations: shifts, stretches, compressions, and reflections.<\/p>\r\n<p id=\"fs-id1165137679088\">We begin with the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Because every logarithmic function of this form is the inverse of an exponential function with the form [latex]y={b}^{x}[\/latex], their graphs will be reflections of each other across the line [latex]y=x[\/latex]. To illustrate this, we can observe the relationship between the input and output values of [latex]y={2}^{x}[\/latex] and its equivalent [latex]x={\\mathrm{log}}_{2}\\left(y\\right)[\/latex] in the table below.<\/p>\r\n\r\n<table id=\"Table_04_04_01\" summary=\"Three rows and eight columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]{2}^{x}=y[\/latex]<\/strong><\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]{\\mathrm{log}}_{2}\\left(y\\right)=x[\/latex]<\/strong><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135509175\">Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/p>\r\n\r\n<table id=\"Table_04_04_02\" summary=\"Two rows and eight columns. The first row is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n<td>[latex]\\left(-3,\\frac{1}{8}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(-2,\\frac{1}{4}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(-1,\\frac{1}{2}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(0,1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(1,2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(3,8\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/strong><\/td>\r\n<td>[latex]\\left(\\frac{1}{8},-3\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(\\frac{1}{4},-2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(\\frac{1}{2},-1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(1,0\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(2,1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(8,3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137761335\">As we\u2019d expect, the <em>x<\/em>- and <em>y<\/em>-coordinates are reversed for the inverse functions. The figure below\u00a0shows the graph of <em>f<\/em>\u00a0and <em>g<\/em>.<\/p>\r\n\r\n<figure class=\"small\"><img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_0022.jpg\" alt=\"Graph of two functions, f(x)=2^x and g(x)=log_2(x), with the line y=x denoting the axis of symmetry.\" \/><\/figure>\r\n<p style=\"text-align: center\"><strong>Figure 2.\u00a0<\/strong>Notice that the graphs of [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] are reflections about the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137406913\">Observe the following from the graph:<\/p>\r\n\r\n<ul id=\"fs-id1165137408405\">\r\n \t<li>[latex]f\\left(x\\right)={2}^{x}[\/latex] has a <em>y<\/em>-intercept at [latex]\\left(0,1\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] has an <em>x<\/em>-intercept at [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(-\\infty ,\\infty \\right)[\/latex], is the same as the range of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(0,\\infty \\right)[\/latex], is the same as the domain of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137780760\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Characteristics of the Graph of the Parent Function, <em>f<\/em>(<em>x<\/em>) = log<sub><em>b<\/em><\/sub>(<em>x<\/em>)<\/h3>\r\n<p id=\"fs-id1165135520250\">For any real number <em>x<\/em>\u00a0and constant <em>b\u00a0<\/em>&gt; 0, [latex]b\\ne 1[\/latex], we can see the following characteristics in the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]:<\/p>\r\n\r\n<ul id=\"fs-id1165137400150\">\r\n \t<li>one-to-one function<\/li>\r\n \t<li>vertical asymptote: <em>x\u00a0<\/em>= 0<\/li>\r\n \t<li>domain: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/li>\r\n \t<li><em>x-<\/em>intercept: [latex]\\left(1,0\\right)[\/latex] and key point [latex]\\left(b,1\\right)[\/latex]<\/li>\r\n \t<li><em>y<\/em>-intercept: none<\/li>\r\n \t<li>increasing if [latex]b&gt;1[\/latex]<\/li>\r\n \t<li>decreasing if 0 &lt; <em>b\u00a0<\/em>&lt; 1<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_04_003\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_003G2.jpg\" alt=\"&quot;Two\" \/><\/figure>\r\nFigure 3\u00a0shows how changing the base <em>b<\/em>\u00a0in [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (<em>Note:<\/em> recall that the function [latex]\\mathrm{ln}\\left(x\\right)[\/latex] has base [latex]e\\approx \\text{2}.\\text{718.)}[\/latex]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0042.jpg\" alt=\"Graph of three equations: y=log_2(x) in blue, y=ln(x) in orange, and y=log(x) in red. The y-axis is the asymptote.\" width=\"487\" height=\"363\" \/> <strong>Figure 4.\u00a0<\/strong>The graphs of three logarithmic functions with different bases, all greater than 1.[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137871937\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137805513\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph the function.<\/h3>\r\n<ol id=\"fs-id1165135435529\">\r\n \t<li>Draw and label the vertical asymptote, <em>x<\/em> = 0.<\/li>\r\n \t<li>Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>Plot the key point [latex]\\left(b,1\\right)[\/latex].<\/li>\r\n \t<li>Draw a smooth curve through the points.<\/li>\r\n \t<li>State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote, <em>x<\/em> = 0.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_03\" class=\"example\">\r\n<div id=\"fs-id1165137550508\" class=\"exercise\">\r\n<div id=\"fs-id1165137550510\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Graphing a Logarithmic Function with the Form\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<\/h3>\r\n<p id=\"fs-id1165137431970\">Graph [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"347847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"347847\"]\r\n<p id=\"fs-id1165137501970\">Before graphing, identify the behavior and key points for the graph.<\/p>\r\n\r\n<ul id=\"fs-id1165135497154\">\r\n \t<li>Since <em>b\u00a0<\/em>= 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote <em>x\u00a0<\/em>= 0, and the right tail will increase slowly without bound.<\/li>\r\n \t<li>The <em>x<\/em>-intercept is [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>The key point [latex]\\left(5,1\\right)[\/latex] is on the graph.<\/li>\r\n \t<li>We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_04_005\" class=\"small\"><span id=\"fs-id1165135508394\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0052.jpg\" alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\" width=\"557\" height=\"419\" \/><\/span><\/figure>\r\n<p id=\"fs-id1165135697920\" style=\"text-align: center\"><strong>Figure 5.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135171582\">Graph [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{5}}\\left(x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"808887\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"808887\"]\r\n\r\nThe domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165134377926\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0062.jpg\" alt=\"Graph of f(x)=log_(1\/5)(x) with labeled points at (1\/5, 1) and (1, 0). The y-axis is the asymptote.\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174289[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Graphing Transformations of Logarithmic Functions<\/h2>\r\n<p id=\"fs-id1165137430986\">As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the <strong>parent function<\/strong> [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] without loss of shape.<\/p>\r\n\r\n<section id=\"fs-id1165137734884\">\r\n<h2>Graphing a Horizontal Shift of\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/h2>\r\nWhen a constant <em>c<\/em>\u00a0is added to the input of the parent function [latex]f\\left(x\\right)=\\text{log}_{b}\\left(x\\right)[\/latex], the result is a <strong>horizontal shift<\/strong> <em>c<\/em>\u00a0units in the <em>opposite<\/em> direction of the sign on <em>c<\/em>. To visualize horizontal shifts, we can observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and for <em>c\u00a0<\/em>&gt; 0 alongside the shift left, [latex]g\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex], and the shift right, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(x-c\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_007n2.jpg\" alt=\"Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x+c) is the translation function with an asymptote at x=-c. This shows the translation of shifting left.\" width=\"900\" height=\"526\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165135296307\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Horizontal Shifts of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165135176174\">For any constant <em>c<\/em>, the function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165135206192\">\r\n \t<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] left <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&gt; 0.<\/li>\r\n \t<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] right <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&lt; 0.<\/li>\r\n \t<li>has the vertical asymptote <em>x\u00a0<\/em>= \u2013<em>c<\/em>.<\/li>\r\n \t<li>has domain [latex]\\left(-c,\\infty \\right)[\/latex].<\/li>\r\n \t<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137641710\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137641715\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex], graph the translation.<\/h3>\r\n<ol id=\"fs-id1165137454284\">\r\n \t<li>Identify the horizontal shift:\r\n<ol id=\"fs-id1165137454288\">\r\n \t<li>If <em>c<\/em> &gt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] left <em>c<\/em>\u00a0units.<\/li>\r\n \t<li>If <em>c\u00a0<\/em>&lt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] right <em>c<\/em>\u00a0units.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Draw the vertical asymptote <em>x\u00a0<\/em>= \u2013<em>c<\/em>.<\/li>\r\n \t<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting <em>c<\/em>\u00a0from the\u00a0<em>x<\/em>\u00a0coordinate.<\/li>\r\n \t<li>Label the three points.<\/li>\r\n \t<li>The Domain is [latex]\\left(-c,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= \u2013c.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_04\" class=\"example\">\r\n<div id=\"fs-id1165137414959\" class=\"exercise\">\r\n<div id=\"fs-id1165137414961\" class=\"problem textbox shaded\">\r\n<h3>Example 4:\u00a0Graphing a Horizontal Shift of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137455420\">Sketch the horizontal shift [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex] alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"785817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785817\"]\r\n<p id=\"fs-id1165137759885\">Since the function is [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex], we notice [latex]x+\\left(-2\\right)=x - 2[\/latex].<\/p>\r\n<p id=\"fs-id1165137784630\">Thus <em>c\u00a0<\/em>= \u20132, so <em>c\u00a0<\/em>&lt; 0. This means we will shift the function [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] right 2 units.<\/p>\r\n<p id=\"fs-id1165137836995\">The vertical asymptote is [latex]x=-\\left(-2\\right)[\/latex] or <em>x\u00a0<\/em>= 2.<\/p>\r\n<p id=\"fs-id1165134042608\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{3},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(3,1\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137475806\">The new coordinates are found by adding 2 to the <em>x<\/em>\u00a0coordinates.<\/p>\r\n<p id=\"fs-id1165137748449\">Label the points [latex]\\left(\\frac{7}{3},-1\\right)[\/latex], [latex]\\left(3,0\\right)[\/latex], and [latex]\\left(5,1\\right)[\/latex].<\/p>\r\nThe domain is [latex]\\left(2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0082.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1\/3, -1), (1, 0), and (3, 1).The translation function f(x)=log_3(x-2) has an asymptote at x=2 and labeled points at (3, 0) and (5, 1).\" width=\"487\" height=\"363\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<span style=\"font-size: 0.9em\">\u00a0<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135329937\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x+4\\right)[\/latex] alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"139882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"139882\"]\r\n\r\nThe domain is [latex]\\left(-4,\\infty \\right)[\/latex], the range [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the asymptote <em>x\u00a0<\/em>= \u20134.<span id=\"fs-id1165135209395\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0092.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1, 0), and (3, 1).The translation function f(x)=log_3(x+4) has an asymptote at x=-4 and labeled points at (-3, 0) and (-1, 1).\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174300[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing a Vertical Shift of\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/span>\r\n\r\n<\/section><section id=\"fs-id1165135403538\">When a constant <em>d<\/em>\u00a0is added to the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], the result is a <strong>vertical shift<\/strong> <em>d<\/em>\u00a0units in the direction of the sign on <em>d<\/em>. To visualize vertical shifts, we can observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the shift up, [latex]g\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex] and the shift down, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)-d[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_010F2.jpg\" alt=\"Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)+d is the translation function with an asymptote at x=0. This shows the translation of shifting up. Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)-d is the translation function with an asymptote at x=0. This shows the translation of shifting down.\" width=\"900\" height=\"684\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165137767601\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Vertical Shifts of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137661370\">For any constant <em>d<\/em>, the function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137803105\">\r\n \t<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] up <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&gt; 0.<\/li>\r\n \t<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] down <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&lt; 0.<\/li>\r\n \t<li>has the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\r\n \t<li>has domain [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137706002\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137706009\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex], graph the translation.<\/h3>\r\n<ol>\r\n \t<li>Identify the vertical shift:\r\n<ol>\r\n \t<li>If <em>d\u00a0<\/em>&gt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] up <em>d<\/em>\u00a0units.<\/li>\r\n \t<li>If <em>d\u00a0<\/em>&lt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] down <em>d\u00a0<\/em>units.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Draw the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\r\n \t<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by adding <em>d<\/em>\u00a0to the <em>y\u00a0<\/em>coordinate.<\/li>\r\n \t<li>Label the three points.<\/li>\r\n \t<li>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_05\" class=\"example\">\r\n<div id=\"fs-id1165137470057\" class=\"exercise\">\r\n<div id=\"fs-id1165137470059\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Graphing a Vertical Shift of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137832038\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)-2[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"657475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"657475\"]\r\n<p id=\"fs-id1165137465913\">Since the function is [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)-2[\/latex], we will notice <em>d\u00a0<\/em>= \u20132. Thus <em>d\u00a0<\/em>&lt; 0.<\/p>\r\n<p id=\"fs-id1165135175015\">This means we will shift the function [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] down 2 units.<\/p>\r\n<p id=\"fs-id1165137644429\">The vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\r\n<p id=\"fs-id1165137408419\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{3},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(3,1\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135503945\">The new coordinates are found by subtracting 2 from the <em>y <\/em>coordinates.<\/p>\r\n<p id=\"fs-id1165135421660\">Label the points [latex]\\left(\\frac{1}{3},-3\\right)[\/latex], [latex]\\left(1,-2\\right)[\/latex], and [latex]\\left(3,-1\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135195524\">The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<span id=\"fs-id1165134393856\">\r\n<img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0112.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1\/3, -1), (1, 0), and (3, 1).The translation function f(x)=log_3(x)-2 has an asymptote at x=0 and labeled points at (1, 0) and (3, 1).\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137698285\" style=\"text-align: center\"><strong>Figure 9.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137760886\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)+2[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"597513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"597513\"]\r\n\r\nThe domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165137874471\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0122.jpg\" alt=\"Graph of two functions. The parent function is y=log_2(x), with an asymptote at x=0 and labeled points at (1, 0), and (2, 1).The translation function f(x)=log_2(x)+2 has an asymptote at x=0 and labeled points at (0.25, 0) and (0.5, 1).\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174304[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing Stretches and Compressions of\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/span>\r\n\r\n<\/section><section id=\"fs-id1165137770245\">When the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by a constant <em>a<\/em> &gt; 0, the result is a <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the original graph. To visualize stretches and compressions, we set <em>a\u00a0<\/em>&gt; 1 and observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the vertical stretch, [latex]g\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and the vertical compression, [latex]h\\left(x\\right)=\\frac{1}{a}{\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_013n2.jpg\" alt=\"&quot;Graph\" \/>\r\n<div id=\"fs-id1165137433996\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Vertical Stretches and Compressions of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137758179\">For any constant <em>a<\/em> &gt; 1, the function [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137428102\">\r\n \t<li>stretches the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if <em>a\u00a0<\/em>&gt; 1.<\/li>\r\n \t<li>compresses the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if 0 &lt; <em>a\u00a0<\/em>&lt; 1.<\/li>\r\n \t<li>has the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\r\n \t<li>has the <em>x<\/em>-intercept [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>has domain [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165135169301\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135169307\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex], [latex]a&gt;0[\/latex], graph the translation.<\/h3>\r\n<ol id=\"fs-id1165137464127\">\r\n \t<li>Identify the vertical stretch or compressions:\r\n<ol id=\"eip-id1165134081434\">\r\n \t<li>If [latex]|a|&gt;1[\/latex], the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is stretched by a factor of <em>a<\/em>\u00a0units.<\/li>\r\n \t<li>If [latex]|a|&lt;1[\/latex], the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is compressed by a factor of <em>a<\/em>\u00a0units.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Draw the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\r\n \t<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the <em>y<\/em>\u00a0coordinates by <em>a<\/em>.<\/li>\r\n \t<li>Label the three points.<\/li>\r\n \t<li>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_06\" class=\"example\">\r\n<div id=\"fs-id1165135309914\" class=\"exercise\">\r\n<div id=\"fs-id1165135309916\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Graphing a Stretch or Compression of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137602128\">Sketch a graph of [latex]f\\left(x\\right)=2{\\mathrm{log}}_{4}\\left(x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"846570\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"846570\"]\r\n<p id=\"fs-id1165135210052\">Since the function is [latex]f\\left(x\\right)=2{\\mathrm{log}}_{4}\\left(x\\right)[\/latex], we will notice <em>a\u00a0<\/em>= 2.<\/p>\r\n<p id=\"fs-id1165135384321\">This means we will stretch the function [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] by a factor of 2.<\/p>\r\n<p id=\"fs-id1165135481989\">The vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\r\n<p id=\"fs-id1165137757801\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{4},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(4,1\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135570058\">The new coordinates are found by multiplying the <em>y<\/em>\u00a0coordinates by 2.<\/p>\r\n<p id=\"fs-id1165137837989\">Label the points [latex]\\left(\\frac{1}{4},-2\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(4,\\text{2}\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135543469\">The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165134059742\">\r\n<img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0142.jpg\" alt=\"Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=2log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (2, 1).\" \/><\/span><\/p>\r\n<p id=\"fs-id1165135566827\" style=\"text-align: center\"><strong>Figure 11.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135471122\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{2}{\\mathrm{log}}_{4}\\left(x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"645251\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"645251\"]\r\n\r\nThe domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165135332505\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0152.jpg\" alt=\"Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=(1\/2)log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (16, 1).\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_04_07\" class=\"example\">\r\n<div id=\"fs-id1165134267814\" class=\"exercise\">\r\n<div id=\"fs-id1165134267816\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Combining a Shift and a Stretch<\/h3>\r\n<p id=\"fs-id1165137863045\">Sketch a graph of [latex]f\\left(x\\right)=5\\mathrm{log}\\left(x+2\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"470378\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470378\"]\r\n<p id=\"fs-id1165137935561\">Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5. The vertical asymptote will be shifted to <em>x\u00a0<\/em>= \u20132. The <em>x<\/em>-intercept will be [latex]\\left(-1,0\\right)[\/latex]. The domain will be [latex]\\left(-2,\\infty \\right)[\/latex]. Two points will help give the shape of the graph: [latex]\\left(-1,0\\right)[\/latex] and [latex]\\left(8,5\\right)[\/latex]. We chose <em>x\u00a0<\/em>= 8 as the <em>x<\/em>-coordinate of one point to graph because when <em>x\u00a0<\/em>= 8, <em>x\u00a0<\/em>+ 2 = 10, the base of the common logarithm.<span id=\"fs-id1165135641650\">\r\n<img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0162.jpg\" alt=\"Graph of three functions. The parent function is y=log(x), with an asymptote at x=0. The first translation function y=5log(x+2) has an asymptote at x=-2. The second translation function y=log(x+2) has an asymptote at x=-2.\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137874883\" style=\"text-align: center\"><strong>Figure 12.\u00a0<\/strong>The domain is [latex]\\left(-2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= \u20132.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137838697\">Sketch a graph of the function [latex]f\\left(x\\right)=3\\mathrm{log}\\left(x - 2\\right)+1[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"793205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"793205\"]\r\n\r\nThe domain is [latex]\\left(2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 2.\r\n<div id=\"fs-id1165137437228\" class=\"solution\">\r\n\r\n<span id=\"fs-id1165135177663\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0172.jpg\" alt=\"Graph of f(x)=3log(x-2)+1 with an asymptote at x=2.\" \/><\/span>\r\n\r\n<span style=\"font-size: 1rem;text-align: initial\">[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174299[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing Reflections of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/span>\r\n\r\n<\/section><section id=\"fs-id1165137629003\">\r\n<p id=\"fs-id1165135169315\">When the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by \u20131, the result is a <strong>reflection<\/strong> about the <em>x<\/em>-axis. When the <em>input<\/em> is multiplied by \u20131, the result is a reflection about the <em>y<\/em>-axis. To visualize reflections, we restrict <em>b\u00a0<\/em>&gt; 1, and observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the reflection about the <em>x<\/em>-axis, [latex]g\\left(x\\right)={\\mathrm{-log}}_{b}\\left(x\\right)[\/latex] and the reflection about the <em>y<\/em>-axis, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex].<\/p>\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_018n2.jpg\" alt=\"&quot;Graph\" \/>\r\n<div id=\"fs-id1165135190744\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Reflections of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\r\n<p id=\"fs-id1165137722409\">The function [latex]f\\left(x\\right)={\\mathrm{-log}}_{b}\\left(x\\right)[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137832285\">\r\n \t<li>reflects the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>x<\/em>-axis.<\/li>\r\n \t<li>has domain, [latex]\\left(0,\\infty \\right)[\/latex], range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and vertical asymptote, <em>x\u00a0<\/em>= 0, which are unchanged from the parent function.<\/li>\r\n<\/ul>\r\nThe function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]\r\n<ul id=\"fs-id1165137734930\">\r\n \t<li>reflects the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>y<\/em>-axis.<\/li>\r\n \t<li>has domain [latex]\\left(-\\infty ,0\\right)[\/latex].<\/li>\r\n \t<li>has range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and vertical asymptote, <em>x\u00a0<\/em>= 0, which are unchanged from the parent function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137638830\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137638837\">How To: Given a logarithmic function with the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph a translation.<\/h3>\r\n<table id=\"Table_04_04_08\" class=\"unnumbered\" summary=\"The first column gives the following instructions of graphing a translation of f(x)=-log_b(x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the x-axis; 4. Draw a smooth curve through the points; 5. State the domain, (0, infinity), the range, (-infinity, infinity), and the vertical asymptote x=0. The second column gives the following instructions of graphing a translation of f(x)=log_b(-x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (-1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the y-axis; 4. Draw a smooth curve through the points; 5. State the domain, (-infinity, 0), the range, (-infinity, infinity), and the vertical asymptote x=0.\">\r\n<thead>\r\n<tr>\r\n<th>[latex]\\text{If }f\\left(x\\right)=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/th>\r\n<th>[latex]\\text{If }f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1. Draw the vertical asymptote, <em>x\u00a0<\/em>= 0.<\/td>\r\n<td>1. Draw the vertical asymptote, <em>x\u00a0<\/em>= 0.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2. Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/td>\r\n<td>2. Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3. Reflect the graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>x<\/em>-axis.<\/td>\r\n<td>3. Reflect the graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>y<\/em>-axis.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4. Draw a smooth curve through the points.<\/td>\r\n<td>4. Draw a smooth curve through the points.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5. State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote <em>x\u00a0<\/em>= 0.<\/td>\r\n<td>5. State the domain, [latex]\\left(-\\infty ,0\\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote <em>x\u00a0<\/em>= 0.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"Example_04_04_08\" class=\"example\">\r\n<div id=\"fs-id1165137697928\" class=\"exercise\">\r\n<div id=\"fs-id1165137849033\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Graphing a Reflection of a Logarithmic Function<\/h3>\r\n<p id=\"fs-id1165137849038\">Sketch a graph of [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"618451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"618451\"]\r\n<p id=\"fs-id1165137836525\">Before graphing [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex], identify the behavior and key points for the graph.<\/p>\r\n\r\n<ul id=\"fs-id1165137769879\">\r\n \t<li>Since <em>b\u00a0<\/em>= 10 is greater than one, we know that the parent function is increasing. Since the <em>input<\/em> value is multiplied by \u20131, <em>f<\/em>\u00a0is a reflection of the parent graph about the <em>y-<\/em>axis. Thus, [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex] will be decreasing as <em>x<\/em>\u00a0moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\r\n \t<li>The <em>x<\/em>-intercept is [latex]\\left(-1,0\\right)[\/latex].<\/li>\r\n \t<li>We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_04_019\" class=\"small\"><span id=\"fs-id1165134042188\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0192.jpg\" alt=\"Graph of two functions. The parent function is y=log(x), with an asymptote at x=0 and labeled points at (1, 0), and (10, 0).The translation function f(x)=log(-x) has an asymptote at x=0 and labeled points at (-1, 0) and (-10, 1).\" \/><\/span><\/figure>\r\n<p id=\"fs-id1165134042202\" style=\"text-align: center\"><strong>Figure 14.\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,0\\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135681852\">Graph [latex]f\\left(x\\right)=-\\mathrm{log}\\left(-x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"485222\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485222\"]\r\n\r\nThe domain is [latex]\\left(-\\infty ,0\\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165137855148\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0202.jpg\" alt=\"Graph of f(x)=-log(-x) with an asymptote at x=0.\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134579621\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165134579627\">How To: Given a logarithmic equation, use a graphing calculator to approximate solutions.<\/h3>\r\n<ol id=\"fs-id1165137431118\">\r\n \t<li>Press <strong>[Y=]<\/strong>. Enter the given logarithm equation or equations as <strong>Y<sub>1<\/sub>=<\/strong> and, if needed, <strong>Y<sub>2<\/sub>=<\/strong>.<\/li>\r\n \t<li>Press <strong>[GRAPH]<\/strong> to observe the graphs of the curves and use <strong>[WINDOW]<\/strong> to find an appropriate view of the graphs, including their point(s) of intersection.<\/li>\r\n \t<li>To find the value of <em>x<\/em>, we compute the point of intersection. Press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select \"intersect\" and press <strong>[ENTER]<\/strong> three times. The point of intersection gives the value of <em>x<\/em>, for the point(s) of intersection.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_04_09\" class=\"example\">\r\n<div id=\"fs-id1165135414229\" class=\"exercise\">\r\n<div id=\"fs-id1165135414231\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Approximating the Solution of a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165135414236\">Solve [latex]4\\mathrm{ln}\\left(x\\right)+1=-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\r\n[reveal-answer q=\"513609\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"513609\"]\r\n<p id=\"fs-id1165135193434\">Press <strong>[Y=]<\/strong> and enter [latex]4\\mathrm{ln}\\left(x\\right)+1[\/latex] next to <strong>Y<sub>1<\/sub><\/strong>=. Then enter [latex]-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] next to <strong>Y<sub>2<\/sub>=<\/strong>. For a window, use the values 0 to 5 for <em>x<\/em>\u00a0and \u201310 to 10 for <em>y<\/em>. Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere a little to right of <em>x\u00a0<\/em>= 1.<\/p>\r\n<p id=\"fs-id1165135245763\">For a better approximation, press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The <em>x<\/em>-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for <strong>Guess?<\/strong>) So, to the nearest thousandth, [latex]x\\approx 1.339[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137639531\">Solve [latex]5\\mathrm{log}\\left(x+2\\right)=4-\\mathrm{log}\\left(x\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\r\n[reveal-answer q=\"862673\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"862673\"]\r\n\r\n[latex]x\\approx 3.049[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165135528930\">\r\n<h2>Summarizing Translations of the Logarithmic Function<\/h2>\r\n<p id=\"fs-id1165135528935\">Now that we have worked with each type of translation for the logarithmic function, we can summarize each in the table below\u00a0to arrive at the general equation for translating exponential functions.<\/p>\r\n\r\n<table id=\"Table_04_04_009\" summary=\"Titled, \">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"2\">Translations of the Parent Function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<th style=\"text-align: center\">Translation<\/th>\r\n<th style=\"text-align: center\">Form<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Shift\r\n<ul id=\"fs-id1165137416971\">\r\n \t<li>Horizontally <em>c<\/em>\u00a0units to the left<\/li>\r\n \t<li>Vertically <em>d<\/em>\u00a0units up<\/li>\r\n<\/ul>\r\n<\/td>\r\n<td>[latex]y={\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Stretch and Compress\r\n<ul id=\"fs-id1165137427553\">\r\n \t<li>Stretch if [latex]|a|&gt;1[\/latex]<\/li>\r\n \t<li>Compression if [latex]|a|&lt;1[\/latex]<\/li>\r\n<\/ul>\r\n<\/td>\r\n<td>[latex]y=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reflect about the <em>x<\/em>-axis<\/td>\r\n<td>[latex]y=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reflect about the <em>y<\/em>-axis<\/td>\r\n<td>[latex]y={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>General equation for all translations<\/td>\r\n<td>[latex]y=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165137414493\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Translations of Logarithmic Functions<\/h3>\r\n<p id=\"fs-id1165137414501\">All translations of the parent logarithmic function, [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], have the form<\/p>\r\n\r\n<div id=\"fs-id1165135408512\" class=\"equation\" style=\"text-align: center\">[latex] f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/div>\r\n<p id=\"fs-id1165137734655\">where the parent function, [latex]y={\\mathrm{log}}_{b}\\left(x\\right),b&gt;1[\/latex], is<\/p>\r\n\r\n<ul id=\"fs-id1165137531610\">\r\n \t<li>shifted vertically up <em>d<\/em>\u00a0units.<\/li>\r\n \t<li>shifted horizontally to the left <em>c<\/em>\u00a0units.<\/li>\r\n \t<li>stretched vertically by a factor of |<em>a<\/em>| if |<em>a<\/em>| &gt; 0.<\/li>\r\n \t<li>compressed vertically by a factor of |<em>a<\/em>| if 0 &lt; |<em>a<\/em>| &lt; 1.<\/li>\r\n \t<li>reflected about the <em>x-<\/em>axis when <em>a\u00a0<\/em>&lt; 0.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137725084\">For [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex], the graph of the parent function is reflected about the <em>y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_04_10\" class=\"example\">\r\n<div id=\"fs-id1165135296269\" class=\"exercise\">\r\n<div id=\"fs-id1165135296271\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Finding the Vertical Asymptote of a Logarithm Graph<\/h3>\r\n<p id=\"fs-id1165135296276\">What is the vertical asymptote of [latex]f\\left(x\\right)=-2{\\mathrm{log}}_{3}\\left(x+4\\right)+5[\/latex]?<\/p>\r\n[reveal-answer q=\"841705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841705\"]\r\n\r\nThe vertical asymptote is at <em>x\u00a0<\/em>= \u20134.\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137871960\">The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to <em>x\u00a0<\/em>= \u20134.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135368433\">What is the vertical asymptote of [latex]f\\left(x\\right)=3+\\mathrm{ln}\\left(x - 1\\right)[\/latex]?<\/p>\r\n[reveal-answer q=\"975284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"975284\"]\r\n\r\n<em>x\u00a0<\/em>= 1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_04_11\" class=\"example\">\r\n<div id=\"fs-id1165137849555\" class=\"exercise\">\r\n<div id=\"fs-id1165137849558\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Finding the Equation from a Graph<\/h3>\r\n<p id=\"fs-id1165137849563\">Find a possible equation for the common logarithmic function graphed in Figure 15.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005323\/CNX_Precalc_Figure_04_04_021.jpg\" alt=\"Graph of a logarithmic function with a vertical asymptote at x=-2, has been vertically reflected, and passes through the points (-1, 1) and (2, -1).\" width=\"487\" height=\"367\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"993624\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"993624\"]\r\n<p id=\"fs-id1165135342979\">This graph has a vertical asymptote at <em>x\u00a0<\/em>= \u20132 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:<\/p>\r\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=-a\\mathrm{log}\\left(x+2\\right)+k[\/latex]<\/p>\r\n<p id=\"fs-id1165135406913\">It appears the graph passes through the points [latex]\\left(-1,1\\right)[\/latex] and [latex]\\left(2,-1\\right)[\/latex]. Substituting [latex]\\left(-1,1\\right)[\/latex],<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;1=-a\\mathrm{log}\\left(-1+2\\right)+k &amp;&amp; \\text{Substitute }\\left(-1,1\\right). \\\\ &amp;1=-a\\mathrm{log}\\left(1\\right)+k &amp;&amp; \\text{Arithmetic}. \\\\ &amp;1=k &amp;&amp; \\text{log(1)}=0. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137628655\">Next, substituting in [latex]\\left(2,-1\\right)[\/latex],<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;-1=-a\\mathrm{log}\\left(2+2\\right)+1 &amp;&amp; \\text{Plug in }\\left(2,-1\\right). \\\\ &amp;-2=-a\\mathrm{log}\\left(4\\right) &amp;&amp; \\text{Arithmetic}. \\\\ &amp;a=\\frac{2}{\\mathrm{log}\\left(4\\right)}&amp;&amp; \\text{Solve for }a. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135192211\">This gives us the equation [latex]f\\left(x\\right)=-\\frac{2}{\\mathrm{log}\\left(4\\right)}\\mathrm{log}\\left(x+2\\right)+1[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137735586\">We can verify this answer by comparing the function values in the table below\u00a0with the points on the graph in Example 11.<\/p>\r\n\r\n<table id=\"Table_04_04_010\" summary=\"..\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u22121<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>f<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<td>\u22120.58496<\/td>\r\n<td>\u22121<\/td>\r\n<td>\u22121.3219<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>f<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\r\n<td>\u22121.5850<\/td>\r\n<td>\u22121.8074<\/td>\r\n<td>\u22122<\/td>\r\n<td>\u22122.1699<\/td>\r\n<td>\u22122.3219<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137665487\">Give the equation of the natural logarithm graphed in Figure 16.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005324\/CNX_Precalc_Figure_04_04_022.jpg\" alt=\"Graph of a logarithmic function with a vertical asymptote at x=-3, has been vertically stretched by 2, and passes through the points (-1, -1).\" width=\"487\" height=\"442\" \/> <b>Figure 16<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"537017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"537017\"]\r\n\r\n[latex]f\\left(x\\right)=2\\mathrm{ln}\\left(x+3\\right)-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137855236\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165137855242\"><strong>Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?<\/strong><\/p>\r\n<p id=\"fs-id1165137827126\"><em>Yes, if we know the function is a general logarithmic function. For example, look at the graph in Try It 11. The graph approaches x = \u20133 (or thereabouts) more and more closely, so x = \u20133 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, [latex]\\left\\{x|x&gt;-3\\right\\}[\/latex]. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as [latex]x\\to -{3}^{+},f\\left(x\\right)\\to -\\infty [\/latex] and as [latex]x\\to \\infty ,f\\left(x\\right)\\to \\infty [\/latex].<\/em><\/p>\r\n\r\n<\/div>\r\n<\/section>\r\n<figure id=\"CNX_Precalc_Figure_04_05_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"244\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_05_001F2.jpg\" alt=\"Testing of the pH of hydrochloric acid.\" width=\"244\" height=\"382\" \/> <b>Figure 1.<\/b> The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137759741\">In chemistry, <strong>pH<\/strong> is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\r\n\r\n<ul id=\"fs-id1165135253210\">\r\n \t<li>Battery acid: 0.8<\/li>\r\n \t<li>Stomach acid: 2.7<\/li>\r\n \t<li>Orange juice: 3.3<\/li>\r\n \t<li>Pure water: 7 (at 25\u00b0 C)<\/li>\r\n \t<li>Human blood: 7.35<\/li>\r\n \t<li>Fresh coconut: 7.8<\/li>\r\n \t<li>Sodium hydroxide (lye): 14<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where <em>a<\/em>\u00a0is the concentration of hydrogen ion in the solution<\/p>\r\n\r\n<div id=\"eip-396\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\text{pH}&amp;=-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right) \\\\ &amp;=\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right) \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137472164\">The equivalence of [latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex] and [latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] is one of the logarithm properties we will examine in this section.<\/p>\r\n\r\n<h2>Use the product rule for logarithms<\/h2>\r\n<p id=\"fs-id1165137405402\">Recall that the logarithmic and exponential functions \"undo\" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\r\n\r\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}1=0\\\\ &amp;{\\mathrm{log}}_{b}b=1\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137628765\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\r\n<p id=\"fs-id1165137772010\">Next, we have the inverse property.<\/p>\r\n\r\n<div id=\"eip-896\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&amp;=x\\hfill \\\\ {b}^{{\\mathrm{log}}_{b}x}&amp;=x,x&gt;0 \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137696455\">For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\r\n<p id=\"fs-id1165134297163\">To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\r\n<p id=\"fs-id1165137592421\">Finally, we have the <strong>one-to-one<\/strong> property.<\/p>\r\n\r\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\r\n<p id=\"fs-id1165137723139\">We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for <em>x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-448\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}3x&amp;=2x+5 &amp;&amp; \\text{Set the arguments equal.} \\\\ x&amp;=5 &amp;&amp; \\text{Subtract 2}x. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135194712\">But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\r\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\r\n<p id=\"fs-id1165137645446\">Given any real number <em>x<\/em>\u00a0and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\r\n\r\n<div id=\"eip-214\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right) &amp;&amp; \\text{Substitute for }M\\text{ and }N. \\\\ &amp; ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right) &amp;&amp; \\text{Apply the product rule for exponents}. \\\\ &amp; =m+n &amp;&amp; \\text{Apply the inverse property of logs}. \\\\ &amp; ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right) &amp;&amp; \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\r\n\r\n<div id=\"eip-502\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\r\n<div id=\"fs-id1165137891324\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b&gt;0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137541378\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165134223340\">How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\r\n<ol id=\"fs-id1165137748303\">\r\n \t<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\r\n \t<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_01\" class=\"example\">\r\n<div id=\"fs-id1165135458651\" class=\"exercise\">\r\n<div id=\"fs-id1165135458654\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using the Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"868590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"868590\"]\r\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137871801\">Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"989002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989002\"]\r\n\r\n[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174314[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use the quotient and power rules for logarithms<\/h2>\r\n<section id=\"fs-id1165135151295\">\r\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{b}}={x}^{a-b}[\/latex]. The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\r\n<p id=\"fs-id1165137431410\">Given any real number <em>x\u00a0<\/em>and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\r\n\r\n<div id=\"eip-589\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-303\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right) &amp;&amp; \\text{Substitute for }M\\text{ and }N. \\\\ &amp; ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right) &amp;&amp; \\text{Apply the quotient rule for exponents}. \\\\ &amp; =m-n &amp;&amp; \\text{Apply the inverse property of logs}. \\\\ &amp; ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right) &amp;&amp; \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\r\n\r\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) &amp; =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right) &amp;&amp; \\text{Factor the numerator and denominator}. \\\\ &amp; =\\mathrm{log}\\left(\\frac{2x}{3}\\right) &amp;&amp; \\text{Cancel the common factors}. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\r\n\r\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&amp;=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right) \\\\ &amp;=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right) \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137733855\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Quotient Rule for Logarithms<\/h3>\r\n<p id=\"eip-id1165135390834\">The <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137834642\" class=\"equation\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137749807\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137749813\">How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\r\n<ol id=\"fs-id1165137749817\">\r\n \t<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\r\n \t<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\r\n \t<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_02\" class=\"example\">\r\n<div id=\"fs-id1165135185904\" class=\"exercise\">\r\n<div id=\"fs-id1165135185906\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using the Quotient Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"47414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"47414\"]\r\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\[1mm] &amp;= \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right] \\\\[1mm] &amp;={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right) \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137629471\">There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and <em>x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that <em>x\u00a0<\/em>&gt; 0, <em>x\u00a0<\/em>&gt; 1, [latex]x&gt;-\\frac{4}{3}[\/latex], and <em>x\u00a0<\/em>&lt; 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137534466\">Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"51755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"51755\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174320[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Using the Power Rule for Logarithms<\/span>\r\n<div id=\"fs-id1165137939600\" class=\"solution\"><section id=\"fs-id1165137627625\">\r\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[\/latex]? One method is as follows:<\/p>\r\n\r\n<div id=\"eip-271\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({x}^{2}\\right) &amp; ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right) \\\\ &amp; ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x \\\\ &amp; =2{\\mathrm{log}}_{b}x \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137534037\">Notice that we used the <strong>product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\r\n\r\n<div id=\"eip-702\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&amp;100={10}^{2} &amp;&amp; \\sqrt{3}={3}^{\\frac{1}{2}} &amp;&amp; \\frac{1}{e}={e}^{-1} \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137676322\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Power Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165137676330\">The <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137639704\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137639709\">How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\r\n<ol id=\"fs-id1165137761651\">\r\n \t<li>Express the argument as a power, if needed.<\/li>\r\n \t<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_03\" class=\"example\">\r\n<div id=\"fs-id1165135593557\" class=\"exercise\">\r\n<div id=\"fs-id1165135593559\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Expanding a Logarithm with Powers<\/h3>\r\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\r\n[reveal-answer q=\"476325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"476325\"]\r\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\r\n[reveal-answer q=\"229979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229979\"]\r\n\r\n[latex]2\\mathrm{ln}x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_04\" class=\"example\">\r\n<div id=\"fs-id1165134163985\" class=\"exercise\">\r\n<div id=\"fs-id1165134163988\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\r\n[reveal-answer q=\"400323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"400323\"]\r\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"153337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153337\"]\r\n\r\n[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_05\" class=\"example\">\r\n<div id=\"fs-id1165134435871\" class=\"exercise\">\r\n<div id=\"fs-id1165134435874\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Using the Power Rule in Reverse<\/h3>\r\n<p id=\"fs-id1165137714060\">Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\n[reveal-answer q=\"851055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"851055\"]\r\n<p id=\"fs-id1165137416106\">Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\r\n<p style=\"text-align: center\">[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137502321\">Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\n[reveal-answer q=\"294897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"294897\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}16[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]45617[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]25581[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Expand logarithmic expressions<\/span>\r\n\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165137657409\">\r\n<p id=\"fs-id1165137558543\">Taken together, the product rule, quotient rule, and power rule are often called \"laws of logs.\" Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\r\n\r\n<div id=\"eip-423\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)&amp; ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y \\\\ &amp;={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135545872\">We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:<\/p>\r\n\r\n<div id=\"eip-622\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right) &amp; ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right) \\\\ &amp; ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right) \\\\ &amp; ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C \\\\ &amp; ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135153099\">We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\r\n<p id=\"fs-id1165135153103\">With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\r\n\r\n<div id=\"Example_04_05_06\" class=\"example\">\r\n<div id=\"fs-id1165135173497\" class=\"exercise\">\r\n<div id=\"fs-id1165135173499\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Expanding Logarithms Using Product, Quotient, and Power Rules<\/h3>\r\n<p id=\"fs-id1165135173504\">Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\r\n[reveal-answer q=\"737111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737111\"]\r\n<p id=\"fs-id1165135253768\">First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137854981\">Then seeing the product in the first term, we use the product rule:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134154611\">Finally, we use the power rule on the first term:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137635162\">Expand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"765910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"765910\"]\r\n\r\n[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_07\" class=\"example\">\r\n<div id=\"fs-id1165137811250\" class=\"exercise\">\r\n<div id=\"fs-id1165137811252\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\r\n<p id=\"fs-id1165135637419\">Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"304354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"304354\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\sqrt{x}\\right) &amp; =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)} \\\\ &amp; =\\frac{1}{2}\\mathrm{log}x \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137804498\">Expand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"410566\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410566\"]\r\n\r\n[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135173426\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1165134197968\"><strong>Can we expand<\/strong> [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/p>\r\n<p id=\"fs-id1165135440437\"><em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_08\" class=\"example\">\r\n<div id=\"fs-id1165135440448\" class=\"exercise\">\r\n<div id=\"fs-id1165135440450\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Expanding Complex Logarithmic Expressions<\/h3>\r\n<p id=\"fs-id1165135150641\">Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"840332\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"840332\"]\r\n<p id=\"fs-id1165134085801\">We can expand by applying the Product and Quotient Rules.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right) &amp; ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; \\text{Apply the Quotient Rule}. \\\\ &amp; ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; {\\text{Simplify by writing 64 as 2}}^{6}. \\\\ &amp; =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; \\text{Apply the Power Rule}.\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135635247\">Expand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"303340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303340\"]\r\n\r\n[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]35034[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174326[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Condense logarithmic expressions<\/span>\r\n\r\n<\/section>\r\n<p id=\"fs-id1165135190860\">We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\r\n\r\n<div id=\"fs-id1165135190866\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135190871\">How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\r\n<ol id=\"fs-id1165137833816\">\r\n \t<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\r\n \t<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\r\n \t<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_09\" class=\"example\">\r\n<div id=\"fs-id1165137833837\" class=\"exercise\">\r\n<div id=\"fs-id1165137833839\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\r\n<p id=\"fs-id1165135484124\">Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"118143\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"118143\"]\r\n<p id=\"fs-id1165135527077\">Using the product and quotient rules<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135400169\">This reduces our original expression to<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137846453\">Then, using the quotient rule<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134047712\">Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\r\n[reveal-answer q=\"144049\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"144049\"]\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_10\" class=\"example\">\r\n<div id=\"fs-id1165134435881\" class=\"exercise\">\r\n<div id=\"fs-id1165134435883\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Condensing Complex Logarithmic Expressions<\/h3>\r\n<p id=\"fs-id1165134435888\">Condense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"788800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788800\"]\r\n<p id=\"fs-id1165135344097\">We apply the power rule first:<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135531546\">Next we apply the product rule to the sum:<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134280852\">Finally, we apply the quotient rule to the difference:<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_05_11\" class=\"example\">\r\n<div id=\"fs-id1165135519254\" class=\"exercise\">\r\n<div id=\"fs-id1165135519256\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Rewriting as a Single Logarithm<\/h3>\r\n<p id=\"fs-id1165134156051\">Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"302963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"302963\"]\r\n<p id=\"fs-id1165135177652\">We apply the power rule first:<\/p>\r\n<p style=\"text-align: center\">[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135195405\">Next we apply the product rule to the sum:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134042304\">Finally, we apply the quotient rule to the difference:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135511375\">Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"235871\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"235871\"]\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135627841\">Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"628183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"628183\"]\r\n\r\n[latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174327[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_12\" class=\"example\">\r\n<div id=\"fs-id1165137400159\" class=\"exercise\">\r\n<div id=\"fs-id1165137400162\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Applying of the Laws of Logs<\/h3>\r\n<p id=\"fs-id1165135530298\">Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\r\n[reveal-answer q=\"197365\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197365\"]\r\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\r\n<p style=\"text-align: center\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\r\n<p style=\"text-align: center\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\r\n<p style=\"text-align: center\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\r\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135251378\">How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\r\n[reveal-answer q=\"371569\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371569\"]\r\n\r\nThe pH increases by about 0.301.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Use the change-of-base formula for logarithms<\/h2>\r\n<section id=\"fs-id1165137675210\">\r\n<p id=\"fs-id1165137675216\">Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the <strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.<\/p>\r\n<p id=\"fs-id1165137855374\">To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\r\n<p id=\"fs-id1165137855378\">Given any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex], we show<\/p>\r\n\r\n<div id=\"eip-643\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/div>\r\n<p id=\"fs-id1165137932683\">Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. By taking the log base [latex]n[\/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-226\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{n}\\left({b}^{y}\\right) &amp; ={\\mathrm{log}}_{n}M &amp;&amp; \\text{Apply the one-to-one property}. \\\\ y{\\mathrm{log}}_{n}b\\hfill &amp; ={\\mathrm{log}}_{n}M &amp;&amp; \\text{Apply the power rule for logarithms}. \\\\ y &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} &amp;&amp; \\text{Isolate }y. \\\\ {\\mathrm{log}}_{b}M &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} &amp;&amp; \\text{Substitute for }y. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135207389\">For example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.<\/p>\r\n\r\n<div id=\"eip-428\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{5}36 &amp; =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)} &amp;&amp; \\text{Apply the change of base formula using base 10.} \\\\ &amp; \\approx 2.2266 &amp;&amp; \\text{Use a calculator to evaluate to 4 decimal places.} \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165134381722\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Change-of-Base Formula<\/h3>\r\n<p id=\"fs-id1165135342066\">The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\r\n<p id=\"fs-id1165135342073\">For any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex],<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\r\n<p id=\"fs-id1165134042184\">It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\r\n<p id=\"fs-id1165137935512\">and<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137893333\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137893339\">How To: Given a logarithm with the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex].<\/h3>\r\n<ol id=\"fs-id1165134172563\">\r\n \t<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10, and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\r\n \t<li>Rewrite the log as a quotient using the change-of-base formula\r\n<ul id=\"fs-id1165134039298\">\r\n \t<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\r\n \t<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_13\" class=\"example\">\r\n<div id=\"fs-id1165134196189\" class=\"exercise\">\r\n<div id=\"fs-id1165134196191\" class=\"problem textbox shaded\">\r\n<h3>Example 13: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\r\n<p id=\"fs-id1165134196197\">Change [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.<\/p>\r\n[reveal-answer q=\"164120\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164120\"]\r\n<p id=\"fs-id1165135444049\">Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base, <em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\r\n<p id=\"fs-id1165135690112\">We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}M &amp; =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b} \\\\[1mm] {\\mathrm{log}}_{5}3 &amp; =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134223311\">Change [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.<\/p>\r\n[reveal-answer q=\"562111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"562111\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135181811\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1172294599410\"><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\r\n<p id=\"fs-id1165135193274\"><em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_14\" class=\"example\">\r\n<div id=\"fs-id1165134084328\" class=\"exercise\">\r\n<div id=\"fs-id1165134084330\" class=\"problem textbox shaded\">\r\n<h3>Example 14: Using the Change-of-Base Formula with a Calculator<\/h3>\r\n<p id=\"fs-id1165134084335\">Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\r\n[reveal-answer q=\"264040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264040\"]\r\n<p id=\"fs-id1165135353038\">According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base <i>e<\/i>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{2}10&amp;=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2} &amp;&amp; \\text{Apply the change of base formula using base }e. \\\\ &amp;\\approx 3.3219 &amp;&amp; \\text{Use a calculator to evaluate to 4 decimal places}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 21<\/h3>\r\n<p id=\"fs-id1165135358918\">Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\r\n[reveal-answer q=\"97931\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"97931\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 22<\/h3>\r\n[ohm_question hide_question_numbers=1]13570[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #1d1d1d;font-size: 1.5em;font-weight: bold;text-align: initial\">Key Equations<\/span>\r\n\r\n<\/section><section id=\"fs-id1165137870892\" class=\"key-equations\">\r\n<table id=\"fs-id1983134\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>Definition of the logarithmic function<\/td>\r\n<td>For [latex]\\text{ } x&gt;0,b&gt;0,b\\ne 1[\/latex],[latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{b}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of the common logarithm<\/td>\r\n<td>For [latex]\\text{ }x&gt;0[\/latex], [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{10}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of the natural logarithm<\/td>\r\n<td>For [latex]\\text{ }x&gt;0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{e}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section>\r\n<table id=\"fs-id1737642\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>General Form for the Translation of the Parent Logarithmic Function [latex]\\text{ }f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\r\n<td>[latex] f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section>\r\n<table id=\"fs-id2922999\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>The Product Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Quotient Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Power Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Change-of-Base Formula<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n&gt;0,n\\ne 1,b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section class=\"key-equations\"><span style=\"color: #1d1d1d;font-size: 1.5em;font-weight: bold\">Key Concepts<\/span><\/section><section id=\"fs-id1165135699130\" class=\"key-concepts\">\r\n<ul id=\"fs-id1165137574258\">\r\n \t<li>The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.<\/li>\r\n \t<li>Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm.<\/li>\r\n \t<li>Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm.<\/li>\r\n \t<li>Logarithmic functions with base <em>b<\/em>\u00a0can be evaluated mentally using previous knowledge of powers of <em>b<\/em>.<\/li>\r\n \t<li>Common logarithms can be evaluated mentally using previous knowledge of powers of 10.<\/li>\r\n \t<li>When common logarithms cannot be evaluated mentally, a calculator can be used.<\/li>\r\n \t<li>Real-world exponential problems with base 10\u00a0can be rewritten as a common logarithm and then evaluated using a calculator.<\/li>\r\n \t<li>Natural logarithms can be evaluated using a calculator.<\/li>\r\n \t<li>To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for <em>x<\/em>.<\/li>\r\n \t<li>The graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] has an <em>x-<\/em>intercept at [latex]\\left(1,0\\right)[\/latex], domain [latex]\\left(0,\\infty \\right)[\/latex], range [latex]\\left(-\\infty ,\\infty \\right)[\/latex], vertical asymptote <em>x\u00a0<\/em>= 0, and\r\n<ul id=\"fs-id1165135441773\">\r\n \t<li>if <em>b\u00a0<\/em>&gt; 1, the function is increasing.<\/li>\r\n \t<li>if 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function is decreasing.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex] shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] horizontally\r\n<ul id=\"fs-id1165135512562\">\r\n \t<li>left <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&gt; 0.<\/li>\r\n \t<li>right <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&lt; 0.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex] shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically\r\n<ul id=\"fs-id1165137761068\">\r\n \t<li>up <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&gt; 0.<\/li>\r\n \t<li>down <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&lt; 0.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>For any constant <em>a\u00a0<\/em>&gt; 0, the equation [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]\r\n<ul id=\"fs-id1165134040579\">\r\n \t<li>stretches the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if |<em>a<\/em>| &gt; 1.<\/li>\r\n \t<li>compresses the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if |<em>a<\/em>| &lt; 1.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>When the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by \u20131, the result is a reflection about the <em>x<\/em>-axis. When the input is multiplied by \u20131, the result is a reflection about the <em>y<\/em>-axis.\r\n<ul id=\"fs-id1165135186594\">\r\n \t<li>The equation [latex]f\\left(x\\right)=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] represents a reflection of the parent function about the <em>x-<\/em>axis.<\/li>\r\n \t<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex] represents a reflection of the parent function about the <em>y-<\/em>axis.<\/li>\r\n<\/ul>\r\n<ul id=\"fs-id1165137834414\">\r\n \t<li>A graphing calculator may be used to approximate solutions to some logarithmic equations.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>All translations of the logarithmic function can be summarized by the general equation [latex] f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex].<\/li>\r\n \t<li>Given an equation with the general form [latex] f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex], we can identify the vertical asymptote <em>x\u00a0<\/em>= \u2013c for the transformation.<\/li>\r\n \t<li>Using the general equation [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex], we can write the equation of a logarithmic function given its graph.<\/li>\r\n \t<li>We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.<\/li>\r\n \t<li>We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.<\/li>\r\n \t<li>We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base.<\/li>\r\n \t<li>We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.<\/li>\r\n \t<li>The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.<\/li>\r\n \t<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\r\n \t<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and <i>e<\/i>\u00a0as the quotient of natural or common logs. That way a calculator can be used to evaluate.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135160066\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137890644\" class=\"definition\">\r\n \t<dt><strong>change-of-base formula<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890649\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890654\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165135160066\" class=\"definition\">\r\n \t<dt><strong>common logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137571387\">the exponent to which 10 must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] is written simply as [latex]\\mathrm{log}\\left(x\\right)[\/latex].<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n \t<dt>\r\n<dl id=\"fs-id1165137780762\" class=\"definition\">\r\n \t<dt><strong>logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137849198\">the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>; written [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137507853\" class=\"definition\">\r\n \t<dt><strong>natural logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134037589\">the exponent to which the number <em>e<\/em>\u00a0must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n \t<dt><strong>power rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890659\">a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890664\" class=\"definition\">\r\n \t<dt><strong>product rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890670\">a rule of logarithms that states that the log of a product is equal to a sum of logarithms<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890674\" class=\"definition\">\r\n \t<dt><strong>quotient rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890679\">a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert between logarithmic to exponential form.<\/li>\n<li>Evaluate logarithms.<\/li>\n<li>Use common logarithms to model real-world problems.<\/li>\n<li>Use natural logarithms to model real-world problems.<\/li>\n<li>Identify the domain of a logarithmic function.<\/li>\n<li>Graph logarithmic functions.<\/li>\n<li>Use the product rule for logarithms.<\/li>\n<li>Use the quotient rule for logarithms.<\/li>\n<li>Use the power rule for logarithms.<\/li>\n<li>Expand logarithmic expressions.<\/li>\n<li>Condense logarithmic expressions.<\/li>\n<li>Use the change-of-base formula for logarithms.<\/li>\n<\/ul>\n<\/div>\n<figure id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137557013\">In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary. Accessed 3\/4\/2013.\" id=\"return-footnote-13703-1\" href=\"#footnote-13703-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary. Accessed 3\/4\/2013.\" id=\"return-footnote-13703-2\" href=\"#footnote-13703-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>\u00a0like those shown in the picture above. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013.\" id=\"return-footnote-13703-3\" href=\"#footnote-13703-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0whereas the Japanese earthquake registered a 9.0.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details. Accessed 3\/4\/2013.\" id=\"return-footnote-13703-4\" href=\"#footnote-13703-4\" aria-label=\"Footnote 4\"><sup class=\"footnote\">[4]<\/sup><\/a><\/p>\n<p id=\"fs-id1165137760714\">The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\n<h2>Convert from logarithmic to exponential form<\/h2>\n<section id=\"fs-id1165137644550\">\n<p id=\"fs-id1165135192781\">In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex], where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?<\/p>\n<p>We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3, since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137662989\">Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\n<p id=\"fs-id1165137404844\">We read a logarithmic expression as, &#8220;The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,&#8221; or, simplified, &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.&#8221; We can also say, &#8220;<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,&#8221; because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as &#8220;log base 2 of 32 is 5.&#8221;<\/p>\n<p id=\"fs-id1165137597501\">We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/div>\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive.<span id=\"fs-id1165137696233\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010820\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/><\/span><\/p>\n<p id=\"fs-id1165137400957\">Because logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex], using parentheses to denote function evaluation, just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<span id=\"fs-id1165137771679\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/><\/span><\/p>\n<p id=\"fs-id1165137575165\">Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.<\/p>\n<div id=\"fs-id1165137472937\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Definition of the Logarithmic Function<\/h3>\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\n<p id=\"fs-id1165137584967\">For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165137433829\" class=\"equation\" style=\"text-align: center\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\n<p id=\"fs-id1165137893373\">where,<\/p>\n<ul id=\"fs-id1165135530561\">\n<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>&#8221; or the &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>.&#8221;<\/li>\n<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\n<\/ul>\n<p id=\"fs-id1165137547773\">Also, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\n<ul id=\"fs-id1165137643167\">\n<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\n<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137677696\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1549475\"><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\n<p id=\"fs-id1165137653864\"><em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165137874700\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137806301\">How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form.<\/h3>\n<ol id=\"fs-id1165137641669\">\n<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\n<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_01\" class=\"example\">\n<div id=\"fs-id1165135570363\" class=\"exercise\">\n<div id=\"fs-id1165137557855\" class=\"problem textbox shaded\">\n<h3>Example 1: Converting from Logarithmic Form to Exponential Form<\/h3>\n<p id=\"fs-id1165137580570\">Write the following logarithmic equations in exponential form.<\/p>\n<ol id=\"fs-id1165137705346\">\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q799633\">Show Solution<\/span><\/p>\n<div id=\"q799633\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137408172\">First, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\n<ol id=\"fs-id1165137705659\">\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]\n<p id=\"fs-id1165137602796\">Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/p>\n<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\n<p id=\"fs-id1165137698078\">Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equivalent to [latex]{3}^{2}=9[\/latex].<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137418681\">Write the following logarithmic equations in exponential form.<\/p>\n<p style=\"padding-left: 60px\">a. [latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/p>\n<p style=\"padding-left: 60px\">b. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q234346\">Show Solution<\/span><\/p>\n<div id=\"q234346\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equivalent to [latex]{10}^{6}=1,000,000[\/latex]<br \/>\nb. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equivalent to [latex]{5}^{2}=25[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm151465\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=151465&theme=oea&iframe_resize_id=ohm151465\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">\u00a0Convert from exponential to logarithmic form<\/span><\/p>\n<\/section>\n<p id=\"fs-id1165137933968\">To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<div id=\"Example_04_03_02\" class=\"example\">\n<div id=\"fs-id1165135168111\" class=\"exercise\">\n<div id=\"fs-id1165137727912\" class=\"problem textbox shaded\">\n<h3>Example 2: Converting from Exponential Form to Logarithmic Form<\/h3>\n<p id=\"fs-id1165137804412\">Write the following exponential equations in logarithmic form.<\/p>\n<ol id=\"fs-id1165135192287\">\n<li>[latex]{2}^{3}=8[\/latex]<\/li>\n<li>[latex]{5}^{2}=25[\/latex]<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q693170\">Show Solution<\/span><\/p>\n<div id=\"q693170\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137474116\">First, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<ol id=\"fs-id1165137573458\">\n<li>[latex]{2}^{3}=8[\/latex]\n<p id=\"fs-id1165137466396\">Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\n<\/li>\n<li>[latex]{5}^{2}=25[\/latex]\n<p id=\"fs-id1165135193035\">Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/p>\n<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\n<p id=\"fs-id1165135187822\">Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equivalent to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137566762\">Write the following exponential equations in logarithmic form.<\/p>\n<p style=\"padding-left: 60px\">a. [latex]{3}^{2}=9[\/latex]<\/p>\n<p style=\"padding-left: 60px\">b. [latex]{5}^{3}=125[\/latex]<\/p>\n<p style=\"padding-left: 60px\">c. [latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932188\">Show Solution<\/span><\/p>\n<div id=\"q932188\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]{3}^{2}=9[\/latex] is equivalent to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<br \/>\nb. [latex]{5}^{3}=125[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]<br \/>\nc. [latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equivalent to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14387\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14387&theme=oea&iframe_resize_id=ohm14387\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Evaluate logarithms<\/h2>\n<section id=\"fs-id1165137530906\">\n<p id=\"fs-id1165137422589\">Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, &#8220;To what exponent must 2\u00a0be raised in order to get 8?&#8221; Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].<\/p>\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137937690\">\n<li>We ask, &#8220;To what exponent must 7 be raised in order to get 49?&#8221; We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex]<\/li>\n<li>We ask, &#8220;To what exponent must 3 be raised in order to get 27?&#8221; We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137456358\">Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137584208\">\n<li>We ask, &#8220;To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? &#8221; We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137455840\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137453770\">How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally.<\/h3>\n<ol id=\"fs-id1165134079724\">\n<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\n<li>Use previous knowledge of powers of <em>b<\/em>\u00a0identify <em>y<\/em>\u00a0by asking, &#8220;To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?&#8221;<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_03\" class=\"example\">\n<div id=\"fs-id1165137732842\" class=\"exercise\">\n<div id=\"fs-id1165135296345\" class=\"problem textbox shaded\">\n<h3>Example 3: Solving Logarithms Mentally<\/h3>\n<p id=\"fs-id1165135393440\">Solve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q960893\">Show Solution<\/span><\/p>\n<div id=\"q960893\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137611276\">First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, &#8220;To what exponent must 4 be raised in order to get 64?&#8221;<\/p>\n<p id=\"fs-id1165137661814\">We know<\/p>\n<p style=\"text-align: center\">[latex]{4}^{3}=64[\/latex]<\/p>\n<p id=\"fs-id1165137619013\">Therefore,<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137745041\">Solve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q650736\">Show Solution<\/span><\/p>\n<div id=\"q650736\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recalling that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex])<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_03_04\" class=\"example\">\n<div id=\"fs-id1165137663658\" class=\"exercise\">\n<div id=\"fs-id1165137680390\" class=\"problem textbox shaded\">\n<h3>Example 4: Evaluating the Logarithm of a Reciprocal<\/h3>\n<p id=\"fs-id1165137938805\">Evaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q177970\">Show Solution<\/span><\/p>\n<div id=\"q177970\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, &#8220;To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]&#8220;?<\/p>\n<p id=\"fs-id1165137552085\">We know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{3}^{-3}&=\\frac{1}{{3}^{3}} \\\\ &=\\frac{1}{27} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135437134\">Evaluate [latex]y={\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q708400\">Show Solution<\/span><\/p>\n<div id=\"q708400\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35042\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35042&theme=oea&iframe_resize_id=ohm35042\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">\u00a0Use common logarithms<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165137405741\">\n<p id=\"fs-id1165137661970\">The most frequently used base for logarithms is <em>e<\/em>. Base <em>e<\/em>\u00a0logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base <em>e<\/em>\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137473872\">Most values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.<\/p>\n<div id=\"fs-id1165137452317\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Definition of the Natural Logarithm<\/h3>\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\n<p id=\"fs-id1165135613642\">For [latex]x>0[\/latex],<\/p>\n<div id=\"fs-id1165137580230\" class=\"equation\" style=\"text-align: center\">[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equivalent to }{e}^{y}=x[\/latex]<\/div>\n<p id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>&#8221; or &#8220;the natural logarithm of <em>x<\/em>.&#8221;<\/p>\n<p id=\"fs-id1165137566720\">The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137705251\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for <em>x\u00a0<\/em>&gt; 0.<\/p>\n<\/div>\n<div id=\"fs-id1165137409558\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137832169\">How To: Given a natural logarithm with the form [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex], evaluate it using a calculator.<\/h3>\n<ol id=\"fs-id1165135407195\">\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter the value given for <em>x<\/em>, followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_08\" class=\"example\">\n<div id=\"fs-id1165137731536\" class=\"exercise\">\n<div id=\"fs-id1165137434974\" class=\"problem textbox shaded\">\n<h3>Example 5: Evaluating a Natural Logarithm Using a Calculator<\/h3>\n<p id=\"fs-id1165137573341\">Evaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q847079\">Show Solution<\/span><\/p>\n<div id=\"q847079\" class=\"hidden-answer\" style=\"display: none\">\n<ul id=\"fs-id1165137563770\">\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter 500, followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ul>\n<p id=\"fs-id1165137645024\">Rounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137435623\">Evaluate [latex]\\mathrm{ln}\\left(-500\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q393229\">Show Solution<\/span><\/p>\n<div id=\"q393229\" class=\"hidden-answer\" style=\"display: none\">\n<p>It is not possible to take the logarithm of a negative number in the set of real numbers.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35022\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35022&theme=oea&iframe_resize_id=ohm35022\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1165135194555\">In <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction-to-graphs-of-exponential-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of Exponential Functions<\/a>, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the <em>cause<\/em> for an <em>effect<\/em>.<\/p>\n<p id=\"fs-id1165137603580\">To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year <em>t<\/em>\u00a0can be found with the equation [latex]A=2500{e}^{0.05t}[\/latex].<\/p>\n<p>But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1\u00a0shows this point on the logarithmic graph.<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_0012.jpg\" alt=\"A graph titled,\" width=\"900\" height=\"459\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165135161452\">In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions.<\/p>\n<h2>Identify the domain of a logarithmic function<\/h2>\n<p id=\"fs-id1165137748716\">Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.<\/p>\n<p id=\"fs-id1165137758495\">Recall that the exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number <em>x<\/em>\u00a0and constant [latex]b>0[\/latex], [latex]b\\ne 1[\/latex], where<\/p>\n<ul id=\"fs-id1165137736024\">\n<li>The domain of <em>y<\/em>\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<li>The range of <em>y<\/em>\u00a0is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165135641666\">In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:<\/p>\n<ul id=\"fs-id1165137656096\">\n<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]:[latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165135245571\">Transformations of the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations\u2014shifts, stretches, compressions, and reflections\u2014to the parent function without loss of shape.<\/p>\n<p id=\"fs-id1165137653624\">In <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction-to-graphs-of-exponential-functions\/\" target=\"_blank\" rel=\"noopener\">Graphs of Exponential Functions<\/a> we saw that certain transformations can change the <em>range<\/em> of [latex]y={b}^{x}[\/latex]. Similarly, applying transformations to the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can change the <em>domain<\/em>. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists <em>only of positive real numbers<\/em>. That is, the argument of the logarithmic function must be greater than zero.<\/p>\n<p id=\"fs-id1165137851584\">For example, consider [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex]. This function is defined for any values of <em>x<\/em>\u00a0such that the argument, in this case [latex]2x - 3[\/latex], is greater than zero. To find the domain, we set up an inequality and solve for\u00a0<em>x<\/em>:<\/p>\n<div id=\"eip-318\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&2x - 3>0 && \\text{Show the argument greater than zero}. \\\\ &2x>3 && \\text{Add 3}. \\\\ &x>1.5 && \\text{Divide by 2}. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137645047\">In interval notation, the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex] is [latex]\\left(1.5,\\infty \\right)[\/latex].<\/p>\n<div id=\"fs-id1165137423048\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135173951\">How To: Given a logarithmic function, identify the domain.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165137823224\">\n<li>Set up an inequality showing the argument greater than zero.<\/li>\n<li>Solve for <em>x<\/em>.<\/li>\n<li>Write the domain in interval notation.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_01\" class=\"example\">\n<div id=\"fs-id1165137846475\" class=\"exercise\">\n<div id=\"fs-id1165137460694\" class=\"problem textbox shaded\">\n<h3>Example 1: Identifying the Domain of a Logarithmic Shift<\/h3>\n<p id=\"fs-id1165135209576\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q174870\">Show Solution<\/span><\/p>\n<div id=\"q174870\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137693442\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]x+3>0[\/latex]. Solving this inequality,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&x+3>0 && \\text{The input must be positive}. \\\\ &x>-3 && \\text{Subtract 3}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137638183\">The domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137645484\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x - 2\\right)+1[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q405290\">Show Solution<\/span><\/p>\n<div id=\"q405290\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(2,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_04_02\" class=\"example\">\n<div id=\"fs-id1165137894615\" class=\"exercise\">\n<div id=\"fs-id1165134108527\" class=\"problem textbox shaded\">\n<h3>Example 2: Identifying the Domain of a Logarithmic Shift and Reflection<\/h3>\n<p id=\"fs-id1165135499558\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q675604\">Show Solution<\/span><\/p>\n<div id=\"q675604\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137780875\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]5 - 2x>0[\/latex]. Solving this inequality,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&5 - 2x>0 && \\text{The input must be positive}. \\\\ &-2x>-5 && \\text{Subtract }5. \\\\ &x<\\frac{5}{2} && \\text{Divide by }-2\\text{ and switch the inequality}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137656879\">The domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137453336\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(x - 5\\right)+2[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q87516\">Show Solution<\/span><\/p>\n<div id=\"q87516\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(5,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174284\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174284&theme=oea&iframe_resize_id=ohm174284\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Graph logarithmic functions<\/h2>\n<p id=\"fs-id1165134104063\">Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] along with all its transformations: shifts, stretches, compressions, and reflections.<\/p>\n<p id=\"fs-id1165137679088\">We begin with the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Because every logarithmic function of this form is the inverse of an exponential function with the form [latex]y={b}^{x}[\/latex], their graphs will be reflections of each other across the line [latex]y=x[\/latex]. To illustrate this, we can observe the relationship between the input and output values of [latex]y={2}^{x}[\/latex] and its equivalent [latex]x={\\mathrm{log}}_{2}\\left(y\\right)[\/latex] in the table below.<\/p>\n<table id=\"Table_04_04_01\" summary=\"Three rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]{2}^{x}=y[\/latex]<\/strong><\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]{\\mathrm{log}}_{2}\\left(y\\right)=x[\/latex]<\/strong><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135509175\">Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/p>\n<table id=\"Table_04_04_02\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]\\left(-3,\\frac{1}{8}\\right)[\/latex]<\/td>\n<td>[latex]\\left(-2,\\frac{1}{4}\\right)[\/latex]<\/td>\n<td>[latex]\\left(-1,\\frac{1}{2}\\right)[\/latex]<\/td>\n<td>[latex]\\left(0,1\\right)[\/latex]<\/td>\n<td>[latex]\\left(1,2\\right)[\/latex]<\/td>\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td>[latex]\\left(3,8\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>[latex]\\left(\\frac{1}{8},-3\\right)[\/latex]<\/td>\n<td>[latex]\\left(\\frac{1}{4},-2\\right)[\/latex]<\/td>\n<td>[latex]\\left(\\frac{1}{2},-1\\right)[\/latex]<\/td>\n<td>[latex]\\left(1,0\\right)[\/latex]<\/td>\n<td>[latex]\\left(2,1\\right)[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\n<td>[latex]\\left(8,3\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137761335\">As we\u2019d expect, the <em>x<\/em>&#8211; and <em>y<\/em>-coordinates are reversed for the inverse functions. The figure below\u00a0shows the graph of <em>f<\/em>\u00a0and <em>g<\/em>.<\/p>\n<figure class=\"small\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_0022.jpg\" alt=\"Graph of two functions, f(x)=2^x and g(x)=log_2(x), with the line y=x denoting the axis of symmetry.\" \/><\/figure>\n<p style=\"text-align: center\"><strong>Figure 2.\u00a0<\/strong>Notice that the graphs of [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] are reflections about the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137406913\">Observe the following from the graph:<\/p>\n<ul id=\"fs-id1165137408405\">\n<li>[latex]f\\left(x\\right)={2}^{x}[\/latex] has a <em>y<\/em>-intercept at [latex]\\left(0,1\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] has an <em>x<\/em>-intercept at [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(-\\infty ,\\infty \\right)[\/latex], is the same as the range of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\n<li>The range of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(0,\\infty \\right)[\/latex], is the same as the domain of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137780760\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Characteristics of the Graph of the Parent Function, <em>f<\/em>(<em>x<\/em>) = log<sub><em>b<\/em><\/sub>(<em>x<\/em>)<\/h3>\n<p id=\"fs-id1165135520250\">For any real number <em>x<\/em>\u00a0and constant <em>b\u00a0<\/em>&gt; 0, [latex]b\\ne 1[\/latex], we can see the following characteristics in the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]:<\/p>\n<ul id=\"fs-id1165137400150\">\n<li>one-to-one function<\/li>\n<li>vertical asymptote: <em>x\u00a0<\/em>= 0<\/li>\n<li>domain: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/li>\n<li><em>x-<\/em>intercept: [latex]\\left(1,0\\right)[\/latex] and key point [latex]\\left(b,1\\right)[\/latex]<\/li>\n<li><em>y<\/em>-intercept: none<\/li>\n<li>increasing if [latex]b>1[\/latex]<\/li>\n<li>decreasing if 0 &lt; <em>b\u00a0<\/em>&lt; 1<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_04_003\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010821\/CNX_Precalc_Figure_04_04_003G2.jpg\" alt=\"&quot;Two\" \/><\/figure>\n<p>Figure 3\u00a0shows how changing the base <em>b<\/em>\u00a0in [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (<em>Note:<\/em> recall that the function [latex]\\mathrm{ln}\\left(x\\right)[\/latex] has base [latex]e\\approx \\text{2}.\\text{718.)}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0042.jpg\" alt=\"Graph of three equations: y=log_2(x) in blue, y=ln(x) in orange, and y=log(x) in red. The y-axis is the asymptote.\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 4.\u00a0<\/strong>The graphs of three logarithmic functions with different bases, all greater than 1.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137871937\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137805513\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph the function.<\/h3>\n<ol id=\"fs-id1165135435529\">\n<li>Draw and label the vertical asymptote, <em>x<\/em> = 0.<\/li>\n<li>Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>Plot the key point [latex]\\left(b,1\\right)[\/latex].<\/li>\n<li>Draw a smooth curve through the points.<\/li>\n<li>State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote, <em>x<\/em> = 0.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_03\" class=\"example\">\n<div id=\"fs-id1165137550508\" class=\"exercise\">\n<div id=\"fs-id1165137550510\" class=\"problem textbox shaded\">\n<h3>Example 3: Graphing a Logarithmic Function with the Form\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<\/h3>\n<p id=\"fs-id1165137431970\">Graph [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q347847\">Show Solution<\/span><\/p>\n<div id=\"q347847\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137501970\">Before graphing, identify the behavior and key points for the graph.<\/p>\n<ul id=\"fs-id1165135497154\">\n<li>Since <em>b\u00a0<\/em>= 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote <em>x\u00a0<\/em>= 0, and the right tail will increase slowly without bound.<\/li>\n<li>The <em>x<\/em>-intercept is [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>The key point [latex]\\left(5,1\\right)[\/latex] is on the graph.<\/li>\n<li>We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_04_005\" class=\"small\"><span id=\"fs-id1165135508394\"> <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0052.jpg\" alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\" width=\"557\" height=\"419\" \/><\/span><\/figure>\n<p id=\"fs-id1165135697920\" style=\"text-align: center\"><strong>Figure 5.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135171582\">Graph [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{5}}\\left(x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808887\">Show Solution<\/span><\/p>\n<div id=\"q808887\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165134377926\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_0062.jpg\" alt=\"Graph of f(x)=log_(1\/5)(x) with labeled points at (1\/5, 1) and (1, 0). The y-axis is the asymptote.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174289\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174289&theme=oea&iframe_resize_id=ohm174289\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Graphing Transformations of Logarithmic Functions<\/h2>\n<p id=\"fs-id1165137430986\">As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the <strong>parent function<\/strong> [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] without loss of shape.<\/p>\n<section id=\"fs-id1165137734884\">\n<h2>Graphing a Horizontal Shift of\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/h2>\n<p>When a constant <em>c<\/em>\u00a0is added to the input of the parent function [latex]f\\left(x\\right)=\\text{log}_{b}\\left(x\\right)[\/latex], the result is a <strong>horizontal shift<\/strong> <em>c<\/em>\u00a0units in the <em>opposite<\/em> direction of the sign on <em>c<\/em>. To visualize horizontal shifts, we can observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and for <em>c\u00a0<\/em>&gt; 0 alongside the shift left, [latex]g\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex], and the shift right, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(x-c\\right)[\/latex].<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010822\/CNX_Precalc_Figure_04_04_007n2.jpg\" alt=\"Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x+c) is the translation function with an asymptote at x=-c. This shows the translation of shifting left.\" width=\"900\" height=\"526\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165135296307\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Horizontal Shifts of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165135176174\">For any constant <em>c<\/em>, the function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex]<\/p>\n<ul id=\"fs-id1165135206192\">\n<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] left <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&gt; 0.<\/li>\n<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] right <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&lt; 0.<\/li>\n<li>has the vertical asymptote <em>x\u00a0<\/em>= \u2013<em>c<\/em>.<\/li>\n<li>has domain [latex]\\left(-c,\\infty \\right)[\/latex].<\/li>\n<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137641710\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137641715\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex], graph the translation.<\/h3>\n<ol id=\"fs-id1165137454284\">\n<li>Identify the horizontal shift:\n<ol id=\"fs-id1165137454288\">\n<li>If <em>c<\/em> &gt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] left <em>c<\/em>\u00a0units.<\/li>\n<li>If <em>c\u00a0<\/em>&lt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] right <em>c<\/em>\u00a0units.<\/li>\n<\/ol>\n<\/li>\n<li>Draw the vertical asymptote <em>x\u00a0<\/em>= \u2013<em>c<\/em>.<\/li>\n<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting <em>c<\/em>\u00a0from the\u00a0<em>x<\/em>\u00a0coordinate.<\/li>\n<li>Label the three points.<\/li>\n<li>The Domain is [latex]\\left(-c,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= \u2013c.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_04\" class=\"example\">\n<div id=\"fs-id1165137414959\" class=\"exercise\">\n<div id=\"fs-id1165137414961\" class=\"problem textbox shaded\">\n<h3>Example 4:\u00a0Graphing a Horizontal Shift of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137455420\">Sketch the horizontal shift [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex] alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q785817\">Show Solution<\/span><\/p>\n<div id=\"q785817\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137759885\">Since the function is [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex], we notice [latex]x+\\left(-2\\right)=x - 2[\/latex].<\/p>\n<p id=\"fs-id1165137784630\">Thus <em>c\u00a0<\/em>= \u20132, so <em>c\u00a0<\/em>&lt; 0. This means we will shift the function [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] right 2 units.<\/p>\n<p id=\"fs-id1165137836995\">The vertical asymptote is [latex]x=-\\left(-2\\right)[\/latex] or <em>x\u00a0<\/em>= 2.<\/p>\n<p id=\"fs-id1165134042608\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{3},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(3,1\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137475806\">The new coordinates are found by adding 2 to the <em>x<\/em>\u00a0coordinates.<\/p>\n<p id=\"fs-id1165137748449\">Label the points [latex]\\left(\\frac{7}{3},-1\\right)[\/latex], [latex]\\left(3,0\\right)[\/latex], and [latex]\\left(5,1\\right)[\/latex].<\/p>\n<p>The domain is [latex]\\left(2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 2.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0082.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1\/3, -1), (1, 0), and (3, 1).The translation function f(x)=log_3(x-2) has an asymptote at x=2 and labeled points at (3, 0) and (5, 1).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"font-size: 0.9em\">\u00a0<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135329937\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x+4\\right)[\/latex] alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q139882\">Show Solution<\/span><\/p>\n<div id=\"q139882\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(-4,\\infty \\right)[\/latex], the range [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the asymptote <em>x\u00a0<\/em>= \u20134.<span id=\"fs-id1165135209395\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0092.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1, 0), and (3, 1).The translation function f(x)=log_3(x+4) has an asymptote at x=-4 and labeled points at (-3, 0) and (-1, 1).\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174300\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174300&theme=oea&iframe_resize_id=ohm174300\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing a Vertical Shift of\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165135403538\">When a constant <em>d<\/em>\u00a0is added to the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], the result is a <strong>vertical shift<\/strong> <em>d<\/em>\u00a0units in the direction of the sign on <em>d<\/em>. To visualize vertical shifts, we can observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the shift up, [latex]g\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex] and the shift down, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)-d[\/latex].<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_010F2.jpg\" alt=\"Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)+d is the translation function with an asymptote at x=0. This shows the translation of shifting up. Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)-d is the translation function with an asymptote at x=0. This shows the translation of shifting down.\" width=\"900\" height=\"684\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165137767601\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Vertical Shifts of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137661370\">For any constant <em>d<\/em>, the function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex]<\/p>\n<ul id=\"fs-id1165137803105\">\n<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] up <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&gt; 0.<\/li>\n<li>shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] down <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&lt; 0.<\/li>\n<li>has the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\n<li>has domain [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137706002\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137706009\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex], graph the translation.<\/h3>\n<ol>\n<li>Identify the vertical shift:\n<ol>\n<li>If <em>d\u00a0<\/em>&gt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] up <em>d<\/em>\u00a0units.<\/li>\n<li>If <em>d\u00a0<\/em>&lt; 0, shift the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] down <em>d\u00a0<\/em>units.<\/li>\n<\/ol>\n<\/li>\n<li>Draw the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\n<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by adding <em>d<\/em>\u00a0to the <em>y\u00a0<\/em>coordinate.<\/li>\n<li>Label the three points.<\/li>\n<li>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_05\" class=\"example\">\n<div id=\"fs-id1165137470057\" class=\"exercise\">\n<div id=\"fs-id1165137470059\" class=\"problem textbox shaded\">\n<h3>Example 5: Graphing a Vertical Shift of the Parent Function\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137832038\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)-2[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q657475\">Show Solution<\/span><\/p>\n<div id=\"q657475\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137465913\">Since the function is [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)-2[\/latex], we will notice <em>d\u00a0<\/em>= \u20132. Thus <em>d\u00a0<\/em>&lt; 0.<\/p>\n<p id=\"fs-id1165135175015\">This means we will shift the function [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] down 2 units.<\/p>\n<p id=\"fs-id1165137644429\">The vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\n<p id=\"fs-id1165137408419\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{3},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(3,1\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135503945\">The new coordinates are found by subtracting 2 from the <em>y <\/em>coordinates.<\/p>\n<p id=\"fs-id1165135421660\">Label the points [latex]\\left(\\frac{1}{3},-3\\right)[\/latex], [latex]\\left(1,-2\\right)[\/latex], and [latex]\\left(3,-1\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135195524\">The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<span id=\"fs-id1165134393856\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010823\/CNX_Precalc_Figure_04_04_0112.jpg\" alt=\"Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1\/3, -1), (1, 0), and (3, 1).The translation function f(x)=log_3(x)-2 has an asymptote at x=0 and labeled points at (1, 0) and (3, 1).\" \/><\/span><\/p>\n<p id=\"fs-id1165137698285\" style=\"text-align: center\"><strong>Figure 9.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137760886\">Sketch a graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)+2[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q597513\">Show Solution<\/span><\/p>\n<div id=\"q597513\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165137874471\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0122.jpg\" alt=\"Graph of two functions. The parent function is y=log_2(x), with an asymptote at x=0 and labeled points at (1, 0), and (2, 1).The translation function f(x)=log_2(x)+2 has an asymptote at x=0 and labeled points at (0.25, 0) and (0.5, 1).\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174304\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174304&theme=oea&iframe_resize_id=ohm174304\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing Stretches and Compressions of\u00a0[latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165137770245\">When the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by a constant <em>a<\/em> &gt; 0, the result is a <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the original graph. To visualize stretches and compressions, we set <em>a\u00a0<\/em>&gt; 1 and observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the vertical stretch, [latex]g\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and the vertical compression, [latex]h\\left(x\\right)=\\frac{1}{a}{\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<img decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_013n2.jpg\" alt=\"&quot;Graph\" \/><\/p>\n<div id=\"fs-id1165137433996\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Vertical Stretches and Compressions of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137758179\">For any constant <em>a<\/em> &gt; 1, the function [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/p>\n<ul id=\"fs-id1165137428102\">\n<li>stretches the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if <em>a\u00a0<\/em>&gt; 1.<\/li>\n<li>compresses the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if 0 &lt; <em>a\u00a0<\/em>&lt; 1.<\/li>\n<li>has the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\n<li>has the <em>x<\/em>-intercept [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>has domain [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<li>has range [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135169301\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135169307\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex], [latex]a>0[\/latex], graph the translation.<\/h3>\n<ol id=\"fs-id1165137464127\">\n<li>Identify the vertical stretch or compressions:\n<ol id=\"eip-id1165134081434\">\n<li>If [latex]|a|>1[\/latex], the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is stretched by a factor of <em>a<\/em>\u00a0units.<\/li>\n<li>If [latex]|a|<1[\/latex], the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is compressed by a factor of <em>a<\/em>\u00a0units.<\/li>\n<\/ol>\n<\/li>\n<li>Draw the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\n<li>Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the <em>y<\/em>\u00a0coordinates by <em>a<\/em>.<\/li>\n<li>Label the three points.<\/li>\n<li>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x<\/em> = 0.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_06\" class=\"example\">\n<div id=\"fs-id1165135309914\" class=\"exercise\">\n<div id=\"fs-id1165135309916\" class=\"problem textbox shaded\">\n<h3>Example 6: Graphing a Stretch or Compression of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137602128\">Sketch a graph of [latex]f\\left(x\\right)=2{\\mathrm{log}}_{4}\\left(x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q846570\">Show Solution<\/span><\/p>\n<div id=\"q846570\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135210052\">Since the function is [latex]f\\left(x\\right)=2{\\mathrm{log}}_{4}\\left(x\\right)[\/latex], we will notice <em>a\u00a0<\/em>= 2.<\/p>\n<p id=\"fs-id1165135384321\">This means we will stretch the function [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] by a factor of 2.<\/p>\n<p id=\"fs-id1165135481989\">The vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\n<p id=\"fs-id1165137757801\">Consider the three key points from the parent function, [latex]\\left(\\frac{1}{4},-1\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(4,1\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135570058\">The new coordinates are found by multiplying the <em>y<\/em>\u00a0coordinates by 2.<\/p>\n<p id=\"fs-id1165137837989\">Label the points [latex]\\left(\\frac{1}{4},-2\\right)[\/latex], [latex]\\left(1,0\\right)[\/latex], and [latex]\\left(4,\\text{2}\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135543469\">The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165134059742\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0142.jpg\" alt=\"Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=2log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (2, 1).\" \/><\/span><\/p>\n<p id=\"fs-id1165135566827\" style=\"text-align: center\"><strong>Figure 11.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135471122\">Sketch a graph of [latex]f\\left(x\\right)=\\frac{1}{2}{\\mathrm{log}}_{4}\\left(x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q645251\">Show Solution<\/span><\/p>\n<div id=\"q645251\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165135332505\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0152.jpg\" alt=\"Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=(1\/2)log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (16, 1).\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_04_07\" class=\"example\">\n<div id=\"fs-id1165134267814\" class=\"exercise\">\n<div id=\"fs-id1165134267816\" class=\"problem textbox shaded\">\n<h3>Example 7: Combining a Shift and a Stretch<\/h3>\n<p id=\"fs-id1165137863045\">Sketch a graph of [latex]f\\left(x\\right)=5\\mathrm{log}\\left(x+2\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q470378\">Show Solution<\/span><\/p>\n<div id=\"q470378\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137935561\">Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5. The vertical asymptote will be shifted to <em>x\u00a0<\/em>= \u20132. The <em>x<\/em>-intercept will be [latex]\\left(-1,0\\right)[\/latex]. The domain will be [latex]\\left(-2,\\infty \\right)[\/latex]. Two points will help give the shape of the graph: [latex]\\left(-1,0\\right)[\/latex] and [latex]\\left(8,5\\right)[\/latex]. We chose <em>x\u00a0<\/em>= 8 as the <em>x<\/em>-coordinate of one point to graph because when <em>x\u00a0<\/em>= 8, <em>x\u00a0<\/em>+ 2 = 10, the base of the common logarithm.<span id=\"fs-id1165135641650\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010824\/CNX_Precalc_Figure_04_04_0162.jpg\" alt=\"Graph of three functions. The parent function is y=log(x), with an asymptote at x=0. The first translation function y=5log(x+2) has an asymptote at x=-2. The second translation function y=log(x+2) has an asymptote at x=-2.\" \/><\/span><\/p>\n<p id=\"fs-id1165137874883\" style=\"text-align: center\"><strong>Figure 12.\u00a0<\/strong>The domain is [latex]\\left(-2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= \u20132.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137838697\">Sketch a graph of the function [latex]f\\left(x\\right)=3\\mathrm{log}\\left(x - 2\\right)+1[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q793205\">Show Solution<\/span><\/p>\n<div id=\"q793205\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(2,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 2.<\/p>\n<div id=\"fs-id1165137437228\" class=\"solution\">\n<p><span id=\"fs-id1165135177663\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0172.jpg\" alt=\"Graph of f(x)=3log(x-2)+1 with an asymptote at x=2.\" \/><\/span><\/p>\n<p><span style=\"font-size: 1rem;text-align: initial\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174299\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174299&theme=oea&iframe_resize_id=ohm174299\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Graphing Reflections of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165137629003\">\n<p id=\"fs-id1165135169315\">When the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by \u20131, the result is a <strong>reflection<\/strong> about the <em>x<\/em>-axis. When the <em>input<\/em> is multiplied by \u20131, the result is a reflection about the <em>y<\/em>-axis. To visualize reflections, we restrict <em>b\u00a0<\/em>&gt; 1, and observe the general graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] alongside the reflection about the <em>x<\/em>-axis, [latex]g\\left(x\\right)={\\mathrm{-log}}_{b}\\left(x\\right)[\/latex] and the reflection about the <em>y<\/em>-axis, [latex]h\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_018n2.jpg\" alt=\"&quot;Graph\" \/><\/p>\n<div id=\"fs-id1165135190744\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Reflections of the Parent Function [latex]y=\\text{log}_{b}\\left(x\\right)[\/latex]<\/h3>\n<p id=\"fs-id1165137722409\">The function [latex]f\\left(x\\right)={\\mathrm{-log}}_{b}\\left(x\\right)[\/latex]<\/p>\n<ul id=\"fs-id1165137832285\">\n<li>reflects the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>x<\/em>-axis.<\/li>\n<li>has domain, [latex]\\left(0,\\infty \\right)[\/latex], range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and vertical asymptote, <em>x\u00a0<\/em>= 0, which are unchanged from the parent function.<\/li>\n<\/ul>\n<p>The function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]<\/p>\n<ul id=\"fs-id1165137734930\">\n<li>reflects the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>y<\/em>-axis.<\/li>\n<li>has domain [latex]\\left(-\\infty ,0\\right)[\/latex].<\/li>\n<li>has range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and vertical asymptote, <em>x\u00a0<\/em>= 0, which are unchanged from the parent function.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137638830\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137638837\">How To: Given a logarithmic function with the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph a translation.<\/h3>\n<table id=\"Table_04_04_08\" class=\"unnumbered\" summary=\"The first column gives the following instructions of graphing a translation of f(x)=-log_b(x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the x-axis; 4. Draw a smooth curve through the points; 5. State the domain, (0, infinity), the range, (-infinity, infinity), and the vertical asymptote x=0. The second column gives the following instructions of graphing a translation of f(x)=log_b(-x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (-1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the y-axis; 4. Draw a smooth curve through the points; 5. State the domain, (-infinity, 0), the range, (-infinity, infinity), and the vertical asymptote x=0.\">\n<thead>\n<tr>\n<th>[latex]\\text{If }f\\left(x\\right)=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/th>\n<th>[latex]\\text{If }f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1. Draw the vertical asymptote, <em>x\u00a0<\/em>= 0.<\/td>\n<td>1. Draw the vertical asymptote, <em>x\u00a0<\/em>= 0.<\/td>\n<\/tr>\n<tr>\n<td>2. Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/td>\n<td>2. Plot the <em>x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>3. Reflect the graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>x<\/em>-axis.<\/td>\n<td>3. Reflect the graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] about the <em>y<\/em>-axis.<\/td>\n<\/tr>\n<tr>\n<td>4. Draw a smooth curve through the points.<\/td>\n<td>4. Draw a smooth curve through the points.<\/td>\n<\/tr>\n<tr>\n<td>5. State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote <em>x\u00a0<\/em>= 0.<\/td>\n<td>5. State the domain, [latex]\\left(-\\infty ,0\\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote <em>x\u00a0<\/em>= 0.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"Example_04_04_08\" class=\"example\">\n<div id=\"fs-id1165137697928\" class=\"exercise\">\n<div id=\"fs-id1165137849033\" class=\"problem textbox shaded\">\n<h3>Example 8: Graphing a Reflection of a Logarithmic Function<\/h3>\n<p id=\"fs-id1165137849038\">Sketch a graph of [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex] alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q618451\">Show Solution<\/span><\/p>\n<div id=\"q618451\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137836525\">Before graphing [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex], identify the behavior and key points for the graph.<\/p>\n<ul id=\"fs-id1165137769879\">\n<li>Since <em>b\u00a0<\/em>= 10 is greater than one, we know that the parent function is increasing. Since the <em>input<\/em> value is multiplied by \u20131, <em>f<\/em>\u00a0is a reflection of the parent graph about the <em>y-<\/em>axis. Thus, [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex] will be decreasing as <em>x<\/em>\u00a0moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote <em>x\u00a0<\/em>= 0.<\/li>\n<li>The <em>x<\/em>-intercept is [latex]\\left(-1,0\\right)[\/latex].<\/li>\n<li>We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_04_019\" class=\"small\"><span id=\"fs-id1165134042188\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0192.jpg\" alt=\"Graph of two functions. The parent function is y=log(x), with an asymptote at x=0 and labeled points at (1, 0), and (10, 0).The translation function f(x)=log(-x) has an asymptote at x=0 and labeled points at (-1, 0) and (-10, 1).\" \/><\/span><\/figure>\n<p id=\"fs-id1165134042202\" style=\"text-align: center\"><strong>Figure 14.\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,0\\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135681852\">Graph [latex]f\\left(x\\right)=-\\mathrm{log}\\left(-x\\right)[\/latex]. State the domain, range, and asymptote.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q485222\">Show Solution<\/span><\/p>\n<div id=\"q485222\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is [latex]\\left(-\\infty ,0\\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex], and the vertical asymptote is <em>x\u00a0<\/em>= 0.<span id=\"fs-id1165137855148\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010825\/CNX_Precalc_Figure_04_04_0202.jpg\" alt=\"Graph of f(x)=-log(-x) with an asymptote at x=0.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134579621\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165134579627\">How To: Given a logarithmic equation, use a graphing calculator to approximate solutions.<\/h3>\n<ol id=\"fs-id1165137431118\">\n<li>Press <strong>[Y=]<\/strong>. Enter the given logarithm equation or equations as <strong>Y<sub>1<\/sub>=<\/strong> and, if needed, <strong>Y<sub>2<\/sub>=<\/strong>.<\/li>\n<li>Press <strong>[GRAPH]<\/strong> to observe the graphs of the curves and use <strong>[WINDOW]<\/strong> to find an appropriate view of the graphs, including their point(s) of intersection.<\/li>\n<li>To find the value of <em>x<\/em>, we compute the point of intersection. Press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select &#8220;intersect&#8221; and press <strong>[ENTER]<\/strong> three times. The point of intersection gives the value of <em>x<\/em>, for the point(s) of intersection.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_04_09\" class=\"example\">\n<div id=\"fs-id1165135414229\" class=\"exercise\">\n<div id=\"fs-id1165135414231\" class=\"problem textbox shaded\">\n<h3>Example 9: Approximating the Solution of a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165135414236\">Solve [latex]4\\mathrm{ln}\\left(x\\right)+1=-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q513609\">Show Solution<\/span><\/p>\n<div id=\"q513609\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135193434\">Press <strong>[Y=]<\/strong> and enter [latex]4\\mathrm{ln}\\left(x\\right)+1[\/latex] next to <strong>Y<sub>1<\/sub><\/strong>=. Then enter [latex]-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] next to <strong>Y<sub>2<\/sub>=<\/strong>. For a window, use the values 0 to 5 for <em>x<\/em>\u00a0and \u201310 to 10 for <em>y<\/em>. Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere a little to right of <em>x\u00a0<\/em>= 1.<\/p>\n<p id=\"fs-id1165135245763\">For a better approximation, press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The <em>x<\/em>-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for <strong>Guess?<\/strong>) So, to the nearest thousandth, [latex]x\\approx 1.339[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137639531\">Solve [latex]5\\mathrm{log}\\left(x+2\\right)=4-\\mathrm{log}\\left(x\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q862673\">Show Solution<\/span><\/p>\n<div id=\"q862673\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx 3.049[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165135528930\">\n<h2>Summarizing Translations of the Logarithmic Function<\/h2>\n<p id=\"fs-id1165135528935\">Now that we have worked with each type of translation for the logarithmic function, we can summarize each in the table below\u00a0to arrive at the general equation for translating exponential functions.<\/p>\n<table id=\"Table_04_04_009\" summary=\"Titled,\">\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"2\">Translations of the Parent Function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/th>\n<\/tr>\n<tr>\n<th style=\"text-align: center\">Translation<\/th>\n<th style=\"text-align: center\">Form<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Shift<\/p>\n<ul id=\"fs-id1165137416971\">\n<li>Horizontally <em>c<\/em>\u00a0units to the left<\/li>\n<li>Vertically <em>d<\/em>\u00a0units up<\/li>\n<\/ul>\n<\/td>\n<td>[latex]y={\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Stretch and Compress<\/p>\n<ul id=\"fs-id1165137427553\">\n<li>Stretch if [latex]|a|>1[\/latex]<\/li>\n<li>Compression if [latex]|a|<1[\/latex]<\/li>\n<\/ul>\n<\/td>\n<td>[latex]y=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reflect about the <em>x<\/em>-axis<\/td>\n<td>[latex]y=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reflect about the <em>y<\/em>-axis<\/td>\n<td>[latex]y={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>General equation for all translations<\/td>\n<td>[latex]y=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137414493\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Translations of Logarithmic Functions<\/h3>\n<p id=\"fs-id1165137414501\">All translations of the parent logarithmic function, [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], have the form<\/p>\n<div id=\"fs-id1165135408512\" class=\"equation\" style=\"text-align: center\">[latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/div>\n<p id=\"fs-id1165137734655\">where the parent function, [latex]y={\\mathrm{log}}_{b}\\left(x\\right),b>1[\/latex], is<\/p>\n<ul id=\"fs-id1165137531610\">\n<li>shifted vertically up <em>d<\/em>\u00a0units.<\/li>\n<li>shifted horizontally to the left <em>c<\/em>\u00a0units.<\/li>\n<li>stretched vertically by a factor of |<em>a<\/em>| if |<em>a<\/em>| &gt; 0.<\/li>\n<li>compressed vertically by a factor of |<em>a<\/em>| if 0 &lt; |<em>a<\/em>| &lt; 1.<\/li>\n<li>reflected about the <em>x-<\/em>axis when <em>a\u00a0<\/em>&lt; 0.<\/li>\n<\/ul>\n<p id=\"fs-id1165137725084\">For [latex]f\\left(x\\right)=\\mathrm{log}\\left(-x\\right)[\/latex], the graph of the parent function is reflected about the <em>y<\/em>-axis.<\/p>\n<\/div>\n<div id=\"Example_04_04_10\" class=\"example\">\n<div id=\"fs-id1165135296269\" class=\"exercise\">\n<div id=\"fs-id1165135296271\" class=\"problem textbox shaded\">\n<h3>Example 10: Finding the Vertical Asymptote of a Logarithm Graph<\/h3>\n<p id=\"fs-id1165135296276\">What is the vertical asymptote of [latex]f\\left(x\\right)=-2{\\mathrm{log}}_{3}\\left(x+4\\right)+5[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q841705\">Show Solution<\/span><\/p>\n<div id=\"q841705\" class=\"hidden-answer\" style=\"display: none\">\n<p>The vertical asymptote is at <em>x\u00a0<\/em>= \u20134.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137871960\">The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to <em>x\u00a0<\/em>= \u20134.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135368433\">What is the vertical asymptote of [latex]f\\left(x\\right)=3+\\mathrm{ln}\\left(x - 1\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q975284\">Show Solution<\/span><\/p>\n<div id=\"q975284\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>x\u00a0<\/em>= 1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_04_11\" class=\"example\">\n<div id=\"fs-id1165137849555\" class=\"exercise\">\n<div id=\"fs-id1165137849558\" class=\"problem textbox shaded\">\n<h3>Example 11: Finding the Equation from a Graph<\/h3>\n<p id=\"fs-id1165137849563\">Find a possible equation for the common logarithmic function graphed in Figure 15.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005323\/CNX_Precalc_Figure_04_04_021.jpg\" alt=\"Graph of a logarithmic function with a vertical asymptote at x=-2, has been vertically reflected, and passes through the points (-1, 1) and (2, -1).\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q993624\">Show Solution<\/span><\/p>\n<div id=\"q993624\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135342979\">This graph has a vertical asymptote at <em>x\u00a0<\/em>= \u20132 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:<\/p>\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=-a\\mathrm{log}\\left(x+2\\right)+k[\/latex]<\/p>\n<p id=\"fs-id1165135406913\">It appears the graph passes through the points [latex]\\left(-1,1\\right)[\/latex] and [latex]\\left(2,-1\\right)[\/latex]. Substituting [latex]\\left(-1,1\\right)[\/latex],<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&1=-a\\mathrm{log}\\left(-1+2\\right)+k && \\text{Substitute }\\left(-1,1\\right). \\\\ &1=-a\\mathrm{log}\\left(1\\right)+k && \\text{Arithmetic}. \\\\ &1=k && \\text{log(1)}=0. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137628655\">Next, substituting in [latex]\\left(2,-1\\right)[\/latex],<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&-1=-a\\mathrm{log}\\left(2+2\\right)+1 && \\text{Plug in }\\left(2,-1\\right). \\\\ &-2=-a\\mathrm{log}\\left(4\\right) && \\text{Arithmetic}. \\\\ &a=\\frac{2}{\\mathrm{log}\\left(4\\right)}&& \\text{Solve for }a. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135192211\">This gives us the equation [latex]f\\left(x\\right)=-\\frac{2}{\\mathrm{log}\\left(4\\right)}\\mathrm{log}\\left(x+2\\right)+1[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137735586\">We can verify this answer by comparing the function values in the table below\u00a0with the points on the graph in Example 11.<\/p>\n<table id=\"Table_04_04_010\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u22121<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><em><strong>f<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\n<td>1<\/td>\n<td>0<\/td>\n<td>\u22120.58496<\/td>\n<td>\u22121<\/td>\n<td>\u22121.3219<\/td>\n<\/tr>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><em><strong>f<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\n<td>\u22121.5850<\/td>\n<td>\u22121.8074<\/td>\n<td>\u22122<\/td>\n<td>\u22122.1699<\/td>\n<td>\u22122.3219<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137665487\">Give the equation of the natural logarithm graphed in Figure 16.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005324\/CNX_Precalc_Figure_04_04_022.jpg\" alt=\"Graph of a logarithmic function with a vertical asymptote at x=-3, has been vertically stretched by 2, and passes through the points (-1, -1).\" width=\"487\" height=\"442\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q537017\">Show Solution<\/span><\/p>\n<div id=\"q537017\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(x\\right)=2\\mathrm{ln}\\left(x+3\\right)-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137855236\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165137855242\"><strong>Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?<\/strong><\/p>\n<p id=\"fs-id1165137827126\"><em>Yes, if we know the function is a general logarithmic function. For example, look at the graph in Try It 11. The graph approaches x = \u20133 (or thereabouts) more and more closely, so x = \u20133 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, [latex]\\left\\{x|x>-3\\right\\}[\/latex]. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as [latex]x\\to -{3}^{+},f\\left(x\\right)\\to -\\infty[\/latex] and as [latex]x\\to \\infty ,f\\left(x\\right)\\to \\infty[\/latex].<\/em><\/p>\n<\/div>\n<\/section>\n<figure id=\"CNX_Precalc_Figure_04_05_001\" class=\"small\">\n<div style=\"width: 254px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_05_001F2.jpg\" alt=\"Testing of the pH of hydrochloric acid.\" width=\"244\" height=\"382\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137759741\">In chemistry, <strong>pH<\/strong> is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\n<ul id=\"fs-id1165135253210\">\n<li>Battery acid: 0.8<\/li>\n<li>Stomach acid: 2.7<\/li>\n<li>Orange juice: 3.3<\/li>\n<li>Pure water: 7 (at 25\u00b0 C)<\/li>\n<li>Human blood: 7.35<\/li>\n<li>Fresh coconut: 7.8<\/li>\n<li>Sodium hydroxide (lye): 14<\/li>\n<\/ul>\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where <em>a<\/em>\u00a0is the concentration of hydrogen ion in the solution<\/p>\n<div id=\"eip-396\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\text{pH}&=-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right) \\\\ &=\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right) \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137472164\">The equivalence of [latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex] and [latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] is one of the logarithm properties we will examine in this section.<\/p>\n<h2>Use the product rule for logarithms<\/h2>\n<p id=\"fs-id1165137405402\">Recall that the logarithmic and exponential functions &#8220;undo&#8221; each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}1=0\\\\ &{\\mathrm{log}}_{b}b=1\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137628765\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\n<p id=\"fs-id1165137772010\">Next, we have the inverse property.<\/p>\n<div id=\"eip-896\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&=x\\hfill \\\\ {b}^{{\\mathrm{log}}_{b}x}&=x,x>0 \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137696455\">For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\n<p id=\"fs-id1165134297163\">To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\n<p id=\"fs-id1165137592421\">Finally, we have the <strong>one-to-one<\/strong> property.<\/p>\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\n<p id=\"fs-id1165137723139\">We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for <em>x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for <em>x<\/em>:<\/p>\n<div id=\"eip-448\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}3x&=2x+5 && \\text{Set the arguments equal.} \\\\ x&=5 && \\text{Subtract 2}x. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135194712\">But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\n<p id=\"fs-id1165137645446\">Given any real number <em>x<\/em>\u00a0and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\n<div id=\"eip-214\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\n<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right) && \\text{Substitute for }M\\text{ and }N. \\\\ & ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right) && \\text{Apply the product rule for exponents}. \\\\ & =m+n && \\text{Apply the inverse property of logs}. \\\\ & ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right) && \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\n<div id=\"eip-502\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\n<div id=\"fs-id1165137891324\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137541378\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165134223340\">How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\n<ol id=\"fs-id1165137748303\">\n<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\n<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_01\" class=\"example\">\n<div id=\"fs-id1165135458651\" class=\"exercise\">\n<div id=\"fs-id1165135458654\" class=\"problem textbox shaded\">\n<h3>Example 1: Using the Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q868590\">Show Solution<\/span><\/p>\n<div id=\"q868590\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137871801\">Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989002\">Show Solution<\/span><\/p>\n<div id=\"q989002\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174314\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174314&theme=oea&iframe_resize_id=ohm174314\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use the quotient and power rules for logarithms<\/h2>\n<section id=\"fs-id1165135151295\">\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{b}}={x}^{a-b}[\/latex]. The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\n<p id=\"fs-id1165137431410\">Given any real number <em>x\u00a0<\/em>and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\n<div id=\"eip-589\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\n<div id=\"eip-303\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right) && \\text{Substitute for }M\\text{ and }N. \\\\ & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right) && \\text{Apply the quotient rule for exponents}. \\\\ & =m-n && \\text{Apply the inverse property of logs}. \\\\ & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right) && \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) & =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right) && \\text{Factor the numerator and denominator}. \\\\ & =\\mathrm{log}\\left(\\frac{2x}{3}\\right) && \\text{Cancel the common factors}. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right) \\\\ &=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right) \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137733855\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Quotient Rule for Logarithms<\/h3>\n<p id=\"eip-id1165135390834\">The <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\n<div id=\"fs-id1165137834642\" class=\"equation\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137749807\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137749813\">How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\n<ol id=\"fs-id1165137749817\">\n<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\n<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\n<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_02\" class=\"example\">\n<div id=\"fs-id1165135185904\" class=\"exercise\">\n<div id=\"fs-id1165135185906\" class=\"problem textbox shaded\">\n<h3>Example 2: Using the Quotient Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q47414\">Show Solution<\/span><\/p>\n<div id=\"q47414\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\[1mm] &= \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right] \\\\[1mm] &={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right) \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137629471\">There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and <em>x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that <em>x\u00a0<\/em>&gt; 0, <em>x\u00a0<\/em>&gt; 1, [latex]x>-\\frac{4}{3}[\/latex], and <em>x\u00a0<\/em>&lt; 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137534466\">Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q51755\">Show Solution<\/span><\/p>\n<div id=\"q51755\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174320\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174320&theme=oea&iframe_resize_id=ohm174320\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Using the Power Rule for Logarithms<\/span><\/p>\n<div id=\"fs-id1165137939600\" class=\"solution\">\n<section id=\"fs-id1165137627625\">\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[\/latex]? One method is as follows:<\/p>\n<div id=\"eip-271\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({x}^{2}\\right) & ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right) \\\\ & ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x \\\\ & =2{\\mathrm{log}}_{b}x \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137534037\">Notice that we used the <strong>product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\n<div id=\"eip-702\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&100={10}^{2} && \\sqrt{3}={3}^{\\frac{1}{2}} && \\frac{1}{e}={e}^{-1} \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137676322\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Power Rule for Logarithms<\/h3>\n<p id=\"fs-id1165137676330\">The <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137639704\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137639709\">How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\n<ol id=\"fs-id1165137761651\">\n<li>Express the argument as a power, if needed.<\/li>\n<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_03\" class=\"example\">\n<div id=\"fs-id1165135593557\" class=\"exercise\">\n<div id=\"fs-id1165135593559\" class=\"problem textbox shaded\">\n<h3>Example 3: Expanding a Logarithm with Powers<\/h3>\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476325\">Show Solution<\/span><\/p>\n<div id=\"q476325\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229979\">Show Solution<\/span><\/p>\n<div id=\"q229979\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{ln}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_04\" class=\"example\">\n<div id=\"fs-id1165134163985\" class=\"exercise\">\n<div id=\"fs-id1165134163988\" class=\"problem textbox shaded\">\n<h3>Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q400323\">Show Solution<\/span><\/p>\n<div id=\"q400323\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q153337\">Show Solution<\/span><\/p>\n<div id=\"q153337\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_05\" class=\"example\">\n<div id=\"fs-id1165134435871\" class=\"exercise\">\n<div id=\"fs-id1165134435874\" class=\"problem textbox shaded\">\n<h3>Example 5: Using the Power Rule in Reverse<\/h3>\n<p id=\"fs-id1165137714060\">Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q851055\">Show Solution<\/span><\/p>\n<div id=\"q851055\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137416106\">Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\n<p style=\"text-align: center\">[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137502321\">Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q294897\">Show Solution<\/span><\/p>\n<div id=\"q294897\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}16[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm45617\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=45617&theme=oea&iframe_resize_id=ohm45617\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm25581\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25581&theme=oea&iframe_resize_id=ohm25581\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Expand logarithmic expressions<\/span><\/p>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137657409\">\n<p id=\"fs-id1165137558543\">Taken together, the product rule, quotient rule, and power rule are often called &#8220;laws of logs.&#8221; Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\n<div id=\"eip-423\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)& ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y \\\\ &={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135545872\">We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:<\/p>\n<div id=\"eip-622\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right) & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right) \\\\ & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right) \\\\ & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C \\\\ & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135153099\">We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n<p id=\"fs-id1165135153103\">With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\n<div id=\"Example_04_05_06\" class=\"example\">\n<div id=\"fs-id1165135173497\" class=\"exercise\">\n<div id=\"fs-id1165135173499\" class=\"problem textbox shaded\">\n<h3>Example 6: Expanding Logarithms Using Product, Quotient, and Power Rules<\/h3>\n<p id=\"fs-id1165135173504\">Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737111\">Show Solution<\/span><\/p>\n<div id=\"q737111\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135253768\">First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137854981\">Then seeing the product in the first term, we use the product rule:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134154611\">Finally, we use the power rule on the first term:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137635162\">Expand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q765910\">Show Solution<\/span><\/p>\n<div id=\"q765910\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_07\" class=\"example\">\n<div id=\"fs-id1165137811250\" class=\"exercise\">\n<div id=\"fs-id1165137811252\" class=\"problem textbox shaded\">\n<h3>Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\n<p id=\"fs-id1165135637419\">Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q304354\">Show Solution<\/span><\/p>\n<div id=\"q304354\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\mathrm{log}\\left(\\sqrt{x}\\right) & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)} \\\\ & =\\frac{1}{2}\\mathrm{log}x \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137804498\">Expand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410566\">Show Solution<\/span><\/p>\n<div id=\"q410566\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135173426\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1165134197968\"><strong>Can we expand<\/strong> [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/p>\n<p id=\"fs-id1165135440437\"><em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em><\/p>\n<\/div>\n<div id=\"Example_04_05_08\" class=\"example\">\n<div id=\"fs-id1165135440448\" class=\"exercise\">\n<div id=\"fs-id1165135440450\" class=\"problem textbox shaded\">\n<h3>Example 8: Expanding Complex Logarithmic Expressions<\/h3>\n<p id=\"fs-id1165135150641\">Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q840332\">Show Solution<\/span><\/p>\n<div id=\"q840332\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134085801\">We can expand by applying the Product and Quotient Rules.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right) & ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && \\text{Apply the Quotient Rule}. \\\\ & ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && {\\text{Simplify by writing 64 as 2}}^{6}. \\\\ & =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && \\text{Apply the Power Rule}.\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135635247\">Expand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303340\">Show Solution<\/span><\/p>\n<div id=\"q303340\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35034\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35034&theme=oea&iframe_resize_id=ohm35034\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174326\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174326&theme=oea&iframe_resize_id=ohm174326\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Condense logarithmic expressions<\/span><\/p>\n<\/section>\n<p id=\"fs-id1165135190860\">We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\n<div id=\"fs-id1165135190866\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135190871\">How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\n<ol id=\"fs-id1165137833816\">\n<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\n<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\n<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_09\" class=\"example\">\n<div id=\"fs-id1165137833837\" class=\"exercise\">\n<div id=\"fs-id1165137833839\" class=\"problem textbox shaded\">\n<h3>Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\n<p id=\"fs-id1165135484124\">Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q118143\">Show Solution<\/span><\/p>\n<div id=\"q118143\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135527077\">Using the product and quotient rules<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135400169\">This reduces our original expression to<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137846453\">Then, using the quotient rule<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134047712\">Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q144049\">Show Solution<\/span><\/p>\n<div id=\"q144049\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_10\" class=\"example\">\n<div id=\"fs-id1165134435881\" class=\"exercise\">\n<div id=\"fs-id1165134435883\" class=\"problem textbox shaded\">\n<h3>Example 10: Condensing Complex Logarithmic Expressions<\/h3>\n<p id=\"fs-id1165134435888\">Condense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788800\">Show Solution<\/span><\/p>\n<div id=\"q788800\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135344097\">We apply the power rule first:<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135531546\">Next we apply the product rule to the sum:<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134280852\">Finally, we apply the quotient rule to the difference:<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_11\" class=\"example\">\n<div id=\"fs-id1165135519254\" class=\"exercise\">\n<div id=\"fs-id1165135519256\" class=\"problem textbox shaded\">\n<h3>Example 11: Rewriting as a Single Logarithm<\/h3>\n<p id=\"fs-id1165134156051\">Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q302963\">Show Solution<\/span><\/p>\n<div id=\"q302963\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135177652\">We apply the power rule first:<\/p>\n<p style=\"text-align: center\">[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135195405\">Next we apply the product rule to the sum:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134042304\">Finally, we apply the quotient rule to the difference:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135511375\">Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q235871\">Show Solution<\/span><\/p>\n<div id=\"q235871\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135627841\">Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q628183\">Show Solution<\/span><\/p>\n<div id=\"q628183\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174327\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174327&theme=oea&iframe_resize_id=ohm174327\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_04_05_12\" class=\"example\">\n<div id=\"fs-id1165137400159\" class=\"exercise\">\n<div id=\"fs-id1165137400162\" class=\"problem textbox shaded\">\n<h3>Example 12: Applying of the Laws of Logs<\/h3>\n<p id=\"fs-id1165135530298\">Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197365\">Show Solution<\/span><\/p>\n<div id=\"q197365\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\n<p style=\"text-align: center\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\n<p style=\"text-align: center\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\n<p style=\"text-align: center\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135251378\">How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371569\">Show Solution<\/span><\/p>\n<div id=\"q371569\" class=\"hidden-answer\" style=\"display: none\">\n<p>The pH increases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Use the change-of-base formula for logarithms<\/h2>\n<section id=\"fs-id1165137675210\">\n<p id=\"fs-id1165137675216\">Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the <strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.<\/p>\n<p id=\"fs-id1165137855374\">To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\n<p id=\"fs-id1165137855378\">Given any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex], we show<\/p>\n<div id=\"eip-643\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/div>\n<p id=\"fs-id1165137932683\">Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. By taking the log base [latex]n[\/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[\/latex]. It follows that<\/p>\n<div id=\"eip-226\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{n}\\left({b}^{y}\\right) & ={\\mathrm{log}}_{n}M && \\text{Apply the one-to-one property}. \\\\ y{\\mathrm{log}}_{n}b\\hfill & ={\\mathrm{log}}_{n}M && \\text{Apply the power rule for logarithms}. \\\\ y & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} && \\text{Isolate }y. \\\\ {\\mathrm{log}}_{b}M & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} && \\text{Substitute for }y. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135207389\">For example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.<\/p>\n<div id=\"eip-428\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{5}36 & =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)} && \\text{Apply the change of base formula using base 10.} \\\\ & \\approx 2.2266 && \\text{Use a calculator to evaluate to 4 decimal places.} \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165134381722\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Change-of-Base Formula<\/h3>\n<p id=\"fs-id1165135342066\">The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\n<p id=\"fs-id1165135342073\">For any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex],<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\n<p id=\"fs-id1165134042184\">It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\n<p id=\"fs-id1165137935512\">and<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137893333\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137893339\">How To: Given a logarithm with the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex].<\/h3>\n<ol id=\"fs-id1165134172563\">\n<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10, and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\n<li>Rewrite the log as a quotient using the change-of-base formula\n<ul id=\"fs-id1165134039298\">\n<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\n<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_13\" class=\"example\">\n<div id=\"fs-id1165134196189\" class=\"exercise\">\n<div id=\"fs-id1165134196191\" class=\"problem textbox shaded\">\n<h3>Example 13: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\n<p id=\"fs-id1165134196197\">Change [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164120\">Show Solution<\/span><\/p>\n<div id=\"q164120\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135444049\">Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base, <em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\n<p id=\"fs-id1165135690112\">We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{b}M & =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b} \\\\[1mm] {\\mathrm{log}}_{5}3 & =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134223311\">Change [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q562111\">Show Solution<\/span><\/p>\n<div id=\"q562111\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135181811\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1172294599410\"><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\n<p id=\"fs-id1165135193274\"><em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em><\/p>\n<\/div>\n<div id=\"Example_04_05_14\" class=\"example\">\n<div id=\"fs-id1165134084328\" class=\"exercise\">\n<div id=\"fs-id1165134084330\" class=\"problem textbox shaded\">\n<h3>Example 14: Using the Change-of-Base Formula with a Calculator<\/h3>\n<p id=\"fs-id1165134084335\">Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q264040\">Show Solution<\/span><\/p>\n<div id=\"q264040\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135353038\">According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base <i>e<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\mathrm{log}}_{2}10&=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2} && \\text{Apply the change of base formula using base }e. \\\\ &\\approx 3.3219 && \\text{Use a calculator to evaluate to 4 decimal places}. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 21<\/h3>\n<p id=\"fs-id1165135358918\">Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q97931\">Show Solution<\/span><\/p>\n<div id=\"q97931\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>try it 22<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm13570\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=13570&theme=oea&iframe_resize_id=ohm13570\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #1d1d1d;font-size: 1.5em;font-weight: bold;text-align: initial\">Key Equations<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165137870892\" class=\"key-equations\">\n<table id=\"fs-id1983134\" summary=\"...\">\n<tbody>\n<tr>\n<td>Definition of the logarithmic function<\/td>\n<td>For [latex]\\text{ } x>0,b>0,b\\ne 1[\/latex],[latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{b}^{y}=x[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Definition of the common logarithm<\/td>\n<td>For [latex]\\text{ }x>0[\/latex], [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{10}^{y}=x[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Definition of the natural logarithm<\/td>\n<td>For [latex]\\text{ }x>0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{e}^{y}=x[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section>\n<table id=\"fs-id1737642\" summary=\"...\">\n<tbody>\n<tr>\n<td>General Form for the Translation of the Parent Logarithmic Function [latex]\\text{ }f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section>\n<table id=\"fs-id2922999\" summary=\"...\">\n<tbody>\n<tr>\n<td>The Product Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Quotient Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Power Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Change-of-Base Formula<\/td>\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n>0,n\\ne 1,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section class=\"key-equations\"><span style=\"color: #1d1d1d;font-size: 1.5em;font-weight: bold\">Key Concepts<\/span><\/section>\n<section id=\"fs-id1165135699130\" class=\"key-concepts\">\n<ul id=\"fs-id1165137574258\">\n<li>The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.<\/li>\n<li>Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm.<\/li>\n<li>Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm.<\/li>\n<li>Logarithmic functions with base <em>b<\/em>\u00a0can be evaluated mentally using previous knowledge of powers of <em>b<\/em>.<\/li>\n<li>Common logarithms can be evaluated mentally using previous knowledge of powers of 10.<\/li>\n<li>When common logarithms cannot be evaluated mentally, a calculator can be used.<\/li>\n<li>Real-world exponential problems with base 10\u00a0can be rewritten as a common logarithm and then evaluated using a calculator.<\/li>\n<li>Natural logarithms can be evaluated using a calculator.<\/li>\n<li>To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for <em>x<\/em>.<\/li>\n<li>The graph of the parent function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] has an <em>x-<\/em>intercept at [latex]\\left(1,0\\right)[\/latex], domain [latex]\\left(0,\\infty \\right)[\/latex], range [latex]\\left(-\\infty ,\\infty \\right)[\/latex], vertical asymptote <em>x\u00a0<\/em>= 0, and\n<ul id=\"fs-id1165135441773\">\n<li>if <em>b\u00a0<\/em>&gt; 1, the function is increasing.<\/li>\n<li>if 0 &lt; <em>b\u00a0<\/em>&lt; 1, the function is decreasing.<\/li>\n<\/ul>\n<\/li>\n<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x+c\\right)[\/latex] shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] horizontally\n<ul id=\"fs-id1165135512562\">\n<li>left <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&gt; 0.<\/li>\n<li>right <em>c<\/em>\u00a0units if <em>c\u00a0<\/em>&lt; 0.<\/li>\n<\/ul>\n<\/li>\n<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)+d[\/latex] shifts the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically\n<ul id=\"fs-id1165137761068\">\n<li>up <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&gt; 0.<\/li>\n<li>down <em>d<\/em>\u00a0units if <em>d\u00a0<\/em>&lt; 0.<\/li>\n<\/ul>\n<\/li>\n<li>For any constant <em>a\u00a0<\/em>&gt; 0, the equation [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)[\/latex]\n<ul id=\"fs-id1165134040579\">\n<li>stretches the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if |<em>a<\/em>| &gt; 1.<\/li>\n<li>compresses the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] vertically by a factor of <em>a<\/em>\u00a0if |<em>a<\/em>| &lt; 1.<\/li>\n<\/ul>\n<\/li>\n<li>When the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is multiplied by \u20131, the result is a reflection about the <em>x<\/em>-axis. When the input is multiplied by \u20131, the result is a reflection about the <em>y<\/em>-axis.\n<ul id=\"fs-id1165135186594\">\n<li>The equation [latex]f\\left(x\\right)=-{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] represents a reflection of the parent function about the <em>x-<\/em>axis.<\/li>\n<li>The equation [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(-x\\right)[\/latex] represents a reflection of the parent function about the <em>y-<\/em>axis.<\/li>\n<\/ul>\n<ul id=\"fs-id1165137834414\">\n<li>A graphing calculator may be used to approximate solutions to some logarithmic equations.<\/li>\n<\/ul>\n<\/li>\n<li>All translations of the logarithmic function can be summarized by the general equation [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex].<\/li>\n<li>Given an equation with the general form [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex], we can identify the vertical asymptote <em>x\u00a0<\/em>= \u2013c for the transformation.<\/li>\n<li>Using the general equation [latex]f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x+c\\right)+d[\/latex], we can write the equation of a logarithmic function given its graph.<\/li>\n<li>We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.<\/li>\n<li>We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.<\/li>\n<li>We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base.<\/li>\n<li>We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.<\/li>\n<li>The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.<\/li>\n<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\n<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and <i>e<\/i>\u00a0as the quotient of natural or common logs. That way a calculator can be used to evaluate.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135160066\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>change-of-base formula<\/strong><\/dt>\n<dd id=\"fs-id1165137890649\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137890654\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>common logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165137571387\">the exponent to which 10 must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] is written simply as [latex]\\mathrm{log}\\left(x\\right)[\/latex].<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137780762\" class=\"definition\">\n<dt><strong>logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165137849198\">the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>; written [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137507853\" class=\"definition\">\n<dt><strong>natural logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165134037589\">the exponent to which the number <em>e<\/em>\u00a0must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/dd>\n<\/dl>\n<p> \t<strong>power rule for logarithms<\/strong><br \/>\n \ta rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base<\/p>\n<dl id=\"fs-id1165137890664\" class=\"definition\">\n<dt><strong>product rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137890670\">a rule of logarithms that states that the log of a product is equal to a sum of logarithms<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137890674\" class=\"definition\">\n<dt><strong>quotient rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137890679\">a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms<\/dd>\n<\/dl>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-13703\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-13703-1\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-13703-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-13703-2\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-13703-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-13703-3\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-13703-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><li id=\"footnote-13703-4\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-13703-4\" class=\"return-footnote\" aria-label=\"Return to footnote 4\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":97803,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-13703","chapter","type-chapter","status-publish","hentry"],"part":15999,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13703","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/users\/97803"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13703\/revisions"}],"predecessor-version":[{"id":15988,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13703\/revisions\/15988"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/parts\/15999"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13703\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/media?parent=13703"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13703"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/contributor?post=13703"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/license?post=13703"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}