{"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":"2021-08-23T06:33:49","modified_gmt":"2021-08-23T06:33:49","slug":"logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/logarithmic-properties\/","title":{"raw":"Section 5.4: Logarithmic Functions","rendered":"Section 5.4: 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.<\/li>\r\n \t<li>Use natural logarithms.<\/li>\r\n \t<li>Identify the domain of a logarithmic function.<\/li>\r\n \t<li>Graph logarithmic functions.<\/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=\"http:\/\/www.hutchmath.com\/Images\/Haiteruins.png\" alt=\"Photo of the aftermath of the earthquake in Haite, 2010\" width=\"488\" height=\"325\" \/> <strong>Figure 1.\u00a0<\/strong>Devastation of January 2010 earthquake in Haiti.[\/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-id1165135161452\">Next 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\">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 6: 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 7: 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 3.\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 4\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 8: 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 9:\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 10: 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 11: 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 12: 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 13: 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&nbsp;\r\n\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 14: 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 15: 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><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Equations<\/span>\r\n\r\n<section id=\"fs-id1165137749167\" class=\"key-equations\">\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 id=\"fs-id1165137863125\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137863132\">\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<\/ul>\r\n<\/section>\r\n<h2>Glossary<\/h2>\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<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<\/section>&nbsp;\r\n<h2 style=\"text-align: center;\">Section 5.4 Homework Exercises<\/h2>\r\n1. What is a base <em>b<\/em>\u00a0logarithm? Discuss the meaning by interpreting each part of the equivalent equations [latex]{b}^{y}=x[\/latex] and [latex]{\\mathrm{log}}_{b}x=y[\/latex] for [latex]b&gt;0,b\\ne 1[\/latex].\r\n\r\n2.\u00a0How is the logarithmic function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}x[\/latex] related to the exponential function [latex]g\\left(x\\right)={b}^{x}[\/latex]? What is the result of composing these two functions?\r\n\r\n3. How can the logarithmic equation [latex]{\\mathrm{log}}_{b}x=y[\/latex] be solved for <em>x<\/em>\u00a0using the properties of exponents?\r\n\r\n4. Discuss the meaning of the common logarithm. What is its relationship to a logarithm with base <em>b<\/em>, and how does the notation differ?\r\n\r\n5. Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with base <em>b<\/em>, and how does the notation differ?\r\n\r\n6.\u00a0What type(s) of translation(s), if any, affect the range of a logarithmic function?\r\n\r\n7. The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each?\r\n\r\n8.\u00a0Consider the general logarithmic function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Why can\u2019t <em>x<\/em>\u00a0be zero?\r\n\r\n9. Does the graph of a general logarithmic function have a horizontal asymptote? Explain.\r\n\r\nFor the following exercises, rewrite each equation in exponential form.\r\n\r\n10. [latex]{\\text{log}}_{4}\\left(q\\right)=m[\/latex]\r\n\r\n11. [latex]{\\text{log}}_{a}\\left(b\\right)=c[\/latex]\r\n\r\n12. [latex]{\\mathrm{log}}_{16}\\left(y\\right)=x[\/latex]\r\n\r\n13. [latex]{\\mathrm{log}}_{x}\\left(64\\right)=y[\/latex]\r\n\r\n14.\u00a0[latex]{\\mathrm{log}}_{y}\\left(x\\right)=-11[\/latex]\r\n\r\n15. [latex]{\\mathrm{log}}_{15}\\left(a\\right)=b[\/latex]\r\n\r\n16.\u00a0[latex]{\\mathrm{log}}_{y}\\left(137\\right)=x[\/latex]\r\n\r\n17. [latex]{\\mathrm{log}}_{13}\\left(142\\right)=a[\/latex]\r\n\r\n18.\u00a0[latex]\\text{log}\\left(v\\right)=t[\/latex]\r\n\r\n19. [latex]\\text{ln}\\left(w\\right)=n[\/latex]\r\n\r\nFor the following exercises, rewrite each equation in logarithmic form.\r\n\r\n20. [latex]{4}^{x}=y[\/latex]\r\n\r\n21. [latex]{c}^{d}=k[\/latex]\r\n\r\n22. [latex]{m}^{-7}=n[\/latex]\r\n\r\n23. [latex]{19}^{x}=y[\/latex]\r\n\r\n24.\u00a0[latex]{x}^{-\\frac{10}{13}}=y[\/latex]\r\n\r\n25. [latex]{n}^{4}=103[\/latex]\r\n\r\n26.\u00a0[latex]{\\left(\\frac{7}{5}\\right)}^{m}=n[\/latex]\r\n\r\n27. [latex]{y}^{x}=\\frac{39}{100}[\/latex]\r\n\r\n28.\u00a0[latex]{10}^{a}=b[\/latex]\r\n\r\n29. [latex]{e}^{k}=h[\/latex]\r\n\r\nFor the following exercises, solve for <em>x<\/em>\u00a0by converting the logarithmic equation to exponential form.\r\n\r\n30. [latex]{\\text{log}}_{3}\\left(x\\right)=2[\/latex]\r\n\r\n31. [latex]{\\text{log}}_{2}\\left(x\\right)=-3[\/latex]\r\n\r\n32.\u00a0[latex]{\\text{log}}_{5}\\left(x\\right)=2[\/latex]\r\n\r\n33. [latex]{\\mathrm{log}}_{3}\\left(x\\right)=3[\/latex]\r\n\r\n34.\u00a0[latex]{\\text{log}}_{2}\\left(x\\right)=6[\/latex]\r\n\r\n35. [latex]{\\text{log}}_{9}\\left(x\\right)=\\frac{1}{2}[\/latex]\r\n\r\n36.\u00a0[latex]{\\text{log}}_{18}\\left(x\\right)=2[\/latex]\r\n\r\n37. [latex]{\\mathrm{log}}_{6}\\left(x\\right)=-3[\/latex]\r\n\r\n38.\u00a0[latex]\\text{log}\\left(x\\right)=3[\/latex]\r\n\r\n39. [latex]\\text{ln}\\left(x\\right)=2[\/latex]\r\n\r\nFor the following exercises, use the definition of common and natural logarithms to simplify.\r\n\r\n40. [latex]\\text{log}\\left({100}^{8}\\right)[\/latex]\r\n\r\n41. [latex]{10}^{\\text{log}\\left(32\\right)}[\/latex]\r\n\r\n42.\u00a0[latex]2\\text{log}\\left(.0001\\right)[\/latex]\r\n\r\n43. [latex]{e}^{\\mathrm{ln}\\left(1.06\\right)}[\/latex]\r\n\r\n44.\u00a0[latex]\\mathrm{ln}\\left({e}^{-5.03}\\right)[\/latex]\r\n\r\n45. [latex]{e}^{\\mathrm{ln}\\left(10.125\\right)}+4[\/latex]\r\n\r\nFor the following exercises, evaluate the base <em>b<\/em>\u00a0logarithmic expression without using a calculator.\r\n\r\n46. [latex]{\\text{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex]\r\n\r\n47. [latex]{\\text{log}}_{6}\\left(\\sqrt{6}\\right)[\/latex]\r\n\r\n48.\u00a0[latex]{\\text{log}}_{2}\\left(\\frac{1}{8}\\right)+4[\/latex]\r\n\r\n49. [latex]6{\\text{log}}_{8}\\left(4\\right)[\/latex]\r\n\r\nFor the following exercises, evaluate the common logarithmic expression without using a calculator.\r\n\r\n50. [latex]\\text{log}\\left(10,000\\right)[\/latex]\r\n\r\n51. [latex]\\text{log}\\left(0.001\\right)[\/latex]\r\n\r\n52.\u00a0[latex]\\text{log}\\left(1\\right)+7[\/latex]\r\n\r\n53. [latex]2\\text{log}\\left({100}^{-3}\\right)[\/latex]\r\n\r\nFor the following exercises, evaluate the natural logarithmic expression without using a calculator.\r\n\r\n54. [latex]\\text{ln}\\left({e}^{\\frac{1}{3}}\\right)[\/latex]\r\n\r\n55. [latex]\\text{ln}\\left(1\\right)[\/latex]\r\n\r\n56.\u00a0[latex]\\text{ln}\\left({e}^{-0.225}\\right)-3[\/latex]\r\n\r\n57. [latex]25\\text{ln}\\left({e}^{\\frac{2}{5}}\\right)[\/latex]\r\n\r\nFor the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.\r\n\r\n58. [latex]\\text{log}\\left(0.04\\right)[\/latex]\r\n\r\n59. [latex]\\text{ln}\\left(15\\right)[\/latex]\r\n\r\n60.\u00a0[latex]\\text{ln}\\left(\\frac{4}{5}\\right)[\/latex]\r\n\r\n61. [latex]\\text{log}\\left(\\sqrt{2}\\right)[\/latex]\r\n\r\nFor the following exercises, state the domain and range of the function.\r\n\r\n62. [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x+4\\right)[\/latex]\r\n\r\n63. [latex]h\\left(x\\right)=\\mathrm{ln}\\left(\\frac{1}{2}-x\\right)[\/latex]\r\n\r\n64.\u00a0[latex]g\\left(x\\right)={\\mathrm{log}}_{5}\\left(2x+9\\right)-2[\/latex]\r\n\r\n65. [latex]h\\left(x\\right)=\\mathrm{ln}\\left(4x+17\\right)-5[\/latex]\r\n\r\n66.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(12 - 3x\\right)-3[\/latex]\r\n\r\nFor the following exercises, state the domain and the vertical asymptote of the function.\r\n\r\n67. [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x - 5\\right)[\/latex]\r\n\r\n68.\u00a0[latex]g\\left(x\\right)=\\mathrm{ln}\\left(3-x\\right)[\/latex]\r\n\r\n69. [latex]f\\left(x\\right)=\\mathrm{log}\\left(3x+1\\right)[\/latex]\r\n\r\n70.\u00a0[latex]f\\left(x\\right)=3\\mathrm{log}\\left(-x\\right)+2[\/latex]\r\n\r\n71. [latex]g\\left(x\\right)=-\\mathrm{ln}\\left(3x+9\\right)-7[\/latex]\r\n\r\nFor the following exercises, state the domain, vertical asymptote, and end behavior of the function.\r\n\r\n72. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(2-x\\right)[\/latex]\r\n\r\n73. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x-\\frac{3}{7}\\right)[\/latex]\r\n\r\n74.\u00a0[latex]h\\left(x\\right)=-\\mathrm{log}\\left(3x - 4\\right)+3[\/latex]\r\n\r\n75. [latex]g\\left(x\\right)=\\mathrm{ln}\\left(2x+6\\right)-5[\/latex]\r\n\r\n76.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(15 - 5x\\right)+6[\/latex]\r\n\r\nFor the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.\r\n\r\n77. [latex]h\\left(x\\right)={\\mathrm{log}}_{4}\\left(x - 1\\right)+1[\/latex]\r\n\r\n78.\u00a0[latex]f\\left(x\\right)=\\mathrm{log}\\left(5x+10\\right)+3[\/latex]\r\n\r\n79. [latex]g\\left(x\\right)=\\mathrm{ln}\\left(-x\\right)-2[\/latex]\r\n\r\n80.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+2\\right)-5[\/latex]\r\n\r\n81. [latex]h\\left(x\\right)=3\\mathrm{ln}\\left(x\\right)-9[\/latex]\r\n\r\nFor the following exercises, match each function in the graph below\u00a0with the letter corresponding to its graph.\r\n<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_201.jpg\" alt=\"Graph of five logarithmic functions.\" \/>\r\n\r\n82. [latex]d\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]\r\n\r\n83. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n84. [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]\r\n\r\n85. [latex]h\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]\r\n\r\n86.\u00a0[latex]j\\left(x\\right)={\\mathrm{log}}_{25}\\left(x\\right)[\/latex]\r\n\r\nFor the following exercises, match each function in the figure below\u00a0with the letter corresponding to its graph.\r\n<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_202.jpg\" alt=\"Graph of three logarithmic functions.\" \/>\r\n\r\n87.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{3}}\\left(x\\right)[\/latex]\r\n\r\n88. [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]\r\n\r\n89. [latex]h\\left(x\\right)={\\mathrm{log}}_{\\frac{3}{4}}\\left(x\\right)[\/latex]\r\n\r\nFor the following exercises, sketch the graphs of each pair of functions on the same axis.\r\n\r\n90. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)={10}^{x}[\/latex]\r\n\r\n91. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)[\/latex]\r\n\r\n92. [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n93. [latex]f\\left(x\\right)={e}^{x}[\/latex] and [latex]g\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\nFor the following exercises, match each function in the graph below\u00a0with the letter corresponding to its graph.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005325\/CNX_PreCalc_Figure_04_04_207.jpg\" alt=\"Graph of three logarithmic functions.\" \/>\r\n94. [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(-x+2\\right)[\/latex]\r\n\r\n95. [latex]g\\left(x\\right)=-{\\mathrm{log}}_{4}\\left(x+2\\right)[\/latex]\r\n\r\n96. [latex]h\\left(x\\right)={\\mathrm{log}}_{4}\\left(x+2\\right)[\/latex]\r\n\r\nFor the following exercises, sketch the graph of the indicated function.\r\n\r\n97. [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+2\\right)[\/latex]\r\n\r\n98. [latex]f\\left(x\\right)=2\\mathrm{log}\\left(x\\right)[\/latex]\r\n\r\n99. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(-x\\right)[\/latex]\r\n\r\n100. [latex]g\\left(x\\right)=\\mathrm{log}\\left(4x+16\\right)+4[\/latex]\r\n\r\n101. [latex]g\\left(x\\right)=\\mathrm{log}\\left(6 - 3x\\right)+1[\/latex]\r\n\r\n102. [latex]h\\left(x\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(x+1\\right)-3[\/latex]\r\n\r\nFor the following exercises, write a logarithmic equation corresponding to the graph shown.\r\n\r\n103. Use [latex]y={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] as the parent function.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005326\/CNX_PreCalc_Figure_04_04_214.jpg\" alt=\"The graph y=log_2(x) has been reflected over the y-axis and shifted to the right by 1.\" \/>\r\n\r\n104.\u00a0Use [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] as the parent function.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005326\/CNX_PreCalc_Figure_04_04_215.jpg\" alt=\"The graph y=log_3(x) has been reflected over the x-axis, vertically stretched by 3, and shifted to the left by 4.\" \/>\r\n\r\n105. Use [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] as the parent function.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005326\/CNX_PreCalc_Figure_04_04_216.jpg\" alt=\"The graph y=log_4(x) has been vertically stretched by 3, and shifted to the left by 2.\" \/>\r\n\r\n106.\u00a0Use [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex] as the parent function.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005326\/CNX_PreCalc_Figure_04_04_217.jpg\" alt=\"The graph y=log_3(x) has been reflected over the x-axis and y-axis, vertically stretched by 2, and shifted to the right by 5.\" \/>\r\n\r\n107. Explore and discuss the graphs of [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)=-{\\mathrm{log}}_{2}\\left(x\\right)[\/latex]. Make a conjecture based on the result.\r\n\r\n108.\u00a0Prove the conjecture made in the previous exercise.\r\n\r\n109. What is the domain of the function [latex]f\\left(x\\right)=\\mathrm{ln}\\left(\\frac{x+2}{x - 4}\\right)[\/latex]? Discuss the result.\r\n\r\n110.\u00a0Use properties of exponents to find the x-intercepts of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left({x}^{2}+4x+4\\right)[\/latex] algebraically. Show the steps for solving, and then verify the result by graphing the function.\r\n\r\n111. Is <em>x<\/em> = 0 in the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]? If so, what is the value of the function when <em>x<\/em> = 0? Verify the result.\r\n\r\n112. Is [latex]f\\left(x\\right)=0[\/latex] in the range of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]? If so, for what value of <em>x<\/em>? Verify the result.\r\n\r\n113. Is there a number <em>x<\/em>\u00a0such that [latex]\\mathrm{ln}x=2[\/latex]? If so, what is that number? Verify the result.\r\n\r\n114.\u00a0Is the following true: [latex]\\frac{{\\mathrm{log}}_{3}\\left(27\\right)}{{\\mathrm{log}}_{4}\\left(\\frac{1}{64}\\right)}=-1[\/latex]? Verify the result.\r\n\r\n115. Is the following true: [latex]\\frac{\\mathrm{ln}\\left({e}^{1.725}\\right)}{\\mathrm{ln}\\left(1\\right)}=1.725[\/latex]? Verify the result.\r\n\r\n116.\u00a0The exposure index <em>EI<\/em> for a 35 millimeter camera is a measurement of the amount of light that hits the film. It is determined by the equation [latex]EI={\\mathrm{log}}_{2}\\left(\\frac{{f}^{2}}{t}\\right)[\/latex], where <em>f<\/em>\u00a0is the \"f-stop\" setting on the camera, and <em>t<\/em>\u00a0is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2\u00a0seconds. What will the resulting exposure index be?\r\n\r\n117. Refer to the previous exercise. Suppose the light meter on a camera indicates an <em>EI<\/em>\u00a0of \u20132, and the desired exposure time is 16 seconds. What should the f-stop setting be?\r\n\r\n118.\u00a0The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula [latex]\\mathrm{log}\\frac{{I}_{1}}{{I}_{2}}={M}_{1}-{M}_{2}[\/latex] where <em>M<\/em>\u00a0is the magnitude given by the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same region experienced yet another, more devastating earthquake, this time with a magnitude of 9.0.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/world\/historical.php\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/world\/historical.php<\/a>. Accessed 3\/4\/2014.[\/footnote] How many times greater was the intensity of the 2011 earthquake? Round to the nearest whole number.","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.<\/li>\n<li>Use natural logarithms.<\/li>\n<li>Identify the domain of a logarithmic function.<\/li>\n<li>Graph logarithmic functions.<\/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=\"http:\/\/www.hutchmath.com\/Images\/Haiteruins.png\" alt=\"Photo of the aftermath of the earthquake in Haite, 2010\" width=\"488\" height=\"325\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>Devastation of January 2010 earthquake in Haiti.<\/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-id1165135161452\">Next 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\">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 6: 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 7: 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 3.\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 4\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 8: 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 9:\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 10: 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 11: 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 12: 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 13: 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<p>&nbsp;<\/p>\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 14: 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 15: 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<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Equations<\/span><\/p>\n<section id=\"fs-id1165137749167\" class=\"key-equations\">\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 id=\"fs-id1165137863125\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137863132\">\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<\/ul>\n<\/section>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135160066\" class=\"definition\">\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<\/section>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Section 5.4 Homework Exercises<\/h2>\n<p>1. What is a base <em>b<\/em>\u00a0logarithm? Discuss the meaning by interpreting each part of the equivalent equations [latex]{b}^{y}=x[\/latex] and [latex]{\\mathrm{log}}_{b}x=y[\/latex] for [latex]b>0,b\\ne 1[\/latex].<\/p>\n<p>2.\u00a0How is the logarithmic function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}x[\/latex] related to the exponential function [latex]g\\left(x\\right)={b}^{x}[\/latex]? What is the result of composing these two functions?<\/p>\n<p>3. How can the logarithmic equation [latex]{\\mathrm{log}}_{b}x=y[\/latex] be solved for <em>x<\/em>\u00a0using the properties of exponents?<\/p>\n<p>4. Discuss the meaning of the common logarithm. What is its relationship to a logarithm with base <em>b<\/em>, and how does the notation differ?<\/p>\n<p>5. Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with base <em>b<\/em>, and how does the notation differ?<\/p>\n<p>6.\u00a0What type(s) of translation(s), if any, affect the range of a logarithmic function?<\/p>\n<p>7. The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each?<\/p>\n<p>8.\u00a0Consider the general logarithmic function [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Why can\u2019t <em>x<\/em>\u00a0be zero?<\/p>\n<p>9. Does the graph of a general logarithmic function have a horizontal asymptote? Explain.<\/p>\n<p>For the following exercises, rewrite each equation in exponential form.<\/p>\n<p>10. [latex]{\\text{log}}_{4}\\left(q\\right)=m[\/latex]<\/p>\n<p>11. [latex]{\\text{log}}_{a}\\left(b\\right)=c[\/latex]<\/p>\n<p>12. [latex]{\\mathrm{log}}_{16}\\left(y\\right)=x[\/latex]<\/p>\n<p>13. [latex]{\\mathrm{log}}_{x}\\left(64\\right)=y[\/latex]<\/p>\n<p>14.\u00a0[latex]{\\mathrm{log}}_{y}\\left(x\\right)=-11[\/latex]<\/p>\n<p>15. [latex]{\\mathrm{log}}_{15}\\left(a\\right)=b[\/latex]<\/p>\n<p>16.\u00a0[latex]{\\mathrm{log}}_{y}\\left(137\\right)=x[\/latex]<\/p>\n<p>17. [latex]{\\mathrm{log}}_{13}\\left(142\\right)=a[\/latex]<\/p>\n<p>18.\u00a0[latex]\\text{log}\\left(v\\right)=t[\/latex]<\/p>\n<p>19. [latex]\\text{ln}\\left(w\\right)=n[\/latex]<\/p>\n<p>For the following exercises, rewrite each equation in logarithmic form.<\/p>\n<p>20. [latex]{4}^{x}=y[\/latex]<\/p>\n<p>21. [latex]{c}^{d}=k[\/latex]<\/p>\n<p>22. [latex]{m}^{-7}=n[\/latex]<\/p>\n<p>23. [latex]{19}^{x}=y[\/latex]<\/p>\n<p>24.\u00a0[latex]{x}^{-\\frac{10}{13}}=y[\/latex]<\/p>\n<p>25. [latex]{n}^{4}=103[\/latex]<\/p>\n<p>26.\u00a0[latex]{\\left(\\frac{7}{5}\\right)}^{m}=n[\/latex]<\/p>\n<p>27. [latex]{y}^{x}=\\frac{39}{100}[\/latex]<\/p>\n<p>28.\u00a0[latex]{10}^{a}=b[\/latex]<\/p>\n<p>29. [latex]{e}^{k}=h[\/latex]<\/p>\n<p>For the following exercises, solve for <em>x<\/em>\u00a0by converting the logarithmic equation to exponential form.<\/p>\n<p>30. [latex]{\\text{log}}_{3}\\left(x\\right)=2[\/latex]<\/p>\n<p>31. [latex]{\\text{log}}_{2}\\left(x\\right)=-3[\/latex]<\/p>\n<p>32.\u00a0[latex]{\\text{log}}_{5}\\left(x\\right)=2[\/latex]<\/p>\n<p>33. [latex]{\\mathrm{log}}_{3}\\left(x\\right)=3[\/latex]<\/p>\n<p>34.\u00a0[latex]{\\text{log}}_{2}\\left(x\\right)=6[\/latex]<\/p>\n<p>35. [latex]{\\text{log}}_{9}\\left(x\\right)=\\frac{1}{2}[\/latex]<\/p>\n<p>36.\u00a0[latex]{\\text{log}}_{18}\\left(x\\right)=2[\/latex]<\/p>\n<p>37. [latex]{\\mathrm{log}}_{6}\\left(x\\right)=-3[\/latex]<\/p>\n<p>38.\u00a0[latex]\\text{log}\\left(x\\right)=3[\/latex]<\/p>\n<p>39. [latex]\\text{ln}\\left(x\\right)=2[\/latex]<\/p>\n<p>For the following exercises, use the definition of common and natural logarithms to simplify.<\/p>\n<p>40. [latex]\\text{log}\\left({100}^{8}\\right)[\/latex]<\/p>\n<p>41. [latex]{10}^{\\text{log}\\left(32\\right)}[\/latex]<\/p>\n<p>42.\u00a0[latex]2\\text{log}\\left(.0001\\right)[\/latex]<\/p>\n<p>43. [latex]{e}^{\\mathrm{ln}\\left(1.06\\right)}[\/latex]<\/p>\n<p>44.\u00a0[latex]\\mathrm{ln}\\left({e}^{-5.03}\\right)[\/latex]<\/p>\n<p>45. [latex]{e}^{\\mathrm{ln}\\left(10.125\\right)}+4[\/latex]<\/p>\n<p>For the following exercises, evaluate the base <em>b<\/em>\u00a0logarithmic expression without using a calculator.<\/p>\n<p>46. [latex]{\\text{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex]<\/p>\n<p>47. [latex]{\\text{log}}_{6}\\left(\\sqrt{6}\\right)[\/latex]<\/p>\n<p>48.\u00a0[latex]{\\text{log}}_{2}\\left(\\frac{1}{8}\\right)+4[\/latex]<\/p>\n<p>49. [latex]6{\\text{log}}_{8}\\left(4\\right)[\/latex]<\/p>\n<p>For the following exercises, evaluate the common logarithmic expression without using a calculator.<\/p>\n<p>50. [latex]\\text{log}\\left(10,000\\right)[\/latex]<\/p>\n<p>51. [latex]\\text{log}\\left(0.001\\right)[\/latex]<\/p>\n<p>52.\u00a0[latex]\\text{log}\\left(1\\right)+7[\/latex]<\/p>\n<p>53. [latex]2\\text{log}\\left({100}^{-3}\\right)[\/latex]<\/p>\n<p>For the following exercises, evaluate the natural logarithmic expression without using a calculator.<\/p>\n<p>54. [latex]\\text{ln}\\left({e}^{\\frac{1}{3}}\\right)[\/latex]<\/p>\n<p>55. [latex]\\text{ln}\\left(1\\right)[\/latex]<\/p>\n<p>56.\u00a0[latex]\\text{ln}\\left({e}^{-0.225}\\right)-3[\/latex]<\/p>\n<p>57. [latex]25\\text{ln}\\left({e}^{\\frac{2}{5}}\\right)[\/latex]<\/p>\n<p>For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.<\/p>\n<p>58. [latex]\\text{log}\\left(0.04\\right)[\/latex]<\/p>\n<p>59. [latex]\\text{ln}\\left(15\\right)[\/latex]<\/p>\n<p>60.\u00a0[latex]\\text{ln}\\left(\\frac{4}{5}\\right)[\/latex]<\/p>\n<p>61. [latex]\\text{log}\\left(\\sqrt{2}\\right)[\/latex]<\/p>\n<p>For the following exercises, state the domain and range of the function.<\/p>\n<p>62. [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x+4\\right)[\/latex]<\/p>\n<p>63. [latex]h\\left(x\\right)=\\mathrm{ln}\\left(\\frac{1}{2}-x\\right)[\/latex]<\/p>\n<p>64.\u00a0[latex]g\\left(x\\right)={\\mathrm{log}}_{5}\\left(2x+9\\right)-2[\/latex]<\/p>\n<p>65. [latex]h\\left(x\\right)=\\mathrm{ln}\\left(4x+17\\right)-5[\/latex]<\/p>\n<p>66.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(12 - 3x\\right)-3[\/latex]<\/p>\n<p>For the following exercises, state the domain and the vertical asymptote of the function.<\/p>\n<p>67. [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x - 5\\right)[\/latex]<\/p>\n<p>68.\u00a0[latex]g\\left(x\\right)=\\mathrm{ln}\\left(3-x\\right)[\/latex]<\/p>\n<p>69. [latex]f\\left(x\\right)=\\mathrm{log}\\left(3x+1\\right)[\/latex]<\/p>\n<p>70.\u00a0[latex]f\\left(x\\right)=3\\mathrm{log}\\left(-x\\right)+2[\/latex]<\/p>\n<p>71. [latex]g\\left(x\\right)=-\\mathrm{ln}\\left(3x+9\\right)-7[\/latex]<\/p>\n<p>For the following exercises, state the domain, vertical asymptote, and end behavior of the function.<\/p>\n<p>72. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(2-x\\right)[\/latex]<\/p>\n<p>73. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x-\\frac{3}{7}\\right)[\/latex]<\/p>\n<p>74.\u00a0[latex]h\\left(x\\right)=-\\mathrm{log}\\left(3x - 4\\right)+3[\/latex]<\/p>\n<p>75. [latex]g\\left(x\\right)=\\mathrm{ln}\\left(2x+6\\right)-5[\/latex]<\/p>\n<p>76.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(15 - 5x\\right)+6[\/latex]<\/p>\n<p>For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.<\/p>\n<p>77. [latex]h\\left(x\\right)={\\mathrm{log}}_{4}\\left(x - 1\\right)+1[\/latex]<\/p>\n<p>78.\u00a0[latex]f\\left(x\\right)=\\mathrm{log}\\left(5x+10\\right)+3[\/latex]<\/p>\n<p>79. [latex]g\\left(x\\right)=\\mathrm{ln}\\left(-x\\right)-2[\/latex]<\/p>\n<p>80.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+2\\right)-5[\/latex]<\/p>\n<p>81. [latex]h\\left(x\\right)=3\\mathrm{ln}\\left(x\\right)-9[\/latex]<\/p>\n<p>For the following exercises, match each function in the graph below\u00a0with the letter corresponding to its graph.<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\/03005324\/CNX_PreCalc_Figure_04_04_201.jpg\" alt=\"Graph of five logarithmic functions.\" \/><\/p>\n<p>82. [latex]d\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]<\/p>\n<p>83. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<p>84. [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/p>\n<p>85. [latex]h\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]<\/p>\n<p>86.\u00a0[latex]j\\left(x\\right)={\\mathrm{log}}_{25}\\left(x\\right)[\/latex]<\/p>\n<p>For the following exercises, match each function in the figure below\u00a0with the letter corresponding to its graph.<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\/03005324\/CNX_PreCalc_Figure_04_04_202.jpg\" alt=\"Graph of three logarithmic functions.\" \/><\/p>\n<p>87.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{3}}\\left(x\\right)[\/latex]<\/p>\n<p>88. [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/p>\n<p>89. [latex]h\\left(x\\right)={\\mathrm{log}}_{\\frac{3}{4}}\\left(x\\right)[\/latex]<\/p>\n<p>For the following exercises, sketch the graphs of each pair of functions on the same axis.<\/p>\n<p>90. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)={10}^{x}[\/latex]<\/p>\n<p>91. [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)[\/latex]<\/p>\n<p>92. [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<p>93. [latex]f\\left(x\\right)={e}^{x}[\/latex] and [latex]g\\left(x\\right)=\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<p>For the following exercises, match each function in the graph below\u00a0with the letter corresponding to its graph.<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\/03005325\/CNX_PreCalc_Figure_04_04_207.jpg\" alt=\"Graph of three logarithmic functions.\" \/><br \/>\n94. [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(-x+2\\right)[\/latex]<\/p>\n<p>95. [latex]g\\left(x\\right)=-{\\mathrm{log}}_{4}\\left(x+2\\right)[\/latex]<\/p>\n<p>96. [latex]h\\left(x\\right)={\\mathrm{log}}_{4}\\left(x+2\\right)[\/latex]<\/p>\n<p>For the following exercises, sketch the graph of the indicated function.<\/p>\n<p>97. [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+2\\right)[\/latex]<\/p>\n<p>98. [latex]f\\left(x\\right)=2\\mathrm{log}\\left(x\\right)[\/latex]<\/p>\n<p>99. [latex]f\\left(x\\right)=\\mathrm{ln}\\left(-x\\right)[\/latex]<\/p>\n<p>100. [latex]g\\left(x\\right)=\\mathrm{log}\\left(4x+16\\right)+4[\/latex]<\/p>\n<p>101. [latex]g\\left(x\\right)=\\mathrm{log}\\left(6 - 3x\\right)+1[\/latex]<\/p>\n<p>102. [latex]h\\left(x\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(x+1\\right)-3[\/latex]<\/p>\n<p>For the following exercises, write a logarithmic equation corresponding to the graph shown.<\/p>\n<p>103. Use [latex]y={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] as the parent function.<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\/03005326\/CNX_PreCalc_Figure_04_04_214.jpg\" alt=\"The graph y=log_2(x) has been reflected over the y-axis and shifted to the right by 1.\" \/><\/p>\n<p>104.\u00a0Use [latex]f\\left(x\\right)={\\mathrm{log}}_{3}\\left(x\\right)[\/latex] as the parent function.<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\/03005326\/CNX_PreCalc_Figure_04_04_215.jpg\" alt=\"The graph y=log_3(x) has been reflected over the x-axis, vertically stretched by 3, and shifted to the left by 4.\" \/><\/p>\n<p>105. Use [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(x\\right)[\/latex] as the parent function.<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\/03005326\/CNX_PreCalc_Figure_04_04_216.jpg\" alt=\"The graph y=log_4(x) has been vertically stretched by 3, and shifted to the left by 2.\" \/><\/p>\n<p>106.\u00a0Use [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex] as the parent function.<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\/03005326\/CNX_PreCalc_Figure_04_04_217.jpg\" alt=\"The graph y=log_3(x) has been reflected over the x-axis and y-axis, vertically stretched by 2, and shifted to the right by 5.\" \/><\/p>\n<p>107. Explore and discuss the graphs of [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)=-{\\mathrm{log}}_{2}\\left(x\\right)[\/latex]. Make a conjecture based on the result.<\/p>\n<p>108.\u00a0Prove the conjecture made in the previous exercise.<\/p>\n<p>109. What is the domain of the function [latex]f\\left(x\\right)=\\mathrm{ln}\\left(\\frac{x+2}{x - 4}\\right)[\/latex]? Discuss the result.<\/p>\n<p>110.\u00a0Use properties of exponents to find the x-intercepts of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left({x}^{2}+4x+4\\right)[\/latex] algebraically. Show the steps for solving, and then verify the result by graphing the function.<\/p>\n<p>111. Is <em>x<\/em> = 0 in the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]? If so, what is the value of the function when <em>x<\/em> = 0? Verify the result.<\/p>\n<p>112. Is [latex]f\\left(x\\right)=0[\/latex] in the range of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)[\/latex]? If so, for what value of <em>x<\/em>? Verify the result.<\/p>\n<p>113. Is there a number <em>x<\/em>\u00a0such that [latex]\\mathrm{ln}x=2[\/latex]? If so, what is that number? Verify the result.<\/p>\n<p>114.\u00a0Is the following true: [latex]\\frac{{\\mathrm{log}}_{3}\\left(27\\right)}{{\\mathrm{log}}_{4}\\left(\\frac{1}{64}\\right)}=-1[\/latex]? Verify the result.<\/p>\n<p>115. Is the following true: [latex]\\frac{\\mathrm{ln}\\left({e}^{1.725}\\right)}{\\mathrm{ln}\\left(1\\right)}=1.725[\/latex]? Verify the result.<\/p>\n<p>116.\u00a0The exposure index <em>EI<\/em> for a 35 millimeter camera is a measurement of the amount of light that hits the film. It is determined by the equation [latex]EI={\\mathrm{log}}_{2}\\left(\\frac{{f}^{2}}{t}\\right)[\/latex], where <em>f<\/em>\u00a0is the &#8220;f-stop&#8221; setting on the camera, and <em>t<\/em>\u00a0is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2\u00a0seconds. What will the resulting exposure index be?<\/p>\n<p>117. Refer to the previous exercise. Suppose the light meter on a camera indicates an <em>EI<\/em>\u00a0of \u20132, and the desired exposure time is 16 seconds. What should the f-stop setting be?<\/p>\n<p>118.\u00a0The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula [latex]\\mathrm{log}\\frac{{I}_{1}}{{I}_{2}}={M}_{1}-{M}_{2}[\/latex] where <em>M<\/em>\u00a0is the magnitude given by the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same region experienced yet another, more devastating earthquake, this time with a magnitude of 9.0.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/world\/historical.php. Accessed 3\/4\/2014.\" id=\"return-footnote-13703-5\" href=\"#footnote-13703-5\" aria-label=\"Footnote 5\"><sup class=\"footnote\">[5]<\/sup><\/a> How many times greater was the intensity of the 2011 earthquake? Round to the nearest whole number.<\/p>\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><li id=\"footnote-13703-5\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/world\/historical.php\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/world\/historical.php<\/a>. Accessed 3\/4\/2014. <a href=\"#return-footnote-13703-5\" class=\"return-footnote\" aria-label=\"Return to footnote 5\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":97803,"menu_order":4,"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":13696,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13703","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/97803"}],"version-history":[{"count":28,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13703\/revisions"}],"predecessor-version":[{"id":17933,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13703\/revisions\/17933"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/13696"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/13703\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=13703"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=13703"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=13703"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=13703"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}