{"id":303,"date":"2019-02-08T21:05:34","date_gmt":"2019-02-08T21:05:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/logarithmic-functions\/"},"modified":"2019-11-18T18:31:22","modified_gmt":"2019-11-18T18:31:22","slug":"logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/logarithmic-functions\/","title":{"raw":"2.4 Logarithmic Functions","rendered":"2.4 Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Convert from logarithmic to exponential form.<\/li>\r\n \t<li>Convert from exponential to logarithmic 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>Use logarithmic properties.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210524\/CNX_Precalc_Figure_04_03_001.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/> Figure 1\u00a0 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137557013\">In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes[footnote]http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary. Accessed 3\/4\/2013.[\/footnote]. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[footnote]http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary. Accessed 3\/4\/2013.[\/footnote]\u00a0like those shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_001\">Figure 1<\/a>. 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\u00a0[footnote]http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013.[\/footnote]\u00a0<sup id=\"footnote-ref3\"><\/sup>whereas the Japanese earthquake registered a 9.0[footnote]http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details. Accessed 3\/4\/2013.[\/footnote]<sup id=\"footnote-ref4\"><\/sup><\/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 section, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\r\n\r\n<div id=\"fs-id1165137644550\" class=\"bc-section section\">\r\n<h3>Converting from Logarithmic to Exponential Form<\/h3>\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 [latex]x[\/latex] represents the difference in magnitudes on the <span class=\"no-emphasis\">Richter Scale<\/span>. How would we solve for [latex]x?[\/latex]<\/p>\r\n<p id=\"fs-id1165135160312\">We have only learned a graphical method for approximating solutions of 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 [latex]x[\/latex] must be some value between 2 and 3, since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph, as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_002\">Figure 2<\/a><strong>,<\/strong> to better estimate the solution.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_03_002\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"257\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210528\/CNX_Precalc_Figure_04_03_002.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"257\" height=\"252\" \/> Figure 2[\/caption]\r\n\r\n<\/div>\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 in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_002\">Figure 2<\/a>\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <span class=\"no-emphasis\">one-to-one<\/span>, so its inverse is also a function. We use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] to describe the inverse.\u00a0 The base [latex]b[\/latex]\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise [latex]b[\/latex] to get that number.<\/p>\r\n<p id=\"fs-id1165137404844\">We read a logarithmic expression as, \u201cThe logarithm with base [latex]b[\/latex] of [latex]x[\/latex] is equal to [latex]y,[\/latex]\u201d or, simplified, \u201clog base [latex]b[\/latex] of [latex]x[\/latex] is [latex]y.[\/latex]\u201d We can also say, \u201c[latex]b[\/latex] raised to the power of [latex]y[\/latex] is [latex]x,[\/latex]\u201d 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 \u201clog base 2 of 32 is 5.\u201d<\/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<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\u21d4{b}^{y}=x,\\text{ }b&gt;0,b\\ne 1[\/latex][latex]\\\\[\/latex]<\/p>\r\n<p id=\"fs-id1165137678993\">Note that the base [latex]b[\/latex] is always positive.<\/p>\r\n<span id=\"fs-id1165137696233\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210530\/CNX_Precalc_Figure_04_03_004.jpg\" alt=\"\" \/><\/span>\r\n\r\nBecause 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 [latex]x.[\/latex]\r\n<p id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<\/p>\r\n<span id=\"fs-id1165137771679\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210533\/CNX_Precalc_Figure_04_03_003.jpg\" alt=\"\" \/><\/span>\r\n<p id=\"fs-id1165137575165\">Notice that, comparing the logarithm function and the exponential function, the input and the output are switched.<\/p>\r\n\r\n<div id=\"fs-id1165137472937\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base [latex]b[\/latex] of a positive number [latex]x[\/latex] satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165137584967\">For [latex]x&gt;0,\\text{ }b&gt;0,\\text{ }b\\ne 1,[\/latex]\u00a0 [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is equivalent to [latex]{b}^{y}=x.[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165135530561\">\r\n \t<li>We read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]b[\/latex] of [latex]x[\/latex]\u201d or the \u201clog base [latex]b[\/latex] of [latex]x.\"[\/latex]<\/li>\r\n \t<li>The logarithm [latex]y[\/latex] is the exponent to which [latex]b[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137547773\">Also, since the logarithmic and exponential functions are inverses of each other, 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[\/latex] is [latex]\\left(0,\\infty \\right),[\/latex] and<\/li>\r\n \t<li>the range of the logarithm function with base [latex]b[\/latex] is [latex]\\left(-\\infty ,\\infty \\right).[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137677696\" class=\"precalculus qa key-takeaways\">\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. We are working with functions of real numbers. Because the base of an exponential function is always a positive real number, no power of that base can ever be a negative real number. We can never take the logarithm of a negative real 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=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137806301\"><strong>Given an equation in logarithmic form\u00a0<\/strong>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y,[\/latex] <strong>convert it to exponential form.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137641669\" type=\"1\">\r\n \t<li>Examine the equation [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and identify [latex]b,\\text{ }y,[\/latex] and [latex]x.[\/latex]<\/li>\r\n \t<li>Rewrite [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] as [latex]{b}^{y}=x.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_01\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135570363\">\r\n<div id=\"fs-id1165137557855\">\r\n<h3>Example 1:\u00a0 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\" type=\"a\">\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165135613330\">[reveal-answer q=\"fs-id1165135613330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135613330\"]\r\n<p id=\"fs-id1165137408172\">First, identify the values of [latex]b,\\text{ }y,[\/latex] and [latex]x[\/latex] in [latex]y= {\\mathrm{log}}_{b}\\left(x\\right).[\/latex] Then, write the equation in the form [latex]{b}^{y}=x.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1165137705659\" type=\"a\">\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex]\r\n<p id=\"fs-id1165137602796\">Here, [latex]b=6,\\text{ }y=\\frac{1}{2},[\/latex] and [latex]x=\\sqrt[\\leftroot{1}\\uproot{2} ]{6}.[\/latex] Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex] is equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt[\\leftroot{1}\\uproot{2} ]{6}.[\/latex]<\/p>\r\n<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex]\r\n<p id=\"fs-id1165137698078\">Here, [latex]b=10,\\text{ }y=2,[\/latex] and [latex]x=100.[\/latex] Therefore, the equation [latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex] is equivalent to [latex]{10}^{2}=100.[\/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 id=\"fs-id1165137640140\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_04_03_01\">\r\n<div id=\"fs-id1165135208926\">\r\n<p id=\"fs-id1165137418681\">Write the following logarithmic equations in exponential form.<\/p>\r\n\r\n<ol id=\"fs-id1165137772342\" type=\"a\">\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165135195688\">[reveal-answer q=\"fs-id1165135195688\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135195688\"]\r\n<ol id=\"fs-id1165137414337\" type=\"a\">\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equivalent to [latex]{10}^{6}=1,000,000.[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equivalent to [latex]{5}^{2}=25.[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137585244\" class=\"bc-section section\">\r\n<h3>Converting from Exponential to Logarithmic Form<\/h3>\r\n<p id=\"fs-id1165137933968\">To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base [latex]b,[\/latex] exponent [latex]x,[\/latex] and output [latex]y[\/latex] in [latex]b^x=y.[\/latex] Then we write\u00a0 [latex]x={\\mathrm{log}}_{b}\\left(y\\right).[\/latex]<\/p>\r\n\r\n<div id=\"Example_04_03_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135168111\">\r\n<div id=\"fs-id1165137727912\">\r\n<h3>Example 2:\u00a0 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\" type=\"a\">\r\n \t<li>[latex]{2}^{3}=8[\/latex]<\/li>\r\n \t<li>[latex]{10}^{4}=10000[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165137702205\">[reveal-answer q=\"fs-id1165137702205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137702205\"]\r\n<p id=\"fs-id1165137474116\">First, identify the values of [latex]b,\\text{ }y,[\/latex] and [latex]x[\/latex] in [latex]b^x=y.[\/latex] 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\" type=\"a\">\r\n \t<li>[latex]{2}^{3}=8[\/latex]\r\n<p id=\"fs-id1165137466396\">Here, [latex]b=2,[\/latex] [latex]x=3,[\/latex] and [latex]y=8.[\/latex] Therefore, the equation [latex]{2}^{3}=8[\/latex] is equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3.[\/latex][latex]\\\\[\/latex]<\/p>\r\n<\/li>\r\n \t<li>[latex]{10}^{4}=10000[\/latex]\r\n<p id=\"fs-id1165135193035\">Here, [latex]b=10,[\/latex] [latex]x=4,[\/latex] and [latex]y=10000.[\/latex] Therefore, the equation [latex]{10}^{4}=10000[\/latex] is equivalent to [latex]{\\mathrm{log}}_{10}\\left(10000\\right)=4.[\/latex][latex]\\\\[\/latex]<\/p>\r\n<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\r\n<p id=\"fs-id1165135187822\">Here, [latex]b=10,[\/latex] [latex]x=-4,[\/latex] 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 id=\"fs-id1165137438165\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_04_03_02\">\r\n<div id=\"fs-id1165135190969\">\r\n<p id=\"fs-id1165137566762\">Write the following exponential equations in logarithmic form.<\/p>\r\n\r\n<ol id=\"fs-id1165137771963\" type=\"a\">\r\n \t<li>[latex]{1.1}^{2}=1.21[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-1}=\\frac{1}{10}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165134065138\">[reveal-answer q=\"fs-id1165134065138\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134065138\"]\r\n<ol id=\"fs-id1165137469846\" type=\"a\">\r\n \t<li>[latex]{1.1}^{2}=1.21[\/latex] is equivalent to [latex]{\\mathrm{log}}_{1.1}\\left(1.21\\right)=2.[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3.[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-1}=\\frac{1}{10}[\/latex] is equivalent to [latex]{\\mathrm{log}}_{10}\\left(\\frac{1}{10}\\right)=-1.[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137530906\" class=\"bc-section section\">\r\n<h3>Evaluating Logarithms<\/h3>\r\nKnowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}\\left(8\\right).[\/latex] We ask, \u201cTo what exponent must [latex]2[\/latex] be raised in order to get 8?\u201d Because we already know [latex]{2}^{3}=8,[\/latex] it follows that [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3.[\/latex]\r\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}\\left(49\\right)[\/latex] and [latex]{\\mathrm{log}}_{3}\\left(27\\right)[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137937690\">\r\n \t<li>We ask, \u201cTo what exponent must 7 be raised in order to get 49?\u201d We know [latex]{7}^{2}=49.[\/latex] Therefore, [latex]{\\mathrm{log}}_{7}\\left(49\\right)=2.[\/latex]<\/li>\r\n \t<li>We ask, \u201cTo what exponent must 3 be raised in order to get 27?\u201d We know [latex]{3}^{3}=27.[\/latex]Therefore, [latex]{\\mathrm{log}}_{3}\\left(27\\right)=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}}\\left(\\frac{4}{9}\\right)[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137584208\">\r\n \t<li>We ask, \u201cTo what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\left(\\frac{4}{9}\\right)?[\/latex]\u201d\u00a0 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[latex]\\\\[\/latex]\r\n<div id=\"fs-id1165137455840\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137453770\"><strong>Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right),[\/latex] evaluate it mentally.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165134079724\" type=\"1\">\r\n \t<li>Rewrite the argument [latex]x[\/latex] as a power of [latex]b:[\/latex]\u00a0 [latex]{b}^{y}=x.[\/latex]<\/li>\r\n \t<li>Use previous knowledge of powers of [latex]b[\/latex] identify [latex]y[\/latex] by asking, \u201cTo what exponent should [latex]b[\/latex] be raised in order to get [latex]x?[\/latex]\u201d<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137732842\">\r\n<div id=\"fs-id1165135296345\">\r\n<h3>Example 3:\u00a0 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\r\n<\/div>\r\n<div id=\"fs-id1165137852123\">[reveal-answer q=\"fs-id1165137852123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137852123\"]\r\n<p id=\"fs-id1165137611276\">First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64.[\/latex] Next, we ask, \u201cTo what exponent must 4 be raised in order to get 64?\u201d<\/p>\r\nWe know [latex]{4}^{3}=64.[\/latex]\u00a0 Therefore, [latex]\\mathrm{log}{}_{4}\\left(64\\right)=3.[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137731430\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_04_03_03\">\r\n<div id=\"fs-id1165137704553\">\r\n<p id=\"fs-id1165137745041\">Solve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137693554\">[reveal-answer q=\"fs-id1165137693554\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137693554\"]\r\n<p id=\"fs-id1165137639199\">[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recalling that [latex]\\sqrt[\\leftroot{1}\\uproot{2} ]{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex])<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_03_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137663658\">\r\n<div id=\"fs-id1165137680390\">\r\n<h3>Example 4:\u00a0 Evaluating the Logarithm of a Reciprocal<\/h3>\r\n<p id=\"fs-id1165137938805\">Evaluate [latex]y={\\mathrm{log}}_{10}\\left(\\frac{1}{100}\\right)[\/latex] without using a calculator.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135526087\">[reveal-answer q=\"fs-id1165135526087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135526087\"]\r\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form:\u00a0 [latex]{10}^{y}=\\frac{1}{100}.[\/latex]\u00a0 Next, we ask, \u201cTo what exponent must 10 be raised in order to get [latex]\\frac{1}{100}?[\/latex]\u201d<\/p>\r\n<p id=\"fs-id1165137552085\">We know [latex]{10}^{2}=100,[\/latex] but what must we do to get the reciprocal, [latex]\\frac{1}{100}?[\/latex] Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}.[\/latex] We use this information to write<\/p>\r\n\r\n<div id=\"eip-id1165137550550\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}{10}^{-2}&amp;=\\frac{1}{{10}^{2}}\\hfill \\\\\\text{ }&amp;=\\frac{1}{100}\\hfill \\end{align*}[\/latex]<\/div>\r\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{10}\\left(\\frac{1}{100}\\right)=-2.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137575754\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_04_03_04\">\r\n<div id=\"fs-id1165137768727\">\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\r\n<\/div>\r\n<div id=\"fs-id1165137812401\">[reveal-answer q=\"fs-id1165137812401\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137812401\"]\r\n<p id=\"fs-id1165137806792\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137547253\" class=\"bc-section section\">\r\n<h3>Using Common Logarithms<\/h3>\r\n<p id=\"fs-id1165137574205\">Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression [latex]\\mathrm{log}\\left(x\\right)[\/latex] means [latex]{\\mathrm{log}}_{10}\\left(x\\right).[\/latex] We call a base-10 logarithm a <strong>common logarithm<\/strong>. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137401037\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165135609332\">A <strong>common logarithm<\/strong> is a logarithm with base [latex]10.[\/latex] We write [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{log}\\left(x\\right).[\/latex] The common logarithm of a positive number [latex]x[\/latex] satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165137601579\">For [latex]x&gt;0,[\/latex] [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] is equivalent to [latex]{10}^{y}=x.[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li id=\"fs-id1165137559681\">We read [latex]\\mathrm{log}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]10[\/latex] of [latex]x[\/latex]\u201d or \u201clog base 10 of [latex]x.[\/latex]\u201d<\/li>\r\n \t<li>The logarithm [latex]y[\/latex] is the exponent to which [latex]10[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137579434\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137810781\"><strong>Given a common logarithm of the form [latex]y=\\mathrm{log}\\left(x\\right),[\/latex] evaluate it mentally.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137828334\" type=\"1\">\r\n \t<li>Rewrite the argument [latex]x[\/latex] as a power of [latex]10:[\/latex]\u00a0 [latex]{10}^{y}=x.[\/latex]<\/li>\r\n \t<li>Use previous knowledge of powers of [latex]10[\/latex] to identify [latex]y[\/latex] by asking, \u201cTo what exponent must [latex]10[\/latex] be raised in order to get [latex]x?[\/latex]\u201d<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_03_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137742366\">\r\n<div id=\"fs-id1165137418239\">\r\n<h3>Example 5:\u00a0 Finding the Value of a Common Logarithm Mentally<\/h3>\r\n<p id=\"fs-id1165137658546\">Evaluate [latex]y=\\mathrm{log}\\left(1000\\right)[\/latex] without using a calculator.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137634154\">[reveal-answer q=\"fs-id1165137634154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137634154\"]\r\n<p id=\"fs-id1165137444192\">First we rewrite the logarithm in exponential form: [latex]{10}^{y}=1000.[\/latex] Next, we ask, \u201cTo what exponent must [latex]10[\/latex] be raised in order to get 1000?\u201d We know [latex]{10}^{3}=1000.[\/latex] Therefore, [latex]\\mathrm{log}\\left(1000\\right)=3.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135503827\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_04_03_05\">\r\n<div id=\"fs-id1165137673696\">\r\n<p id=\"fs-id1165137393877\">Evaluate [latex]y=\\mathrm{log}\\left(1,000,000\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137768485\">[reveal-answer q=\"fs-id1165137768485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137768485\"]\r\n<p id=\"fs-id1165137436094\">[latex]\\mathrm{log}\\left(1,000,000\\right)=6[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137552804\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137827812\"><strong>Given a common logarithm with the form [latex]y=\\mathrm{log}\\left(x\\right),[\/latex] evaluate it using a calculator.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137418685\" type=\"1\">\r\n \t<li>Press <strong>[LOG]<\/strong>.<\/li>\r\n \t<li>Enter the value given for [latex]x,[\/latex] 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_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137793928\">\r\n<div id=\"fs-id1165137892249\">\r\n<h3>Example 6:\u00a0 Finding the Value of a Common Logarithm Using a Calculator<\/h3>\r\n<p id=\"fs-id1165137667877\">Evaluate [latex]y=\\mathrm{log}\\left(321\\right)[\/latex] to four decimal places using a calculator.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137404714\">[reveal-answer q=\"fs-id1165137404714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137404714\"]\r\n<ul id=\"fs-id1165137786486\">\r\n \t<li>Press <strong>[LOG]<\/strong>.<\/li>\r\n \t<li>Enter 321<em>,<\/em> followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137735413\">Rounding to four decimal places, [latex]\\mathrm{log}\\left(321\\right)\\approx 2.5065.[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1165137789015\">Note that [latex]{10}^{2}=100[\/latex] and that [latex]{10}^{3}=1000.[\/latex] Since 321 is between 100 and 1000, we know that [latex]\\mathrm{log}\\left(321\\right)[\/latex] must be between [latex]\\mathrm{log}\\left(100\\right)[\/latex] and [latex]\\mathrm{log}\\left(1000\\right).[\/latex] This gives us the following:<\/p>\r\n\r\n<div id=\"eip-id1165134280435\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ccccc}100&amp; &lt;&amp; 321&amp; &lt;&amp; 1000\\\\ 2&amp; &lt;&amp; 2.5065&amp; &lt;&amp; 3\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<div id=\"eip-id1165134280435\" class=\"unnumbered\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137780842\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"ti_04_03_06\">\r\n<div id=\"fs-id1165135241210\">\r\n<p id=\"fs-id1165137735373\">Evaluate [latex]y=\\mathrm{log}\\left(123\\right)[\/latex] to four decimal places using a calculator.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137550190\">[reveal-answer q=\"fs-id1165137550190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137550190\"]\r\n<p id=\"fs-id1165137844052\">[latex]\\mathrm{log}\\left(123\\right)\\approx 2.0899[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_03_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137603561\">\r\n<div id=\"fs-id1165135704023\">\r\n<h3>Example 7:\u00a0 Rewriting and Solving a Real-World Exponential Model<\/h3>\r\n<p id=\"fs-id1165135194300\">The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation [latex]{10}^{x}=500[\/latex] represents this situation, where [latex]x[\/latex] is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137784516\">[reveal-answer q=\"fs-id1165137784516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137784516\"]\r\n<p id=\"fs-id1165137827621\">We begin by rewriting the exponential equation in logarithmic form.<\/p>\r\n\r\n<div id=\"eip-id1165134048114\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}{10}^{x}\\hfill &amp; =500\\hfill &amp;&amp; \\hfill \\\\ \\mathrm{log}\\left(500\\right)\\hfill &amp; =x\\hfill &amp; \\textrm{Use the definition of the common logarithm}\\text{.}\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165137419444\">Next we evaluate the logarithm using a calculator:<\/p>\r\n\r\n<ul id=\"fs-id1165137736356\">\r\n \t<li>Press <strong>[LOG]<\/strong>.<\/li>\r\n \t<li>Enter [latex]500,[\/latex] followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n \t<li>To the nearest thousandth, [latex]\\mathrm{log}\\left(500\\right)\\approx 2.699.[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137422793\">The difference in magnitudes was about [latex]2.699.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137749635\" class=\"precalculus tryit\">\r\n<h3>Try it #7<\/h3>\r\n<div id=\"ti_04_03_07\">\r\n<div id=\"fs-id1165135195254\">\r\n<p id=\"fs-id1165137736970\">The amount of energy released from one earthquake was [latex]\\text{8,500}[\/latex] times greater than the amount of energy released from another. The equation [latex]{10}^{x}=8500[\/latex] represents this situation, where [latex]x[\/latex] is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137656499\">[reveal-answer q=\"fs-id1165137656499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137656499\"]\r\n<p id=\"fs-id1165137438675\">The difference in magnitudes was about [latex]3.929.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137405741\" class=\"bc-section section\">\r\n<h3>Using Natural Logarithms<\/h3>\r\n<p id=\"fs-id1165137661970\">The most frequently used base for logarithms is [latex]e.[\/latex] Base [latex]e[\/latex] logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base [latex]e[\/latex] logarithm, [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}\\left(1\\right)=0.[\/latex] For other natural logarithms, we can use the <strong>[LN]<\/strong> key that can be found on most scientific calculators. We can also find the natural logarithm of any power of [latex]e[\/latex] using the inverse property of logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137452317\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base [latex]e.[\/latex] 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 [latex]x[\/latex] satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165135613642\">For [latex]x&gt;0,[\/latex] [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] is equivalent to [latex]{e}^{y}=x.[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]e[\/latex] of [latex]x[\/latex]\u201d or \u201cthe natural logarithm of [latex]x.[\/latex]\u201d<\/li>\r\n \t<li id=\"fs-id1165137566720\">The logarithm [latex]y[\/latex] is the exponent to which [latex]e[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137409558\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137832169\"><strong>Given a natural logarithm with the form [latex]y=\\mathrm{ln}\\left(x\\right),[\/latex] evaluate it using a calculator.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135407195\" type=\"1\">\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter the value given for [latex]x,[\/latex] 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=\"textbox examples\">\r\n<div id=\"fs-id1165137731536\">\r\n<div id=\"fs-id1165137434974\">\r\n<h3>Example 8:\u00a0 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\r\n<\/div>\r\n<div id=\"fs-id1165137702133\">[reveal-answer q=\"fs-id1165137702133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137702133\"]\r\n<ul id=\"fs-id1165137563770\">\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter [latex]500,[\/latex] 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 id=\"fs-id1165137676028\" class=\"precalculus tryit\">\r\n<h3>Try it #8<\/h3>\r\n<div id=\"ti_04_03_08\">\r\n<div id=\"fs-id1165137431140\">\r\n<p id=\"fs-id1165137435623\">Evaluate:\u00a0 a. [latex]\\mathrm{ln}\\left(-500\\right).[\/latex]\u00a0 b.\u00a0[latex]\\mathrm{ln}\\left(8\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137737001\">[reveal-answer q=\"fs-id1165137737001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137737001\"]\r\n<p id=\"fs-id1165137639598\">a. It is not possible to take the logarithm of a negative number in the set of real numbers.<\/p>\r\nb. Approximately 2.0794.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137648012\" class=\"precalculus media\">\r\n<h3>Properties of Logarithms<\/h3>\r\nIn applications, equation solving and advance mathematics, it is often easier to work with simplified logarithmic expressions.\u00a0 We will next study the properties of logarithms which allow logarithmic expressions to be written in multiple ways so that they can be interpreted from multiple view points.\r\n\r\n<span style=\"font-size: 1rem;text-align: initial\">The logarithmic and exponential functions with the same base \u201cundo\u201d each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/span>\r\n<div id=\"fs-id1165137737533\" class=\"bc-section section\">\r\n<div id=\"eip-id1165135349439\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\ {\\mathrm{log}}_{b}b=1\\\\\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137628765\" style=\"text-align: left\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1,[\/latex] and [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5.[\/latex]<\/p>\r\n<p id=\"fs-id1165137772010\">Next, we have the <em><strong>inverse property.<\/strong><\/em><\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align*}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&amp;=x\\hfill \\\\{b}^{{\\mathrm{log}}_{b}x}&amp;=x,\\text{ }x&gt;0\\hfill \\end{align*}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, we know that the composition of the functions produce the identity over the appropriate domain.\u00a0 Therefore, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all [latex]x[\/latex] and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for [latex]x&gt;0.[\/latex] Similarly,\u00a0[latex]\\mathrm{log}\\left({10}^{x}\\right)=x[\/latex] for all [latex]x[\/latex] and [latex]10{}^{\\mathrm{log}\\left(x\\right)}=x[\/latex] for [latex]x&gt;0.[\/latex]\u00a0For example, to evaluate [latex]\\mathrm{log}\\left(100\\right),[\/latex] we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right),[\/latex] and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2.[\/latex] To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)},[\/latex] we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7},[\/latex] and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137737533\" class=\"bc-section section\">\r\n<h4>Product Rule of Logarithms<\/h4>\r\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}.[\/latex] We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\r\n<p id=\"fs-id1165137645446\">Given any real number [latex]x[\/latex] and positive real numbers [latex]M,\\text{ }N,[\/latex] and [latex]b,[\/latex] where [latex]b\\ne 1,[\/latex] we will show<\/p>\r\n\r\n<div id=\"eip-214\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right).[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}\\left(M\\right)[\/latex] and [latex]n={\\mathrm{log}}_{b}\\left(N\\right).[\/latex] In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N.[\/latex] It follows that<\/p>\r\n\r\n<div id=\"eip-54\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill &amp; \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right)\\hfill &amp; \\text{Apply the product rule for exponents}.\\hfill \\\\ \\hfill &amp; =m+n\\hfill &amp; \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\hfill &amp; \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right).[\/latex] Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\r\n\r\n<div id=\"eip-502\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}\\left(w\\right)+{\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(y\\right)+{\\mathrm{log}}_{b}\\left(z\\right)[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165137891324\">\r\n<div id=\"fs-id1165134191558\" style=\"text-align: center\"><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137541378\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165134223340\"><strong>Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137748303\" type=\"1\">\r\n \t<li>Factor the argument completely, expressing each whole number factor as a product of prime numbers.<\/li>\r\n \t<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_01\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135458651\">\r\n<div id=\"fs-id1165135458654\">\r\n<h3>Example 9:\u00a0 Using the Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137676248\">[reveal-answer q=\"fs-id1165137676248\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137676248\"]\r\n<div>\r\n<div><\/div>\r\n<div>The input expression is the product of the three factors [latex]30[\/latex], [latex]x[\/latex] and [latex]3x+4.[\/latex]\u00a0 Using the product rule of logarithms we write the equivalent equation by summing the logarithms of each factor.<\/div>\r\n<div><\/div>\r\n<div id=\"eip-id1165137836500\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135525909\" class=\"precalculus tryit\">\r\n<h3>Try it #9<\/h3>\r\n<div id=\"ti_04_05_01\">\r\n<div id=\"fs-id1165137871799\">\r\n\r\nExpand [latex]{\\mathrm{log}}_{b}\\left(10k\\right).[\/latex]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137849381\">[reveal-answer q=\"fs-id1165137849381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137849381\"]\r\n<p id=\"fs-id1165137849383\">[latex]{\\mathrm{log}}_{b}\\left(10\\right)+{\\mathrm{log}}_{b}\\left(k\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135151295\" class=\"bc-section section\">\r\n<h4>Quotient Rule for Logarithms<\/h4>\r\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]\\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}.[\/latex] The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\r\n<p id=\"fs-id1165137431410\">Given any real number [latex]x[\/latex] and positive real numbers [latex]M,[\/latex] [latex]N,[\/latex] and\u00a0 [latex]b,[\/latex] where [latex]b\\ne 1,[\/latex] we will show<\/p>\r\n\r\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right).[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}\\left(M\\right)[\/latex] and [latex]n={\\mathrm{log}}_{b}\\left(N\\right).[\/latex] In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N.[\/latex] It follows that<\/p>\r\n\r\n<div id=\"eip-303\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill &amp; \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill &amp; \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill &amp; =m-n\\hfill &amp; \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill &amp; \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right),[\/latex] we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\r\n\r\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)&amp;=\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill &amp;&amp; \\textrm{Factor}.\\hfill \\\\ \\text{ }&amp;=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill &amp;&amp; \\textrm{Cancel the common factors}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>When the common factors are canceled, keep in mind that [latex]x=-3[\/latex] is not in the domain of the original function and should be noted in the work as the expression is simplified.<\/div>\r\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule to the first term.<\/p>\r\n\r\n<div id=\"eip-46\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&amp;=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{ }&amp;=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill\\\\ \\end{align*}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Finally we notice that\u00a0[latex]x=-3[\/latex] is not in the domain of the simplified expression either.<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165137405741\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137648012\" class=\"precalculus media\">\r\n<div style=\"text-align: center\"><\/div>\r\n<div id=\"fs-id1165135151295\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137749807\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137749813\"><strong>Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137749817\" type=\"1\">\r\n \t<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\r\n \t<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\r\n \t<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135185904\">\r\n<div id=\"fs-id1165135185906\">\r\n<h3>Example 10:\u00a0 Using the Quotient Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x-1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137534099\">[reveal-answer q=\"fs-id1165137534099\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137534099\"]\r\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\r\n\r\n<div id=\"eip-id1165135241002\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x-1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x-1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.<\/p>\r\n\r\n<div id=\"eip-id1165137930288\" class=\"unnumbered\" style=\"text-align: left\">[latex]{\\mathrm{log}}_{2}\\left(15x\\left(x-1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\text{ }&amp;=\\left[{\\mathrm{log}}_{2}\\left(15\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x-1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right]\\hfill\\\\\\text{ }&amp;={\\mathrm{log}}_{2}\\left(15\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x-1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right)\\hfill\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<h3>Analysis<\/h3>\r\n<div class=\"unnumbered\">There are exceptions to consider in this and many other examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and [latex]x=2.[\/latex] Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that [latex]x&gt;0,[\/latex] [latex]x&gt;1,[\/latex] [latex]x&gt;-\\frac{4}{3},[\/latex] and [latex]x&lt;2.[\/latex] Combining these conditions is beyond the scope of this section, and we will not consider them here.<\/div>\r\n<div><\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137534455\" class=\"precalculus tryit\">\r\n<h3>Try it #10<\/h3>\r\n<div id=\"ti_04_05_02\">\r\n<div id=\"fs-id1165137534464\">\r\n<p id=\"fs-id1165137534466\">Expand [latex]\\mathrm{log}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x-1\\right)\\left(x-2\\right)}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137939600\">[reveal-answer q=\"fs-id1165137939600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137939600\"]\r\n<p id=\"fs-id1165137939602\">[latex]\\mathrm{log}\\left(x+3\\right)-\\mathrm{log}\\left(x-1\\right)-\\mathrm{log}\\left(x-2\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137627625\" class=\"bc-section section\">\r\n<h4>Power Rule for Logarithms<\/h4>\r\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}?[\/latex]\u00a0 One method is as follows:<\/p>\r\n\r\n<div id=\"eip-271\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ll}{\\mathrm{log}}_{b}\\left({x}^{2}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(x\\right)\\hfill \\\\ \\hfill &amp; =2{\\mathrm{log}}_{b}\\left(x\\right)\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137534037\">Notice that we used the <span class=\"no-emphasis\">product rule for logarithms<\/span> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\r\n\r\n<div id=\"eip-702\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}100={10}^{2}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill &amp; \\sqrt{3}={3}^{\\frac{1}{2}}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill &amp; \\frac{1}{e}={e}^{-1}\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div id=\"fs-id1165137676322\">\r\n<p id=\"fs-id1165137676330\">The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\r\n\r\n<div style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}\\left(M\\right)[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165137627625\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137639704\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137639709\"><strong>Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137761651\" type=\"1\">\r\n \t<li>Express the argument as a power, if needed.<\/li>\r\n \t<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135593557\">\r\n<div id=\"fs-id1165135593559\">\r\n<h3>example 11:\u00a0 Expanding a Logarithm with Powers<\/h3>\r\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137767301\">[reveal-answer q=\"fs-id1165137767301\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137767301\"]\r\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, [latex]x,[\/latex] and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n\r\n<div id=\"eip-id1165132032525\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}\\left(x\\right)[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135508373\" class=\"precalculus tryit\">\r\n<h3>Try it #11<\/h3>\r\n<div id=\"ti_04_05_03\">\r\n<div id=\"fs-id1165135508382\">\r\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}\\left({x}^{2}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135180118\">[reveal-answer q=\"fs-id1165135180118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135180118\"]\r\n<p id=\"fs-id1165135180121\">[latex]2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_05_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134163985\">\r\n<div id=\"fs-id1165134163988\">\r\n<h3>example 12:\u00a0 Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137827656\">[reveal-answer q=\"fs-id1165137827656\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137827656\"]\r\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right).[\/latex]<\/p>\r\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n\r\n<div id=\"eip-id1165133276220\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137643493\" class=\"precalculus tryit\">\r\n<h3>Try it #12<\/h3>\r\n<div id=\"ti_04_05_04\">\r\n<div id=\"fs-id1165137643502\">\r\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137705119\">[reveal-answer q=\"fs-id1165137705119\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137705119\"]\r\n<p id=\"fs-id1165137705121\">[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137451079\">Access this online resource for additional instruction and practice with logarithms.<\/p>\r\n\r\n<ul id=\"fs-id1165137732886\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/intrologarithms\">Introduction to Logarithms<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137870892\" class=\"key-equations\">\r\n<h3>Key Equations<\/h3>\r\n<table id=\"fs-id1983134\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 126.5px\">Definition of the logarithmic function<\/td>\r\n<td class=\"border\" style=\"width: 584.5px\">For [latex] x&gt;0,b&gt;0,b\\ne 1,[\/latex]\r\n<div><\/div>\r\n[latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]{b}^{y}=x.[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 126.5px\">Definition of the common logarithm<\/td>\r\n<td class=\"border\" style=\"width: 584.5px\">For [latex]x&gt;0,[\/latex] [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]{10}^{y}=x.[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 126.5px\">Definition of the natural logarithm<\/td>\r\n<td class=\"border\" style=\"width: 584.5px\">For [latex]x&gt;0,[\/latex] [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]{e}^{y}=x.[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 126.5px\">Properties of\u00a0 logarithms<\/td>\r\n<td class=\"border\" style=\"width: 584.5px\">[latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex]\r\n\r\n[latex]{b}^{{\\mathrm{log}}_{b}x}=x,\\text{ }x&gt;0[\/latex]\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex] for [latex]b&gt;0[\/latex]\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}\\left(M\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165135699130\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165137574258\">\r\n \t<li>The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.<\/li>\r\n \t<li>Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm.<\/li>\r\n \t<li>Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm.<\/li>\r\n \t<li>Logarithmic functions with base [latex]b[\/latex] can be evaluated mentally using previous knowledge of powers of [latex]b.[\/latex]<\/li>\r\n \t<li>Common logarithms can be evaluated mentally using previous knowledge of powers of [latex]10.[\/latex]<\/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 [latex]10[\/latex] can 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<strong>.<\/strong><\/li>\r\n \t<li>Properties of logarithms can be used to simplify expressions and expand them into sums and differences.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165135160066\">\r\n \t<dt>common logarithm<\/dt>\r\n \t<dd id=\"fs-id1165137571387\">the exponent to which 10 must be raised to get [latex]x;[\/latex] [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\">\r\n \t<dt>logarithm<\/dt>\r\n \t<dd id=\"fs-id1165137849198\">the exponent to which [latex]b[\/latex] must be raised to get [latex]x;[\/latex] written [latex]y={\\mathrm{log}}_{b}\\left(x\\right).[\/latex]<\/dd>\r\n<\/dl>\r\n<dl>\r\n \t<dt>natural logarithm<\/dt>\r\n \t<dd>the exponent to which the number [latex]e[\/latex] must be raised to get [latex]x;[\/latex] [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right).[\/latex]<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Convert from logarithmic to exponential form.<\/li>\n<li>Convert from exponential to logarithmic form.<\/li>\n<li>Evaluate logarithms.<\/li>\n<li>Use common logarithms.<\/li>\n<li>Use natural logarithms.<\/li>\n<li>Use logarithmic properties.<\/li>\n<\/ul>\n<\/div>\n<div id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210524\/CNX_Precalc_Figure_04_03_001.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1\u00a0 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)<\/p>\n<\/div>\n<\/div>\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-303-1\" href=\"#footnote-303-1\" 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-303-2\" href=\"#footnote-303-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>\u00a0like those shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_001\">Figure 1<\/a>. 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\u00a0<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013.\" id=\"return-footnote-303-3\" href=\"#footnote-303-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0<sup id=\"footnote-ref3\"><\/sup>whereas 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-303-4\" href=\"#footnote-303-4\" aria-label=\"Footnote 4\"><sup class=\"footnote\">[4]<\/sup><\/a><sup id=\"footnote-ref4\"><\/sup><\/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 section, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\n<div id=\"fs-id1165137644550\" class=\"bc-section section\">\n<h3>Converting from Logarithmic to Exponential Form<\/h3>\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 [latex]x[\/latex] represents the difference in magnitudes on the <span class=\"no-emphasis\">Richter Scale<\/span>. How would we solve for [latex]x?[\/latex]<\/p>\n<p id=\"fs-id1165135160312\">We have only learned a graphical method for approximating solutions of 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 [latex]x[\/latex] must be some value between 2 and 3, since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph, as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_002\">Figure 2<\/a><strong>,<\/strong> to better estimate the solution.<\/p>\n<div id=\"CNX_Precalc_Figure_04_03_002\" class=\"small\">\n<div style=\"width: 267px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210528\/CNX_Precalc_Figure_04_03_002.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"257\" height=\"252\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2<\/p>\n<\/div>\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 in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_03_002\">Figure 2<\/a>\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <span class=\"no-emphasis\">one-to-one<\/span>, so its inverse is also a function. We use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] to describe the inverse.\u00a0 The base [latex]b[\/latex]\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise [latex]b[\/latex] to get that number.<\/p>\n<p id=\"fs-id1165137404844\">We read a logarithmic expression as, \u201cThe logarithm with base [latex]b[\/latex] of [latex]x[\/latex] is equal to [latex]y,[\/latex]\u201d or, simplified, \u201clog base [latex]b[\/latex] of [latex]x[\/latex] is [latex]y.[\/latex]\u201d We can also say, \u201c[latex]b[\/latex] raised to the power of [latex]y[\/latex] is [latex]x,[\/latex]\u201d 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 \u201clog base 2 of 32 is 5.\u201d<\/p>\n<p id=\"fs-id1165137597501\">We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\u21d4{b}^{y}=x,\\text{ }b>0,b\\ne 1[\/latex][latex]\\\\[\/latex]<\/p>\n<p id=\"fs-id1165137678993\">Note that the base [latex]b[\/latex] is always positive.<\/p>\n<p><span id=\"fs-id1165137696233\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210530\/CNX_Precalc_Figure_04_03_004.jpg\" alt=\"\" \/><\/span><\/p>\n<p>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 [latex]x.[\/latex]<\/p>\n<p id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<\/p>\n<p><span id=\"fs-id1165137771679\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210533\/CNX_Precalc_Figure_04_03_003.jpg\" alt=\"\" \/><\/span><\/p>\n<p id=\"fs-id1165137575165\">Notice that, comparing the logarithm function and the exponential function, the input and the output are switched.<\/p>\n<div id=\"fs-id1165137472937\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base [latex]b[\/latex] of a positive number [latex]x[\/latex] satisfies the following definition.<\/p>\n<p id=\"fs-id1165137584967\">For [latex]x>0,\\text{ }b>0,\\text{ }b\\ne 1,[\/latex]\u00a0 [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is equivalent to [latex]{b}^{y}=x.[\/latex]<\/p>\n<ul id=\"fs-id1165135530561\">\n<li>We read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]b[\/latex] of [latex]x[\/latex]\u201d or the \u201clog base [latex]b[\/latex] of [latex]x.\"[\/latex]<\/li>\n<li>The logarithm [latex]y[\/latex] is the exponent to which [latex]b[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137547773\">Also, since the logarithmic and exponential functions are inverses of each other, 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[\/latex] is [latex]\\left(0,\\infty \\right),[\/latex] and<\/li>\n<li>the range of the logarithm function with base [latex]b[\/latex] is [latex]\\left(-\\infty ,\\infty \\right).[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165137677696\" class=\"precalculus qa key-takeaways\">\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. We are working with functions of real numbers. Because the base of an exponential function is always a positive real number, no power of that base can ever be a negative real number. We can never take the logarithm of a negative real 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=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137806301\"><strong>Given an equation in logarithmic form\u00a0<\/strong>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y,[\/latex] <strong>convert it to exponential form.<\/strong><\/p>\n<ol id=\"fs-id1165137641669\" type=\"1\">\n<li>Examine the equation [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and identify [latex]b,\\text{ }y,[\/latex] and [latex]x.[\/latex]<\/li>\n<li>Rewrite [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] as [latex]{b}^{y}=x.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_01\" class=\"textbox examples\">\n<div id=\"fs-id1165135570363\">\n<div id=\"fs-id1165137557855\">\n<h3>Example 1:\u00a0 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\" type=\"a\">\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165135613330\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135613330\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135613330\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137408172\">First, identify the values of [latex]b,\\text{ }y,[\/latex] and [latex]x[\/latex] in [latex]y= {\\mathrm{log}}_{b}\\left(x\\right).[\/latex] Then, write the equation in the form [latex]{b}^{y}=x.[\/latex]<\/p>\n<ol id=\"fs-id1165137705659\" type=\"a\">\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex]\n<p id=\"fs-id1165137602796\">Here, [latex]b=6,\\text{ }y=\\frac{1}{2},[\/latex] and [latex]x=\\sqrt[\\leftroot{1}\\uproot{2} ]{6}.[\/latex] Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt[\\leftroot{1}\\uproot{2} ]{6}\\right)=\\frac{1}{2}[\/latex] is equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt[\\leftroot{1}\\uproot{2} ]{6}.[\/latex]<\/p>\n<\/li>\n<li>[latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex]\n<p id=\"fs-id1165137698078\">Here, [latex]b=10,\\text{ }y=2,[\/latex] and [latex]x=100.[\/latex] Therefore, the equation [latex]{\\mathrm{log}}_{10}\\left(100\\right)=2[\/latex] is equivalent to [latex]{10}^{2}=100.[\/latex]<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137640140\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_04_03_01\">\n<div id=\"fs-id1165135208926\">\n<p id=\"fs-id1165137418681\">Write the following logarithmic equations in exponential form.<\/p>\n<ol id=\"fs-id1165137772342\" type=\"a\">\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165135195688\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135195688\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135195688\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165137414337\" type=\"a\">\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equivalent to [latex]{10}^{6}=1,000,000.[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equivalent to [latex]{5}^{2}=25.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137585244\" class=\"bc-section section\">\n<h3>Converting from Exponential to Logarithmic Form<\/h3>\n<p id=\"fs-id1165137933968\">To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base [latex]b,[\/latex] exponent [latex]x,[\/latex] and output [latex]y[\/latex] in [latex]b^x=y.[\/latex] Then we write\u00a0 [latex]x={\\mathrm{log}}_{b}\\left(y\\right).[\/latex]<\/p>\n<div id=\"Example_04_03_02\" class=\"textbox examples\">\n<div id=\"fs-id1165135168111\">\n<div id=\"fs-id1165137727912\">\n<h3>Example 2:\u00a0 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\" type=\"a\">\n<li>[latex]{2}^{3}=8[\/latex]<\/li>\n<li>[latex]{10}^{4}=10000[\/latex]<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137702205\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137702205\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137702205\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137474116\">First, identify the values of [latex]b,\\text{ }y,[\/latex] and [latex]x[\/latex] in [latex]b^x=y.[\/latex] Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right).[\/latex]<\/p>\n<ol id=\"fs-id1165137573458\" type=\"a\">\n<li>[latex]{2}^{3}=8[\/latex]\n<p id=\"fs-id1165137466396\">Here, [latex]b=2,[\/latex] [latex]x=3,[\/latex] and [latex]y=8.[\/latex] Therefore, the equation [latex]{2}^{3}=8[\/latex] is equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3.[\/latex][latex]\\\\[\/latex]<\/p>\n<\/li>\n<li>[latex]{10}^{4}=10000[\/latex]\n<p id=\"fs-id1165135193035\">Here, [latex]b=10,[\/latex] [latex]x=4,[\/latex] and [latex]y=10000.[\/latex] Therefore, the equation [latex]{10}^{4}=10000[\/latex] is equivalent to [latex]{\\mathrm{log}}_{10}\\left(10000\\right)=4.[\/latex][latex]\\\\[\/latex]<\/p>\n<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\n<p id=\"fs-id1165135187822\">Here, [latex]b=10,[\/latex] [latex]x=-4,[\/latex] 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 id=\"fs-id1165137438165\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_04_03_02\">\n<div id=\"fs-id1165135190969\">\n<p id=\"fs-id1165137566762\">Write the following exponential equations in logarithmic form.<\/p>\n<ol id=\"fs-id1165137771963\" type=\"a\">\n<li>[latex]{1.1}^{2}=1.21[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex]<\/li>\n<li>[latex]{10}^{-1}=\\frac{1}{10}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165134065138\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134065138\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134065138\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165137469846\" type=\"a\">\n<li>[latex]{1.1}^{2}=1.21[\/latex] is equivalent to [latex]{\\mathrm{log}}_{1.1}\\left(1.21\\right)=2.[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3.[\/latex]<\/li>\n<li>[latex]{10}^{-1}=\\frac{1}{10}[\/latex] is equivalent to [latex]{\\mathrm{log}}_{10}\\left(\\frac{1}{10}\\right)=-1.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137530906\" class=\"bc-section section\">\n<h3>Evaluating Logarithms<\/h3>\n<p>Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}\\left(8\\right).[\/latex] We ask, \u201cTo what exponent must [latex]2[\/latex] be raised in order to get 8?\u201d Because we already know [latex]{2}^{3}=8,[\/latex] it follows that [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3.[\/latex]<\/p>\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}\\left(49\\right)[\/latex] and [latex]{\\mathrm{log}}_{3}\\left(27\\right)[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137937690\">\n<li>We ask, \u201cTo what exponent must 7 be raised in order to get 49?\u201d We know [latex]{7}^{2}=49.[\/latex] Therefore, [latex]{\\mathrm{log}}_{7}\\left(49\\right)=2.[\/latex]<\/li>\n<li>We ask, \u201cTo what exponent must 3 be raised in order to get 27?\u201d We know [latex]{3}^{3}=27.[\/latex]Therefore, [latex]{\\mathrm{log}}_{3}\\left(27\\right)=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}}\\left(\\frac{4}{9}\\right)[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137584208\">\n<li>We ask, \u201cTo what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\left(\\frac{4}{9}\\right)?[\/latex]\u201d\u00a0 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<p>[latex]\\\\[\/latex]<\/p>\n<div id=\"fs-id1165137455840\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137453770\"><strong>Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right),[\/latex] evaluate it mentally.<\/strong><\/p>\n<ol id=\"fs-id1165134079724\" type=\"1\">\n<li>Rewrite the argument [latex]x[\/latex] as a power of [latex]b:[\/latex]\u00a0 [latex]{b}^{y}=x.[\/latex]<\/li>\n<li>Use previous knowledge of powers of [latex]b[\/latex] identify [latex]y[\/latex] by asking, \u201cTo what exponent should [latex]b[\/latex] be raised in order to get [latex]x?[\/latex]\u201d<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_03\" class=\"textbox examples\">\n<div id=\"fs-id1165137732842\">\n<div id=\"fs-id1165135296345\">\n<h3>Example 3:\u00a0 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>\n<div id=\"fs-id1165137852123\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137852123\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137852123\" 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, \u201cTo what exponent must 4 be raised in order to get 64?\u201d<\/p>\n<p>We know [latex]{4}^{3}=64.[\/latex]\u00a0 Therefore, [latex]\\mathrm{log}{}_{4}\\left(64\\right)=3.[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137731430\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_04_03_03\">\n<div id=\"fs-id1165137704553\">\n<p id=\"fs-id1165137745041\">Solve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165137693554\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137693554\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137693554\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137639199\">[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recalling that [latex]\\sqrt[\\leftroot{1}\\uproot{2} ]{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex])<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_03_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137663658\">\n<div id=\"fs-id1165137680390\">\n<h3>Example 4:\u00a0 Evaluating the Logarithm of a Reciprocal<\/h3>\n<p id=\"fs-id1165137938805\">Evaluate [latex]y={\\mathrm{log}}_{10}\\left(\\frac{1}{100}\\right)[\/latex] without using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165135526087\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135526087\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135526087\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form:\u00a0 [latex]{10}^{y}=\\frac{1}{100}.[\/latex]\u00a0 Next, we ask, \u201cTo what exponent must 10 be raised in order to get [latex]\\frac{1}{100}?[\/latex]\u201d<\/p>\n<p id=\"fs-id1165137552085\">We know [latex]{10}^{2}=100,[\/latex] but what must we do to get the reciprocal, [latex]\\frac{1}{100}?[\/latex] Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}.[\/latex] We use this information to write<\/p>\n<div id=\"eip-id1165137550550\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}{10}^{-2}&=\\frac{1}{{10}^{2}}\\hfill \\\\\\text{ }&=\\frac{1}{100}\\hfill \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{10}\\left(\\frac{1}{100}\\right)=-2.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137575754\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_04_03_04\">\n<div id=\"fs-id1165137768727\">\n<p id=\"fs-id1165135437134\">Evaluate [latex]y={\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)[\/latex] without using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165137812401\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137812401\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137812401\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137806792\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137547253\" class=\"bc-section section\">\n<h3>Using Common Logarithms<\/h3>\n<p id=\"fs-id1165137574205\">Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression [latex]\\mathrm{log}\\left(x\\right)[\/latex] means [latex]{\\mathrm{log}}_{10}\\left(x\\right).[\/latex] We call a base-10 logarithm a <strong>common logarithm<\/strong>. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.<\/p>\n<div id=\"fs-id1165137401037\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165135609332\">A <strong>common logarithm<\/strong> is a logarithm with base [latex]10.[\/latex] We write [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{log}\\left(x\\right).[\/latex] The common logarithm of a positive number [latex]x[\/latex] satisfies the following definition.<\/p>\n<p id=\"fs-id1165137601579\">For [latex]x>0,[\/latex] [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] is equivalent to [latex]{10}^{y}=x.[\/latex]<\/p>\n<ul>\n<li id=\"fs-id1165137559681\">We read [latex]\\mathrm{log}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]10[\/latex] of [latex]x[\/latex]\u201d or \u201clog base 10 of [latex]x.[\/latex]\u201d<\/li>\n<li>The logarithm [latex]y[\/latex] is the exponent to which [latex]10[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137579434\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137810781\"><strong>Given a common logarithm of the form [latex]y=\\mathrm{log}\\left(x\\right),[\/latex] evaluate it mentally.<\/strong><\/p>\n<ol id=\"fs-id1165137828334\" type=\"1\">\n<li>Rewrite the argument [latex]x[\/latex] as a power of [latex]10:[\/latex]\u00a0 [latex]{10}^{y}=x.[\/latex]<\/li>\n<li>Use previous knowledge of powers of [latex]10[\/latex] to identify [latex]y[\/latex] by asking, \u201cTo what exponent must [latex]10[\/latex] be raised in order to get [latex]x?[\/latex]\u201d<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_05\" class=\"textbox examples\">\n<div id=\"fs-id1165137742366\">\n<div id=\"fs-id1165137418239\">\n<h3>Example 5:\u00a0 Finding the Value of a Common Logarithm Mentally<\/h3>\n<p id=\"fs-id1165137658546\">Evaluate [latex]y=\\mathrm{log}\\left(1000\\right)[\/latex] without using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165137634154\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137634154\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137634154\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137444192\">First we rewrite the logarithm in exponential form: [latex]{10}^{y}=1000.[\/latex] Next, we ask, \u201cTo what exponent must [latex]10[\/latex] be raised in order to get 1000?\u201d We know [latex]{10}^{3}=1000.[\/latex] Therefore, [latex]\\mathrm{log}\\left(1000\\right)=3.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135503827\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_04_03_05\">\n<div id=\"fs-id1165137673696\">\n<p id=\"fs-id1165137393877\">Evaluate [latex]y=\\mathrm{log}\\left(1,000,000\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137768485\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137768485\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137768485\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137436094\">[latex]\\mathrm{log}\\left(1,000,000\\right)=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137552804\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137827812\"><strong>Given a common logarithm with the form [latex]y=\\mathrm{log}\\left(x\\right),[\/latex] evaluate it using a calculator.<\/strong><\/p>\n<ol id=\"fs-id1165137418685\" type=\"1\">\n<li>Press <strong>[LOG]<\/strong>.<\/li>\n<li>Enter the value given for [latex]x,[\/latex] followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_06\" class=\"textbox examples\">\n<div id=\"fs-id1165137793928\">\n<div id=\"fs-id1165137892249\">\n<h3>Example 6:\u00a0 Finding the Value of a Common Logarithm Using a Calculator<\/h3>\n<p id=\"fs-id1165137667877\">Evaluate [latex]y=\\mathrm{log}\\left(321\\right)[\/latex] to four decimal places using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165137404714\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137404714\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137404714\" class=\"hidden-answer\" style=\"display: none\">\n<ul id=\"fs-id1165137786486\">\n<li>Press <strong>[LOG]<\/strong>.<\/li>\n<li>Enter 321<em>,<\/em> followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ul>\n<p id=\"fs-id1165137735413\">Rounding to four decimal places, [latex]\\mathrm{log}\\left(321\\right)\\approx 2.5065.[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p id=\"fs-id1165137789015\">Note that [latex]{10}^{2}=100[\/latex] and that [latex]{10}^{3}=1000.[\/latex] Since 321 is between 100 and 1000, we know that [latex]\\mathrm{log}\\left(321\\right)[\/latex] must be between [latex]\\mathrm{log}\\left(100\\right)[\/latex] and [latex]\\mathrm{log}\\left(1000\\right).[\/latex] This gives us the following:<\/p>\n<div id=\"eip-id1165134280435\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ccccc}100& <& 321& <& 1000\\\\ 2& <& 2.5065& <& 3\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div id=\"eip-id1165134280435\" class=\"unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137780842\" class=\"precalculus tryit\">\n<h3>Try it #6<\/h3>\n<div id=\"ti_04_03_06\">\n<div id=\"fs-id1165135241210\">\n<p id=\"fs-id1165137735373\">Evaluate [latex]y=\\mathrm{log}\\left(123\\right)[\/latex] to four decimal places using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165137550190\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137550190\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137550190\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137844052\">[latex]\\mathrm{log}\\left(123\\right)\\approx 2.0899[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_03_07\" class=\"textbox examples\">\n<div id=\"fs-id1165137603561\">\n<div id=\"fs-id1165135704023\">\n<h3>Example 7:\u00a0 Rewriting and Solving a Real-World Exponential Model<\/h3>\n<p id=\"fs-id1165135194300\">The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation [latex]{10}^{x}=500[\/latex] represents this situation, where [latex]x[\/latex] is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\n<\/div>\n<div id=\"fs-id1165137784516\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137784516\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137784516\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827621\">We begin by rewriting the exponential equation in logarithmic form.<\/p>\n<div id=\"eip-id1165134048114\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}{10}^{x}\\hfill & =500\\hfill && \\hfill \\\\ \\mathrm{log}\\left(500\\right)\\hfill & =x\\hfill & \\textrm{Use the definition of the common logarithm}\\text{.}\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165137419444\">Next we evaluate the logarithm using a calculator:<\/p>\n<ul id=\"fs-id1165137736356\">\n<li>Press <strong>[LOG]<\/strong>.<\/li>\n<li>Enter [latex]500,[\/latex] followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<li>To the nearest thousandth, [latex]\\mathrm{log}\\left(500\\right)\\approx 2.699.[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137422793\">The difference in magnitudes was about [latex]2.699.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137749635\" class=\"precalculus tryit\">\n<h3>Try it #7<\/h3>\n<div id=\"ti_04_03_07\">\n<div id=\"fs-id1165135195254\">\n<p id=\"fs-id1165137736970\">The amount of energy released from one earthquake was [latex]\\text{8,500}[\/latex] times greater than the amount of energy released from another. The equation [latex]{10}^{x}=8500[\/latex] represents this situation, where [latex]x[\/latex] is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\n<\/div>\n<div id=\"fs-id1165137656499\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137656499\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137656499\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137438675\">The difference in magnitudes was about [latex]3.929.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137405741\" class=\"bc-section section\">\n<h3>Using Natural Logarithms<\/h3>\n<p id=\"fs-id1165137661970\">The most frequently used base for logarithms is [latex]e.[\/latex] Base [latex]e[\/latex] logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base [latex]e[\/latex] logarithm, [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}\\left(1\\right)=0.[\/latex] For other natural logarithms, we can use the <strong>[LN]<\/strong> key that can be found on most scientific calculators. We can also find the natural logarithm of any power of [latex]e[\/latex] using the inverse property of logarithms.<\/p>\n<div id=\"fs-id1165137452317\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base [latex]e.[\/latex] 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 [latex]x[\/latex] satisfies the following definition.<\/p>\n<p id=\"fs-id1165135613642\">For [latex]x>0,[\/latex] [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] is equivalent to [latex]{e}^{y}=x.[\/latex]<\/p>\n<ul>\n<li id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \u201cthe logarithm with base [latex]e[\/latex] of [latex]x[\/latex]\u201d or \u201cthe natural logarithm of [latex]x.[\/latex]\u201d<\/li>\n<li id=\"fs-id1165137566720\">The logarithm [latex]y[\/latex] is the exponent to which [latex]e[\/latex] must be raised to get [latex]x.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137409558\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137832169\"><strong>Given a natural logarithm with the form [latex]y=\\mathrm{ln}\\left(x\\right),[\/latex] evaluate it using a calculator.<\/strong><\/p>\n<ol id=\"fs-id1165135407195\" type=\"1\">\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter the value given for [latex]x,[\/latex] followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_03_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137731536\">\n<div id=\"fs-id1165137434974\">\n<h3>Example 8:\u00a0 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>\n<div id=\"fs-id1165137702133\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137702133\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137702133\" class=\"hidden-answer\" style=\"display: none\">\n<ul id=\"fs-id1165137563770\">\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter [latex]500,[\/latex] 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 id=\"fs-id1165137676028\" class=\"precalculus tryit\">\n<h3>Try it #8<\/h3>\n<div id=\"ti_04_03_08\">\n<div id=\"fs-id1165137431140\">\n<p id=\"fs-id1165137435623\">Evaluate:\u00a0 a. [latex]\\mathrm{ln}\\left(-500\\right).[\/latex]\u00a0 b.\u00a0[latex]\\mathrm{ln}\\left(8\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737001\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137737001\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137737001\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137639598\">a. It is not possible to take the logarithm of a negative number in the set of real numbers.<\/p>\n<p>b. Approximately 2.0794.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137648012\" class=\"precalculus media\">\n<h3>Properties of Logarithms<\/h3>\n<p>In applications, equation solving and advance mathematics, it is often easier to work with simplified logarithmic expressions.\u00a0 We will next study the properties of logarithms which allow logarithmic expressions to be written in multiple ways so that they can be interpreted from multiple view points.<\/p>\n<p><span style=\"font-size: 1rem;text-align: initial\">The logarithmic and exponential functions with the same base \u201cundo\u201d each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/span><\/p>\n<div id=\"fs-id1165137737533\" class=\"bc-section section\">\n<div id=\"eip-id1165135349439\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\ {\\mathrm{log}}_{b}b=1\\\\\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137628765\" style=\"text-align: left\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1,[\/latex] and [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5.[\/latex]<\/p>\n<p id=\"fs-id1165137772010\">Next, we have the <em><strong>inverse property.<\/strong><\/em><\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&=x\\hfill \\\\{b}^{{\\mathrm{log}}_{b}x}&=x,\\text{ }x>0\\hfill \\end{align*}[\/latex]<\/p>\n<p style=\"text-align: left\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, we know that the composition of the functions produce the identity over the appropriate domain.\u00a0 Therefore, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all [latex]x[\/latex] and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for [latex]x>0.[\/latex] Similarly,\u00a0[latex]\\mathrm{log}\\left({10}^{x}\\right)=x[\/latex] for all [latex]x[\/latex] and [latex]10{}^{\\mathrm{log}\\left(x\\right)}=x[\/latex] for [latex]x>0.[\/latex]\u00a0For example, to evaluate [latex]\\mathrm{log}\\left(100\\right),[\/latex] we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right),[\/latex] and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2.[\/latex] To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)},[\/latex] we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7},[\/latex] and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737533\" class=\"bc-section section\">\n<h4>Product Rule of Logarithms<\/h4>\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}.[\/latex] We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\n<p id=\"fs-id1165137645446\">Given any real number [latex]x[\/latex] and positive real numbers [latex]M,\\text{ }N,[\/latex] and [latex]b,[\/latex] where [latex]b\\ne 1,[\/latex] we will show<\/p>\n<div id=\"eip-214\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right).[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}\\left(M\\right)[\/latex] and [latex]n={\\mathrm{log}}_{b}\\left(N\\right).[\/latex] In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N.[\/latex] It follows that<\/p>\n<div id=\"eip-54\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right)\\hfill & \\text{Apply the product rule for exponents}.\\hfill \\\\ \\hfill & =m+n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right).[\/latex] Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\n<div id=\"eip-502\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}\\left(w\\right)+{\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(y\\right)+{\\mathrm{log}}_{b}\\left(z\\right)[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1165137891324\">\n<div id=\"fs-id1165134191558\" style=\"text-align: center\"><\/div>\n<\/div>\n<div id=\"fs-id1165137541378\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165134223340\"><strong>Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/strong><\/p>\n<ol id=\"fs-id1165137748303\" type=\"1\">\n<li>Factor the argument completely, expressing each whole number factor as a product of prime numbers.<\/li>\n<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_01\" class=\"textbox examples\">\n<div id=\"fs-id1165135458651\">\n<div id=\"fs-id1165135458654\">\n<h3>Example 9:\u00a0 Using the Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137676248\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137676248\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137676248\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<div><\/div>\n<div>The input expression is the product of the three factors [latex]30[\/latex], [latex]x[\/latex] and [latex]3x+4.[\/latex]\u00a0 Using the product rule of logarithms we write the equivalent equation by summing the logarithms of each factor.<\/div>\n<div><\/div>\n<div id=\"eip-id1165137836500\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135525909\" class=\"precalculus tryit\">\n<h3>Try it #9<\/h3>\n<div id=\"ti_04_05_01\">\n<div id=\"fs-id1165137871799\">\n<p>Expand [latex]{\\mathrm{log}}_{b}\\left(10k\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137849381\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137849381\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137849381\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137849383\">[latex]{\\mathrm{log}}_{b}\\left(10\\right)+{\\mathrm{log}}_{b}\\left(k\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135151295\" class=\"bc-section section\">\n<h4>Quotient Rule for Logarithms<\/h4>\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]\\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}.[\/latex] The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\n<p id=\"fs-id1165137431410\">Given any real number [latex]x[\/latex] and positive real numbers [latex]M,[\/latex] [latex]N,[\/latex] and\u00a0 [latex]b,[\/latex] where [latex]b\\ne 1,[\/latex] we will show<\/p>\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right).[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}\\left(M\\right)[\/latex] and [latex]n={\\mathrm{log}}_{b}\\left(N\\right).[\/latex] In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N.[\/latex] It follows that<\/p>\n<div id=\"eip-303\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill & \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill & =m-n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right),[\/latex] we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)&=\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill && \\textrm{Factor}.\\hfill \\\\ \\text{ }&=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill && \\textrm{Cancel the common factors}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>When the common factors are canceled, keep in mind that [latex]x=-3[\/latex] is not in the domain of the original function and should be noted in the work as the expression is simplified.<\/div>\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule to the first term.<\/p>\n<div id=\"eip-46\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{ }&=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill\\\\ \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Finally we notice that\u00a0[latex]x=-3[\/latex] is not in the domain of the simplified expression either.<\/div>\n<div><\/div>\n<div id=\"fs-id1165137405741\" class=\"bc-section section\">\n<div id=\"fs-id1165137648012\" class=\"precalculus media\">\n<div style=\"text-align: center\"><\/div>\n<div id=\"fs-id1165135151295\" class=\"bc-section section\">\n<div id=\"fs-id1165137749807\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137749813\"><strong>Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/strong><\/p>\n<ol id=\"fs-id1165137749817\" type=\"1\">\n<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\n<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\n<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_02\" class=\"textbox examples\">\n<div id=\"fs-id1165135185904\">\n<div id=\"fs-id1165135185906\">\n<h3>Example 10:\u00a0 Using the Quotient Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x-1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137534099\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137534099\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137534099\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\n<div id=\"eip-id1165135241002\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x-1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x-1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.<\/p>\n<div id=\"eip-id1165137930288\" class=\"unnumbered\" style=\"text-align: left\">[latex]{\\mathrm{log}}_{2}\\left(15x\\left(x-1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/div>\n<div><\/div>\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\text{ }&=\\left[{\\mathrm{log}}_{2}\\left(15\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x-1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right]\\hfill\\\\\\text{ }&={\\mathrm{log}}_{2}\\left(15\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x-1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right)\\hfill\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<h3>Analysis<\/h3>\n<div class=\"unnumbered\">There are exceptions to consider in this and many other examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and [latex]x=2.[\/latex] Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that [latex]x>0,[\/latex] [latex]x>1,[\/latex] [latex]x>-\\frac{4}{3},[\/latex] and [latex]x<2.[\/latex] Combining these conditions is beyond the scope of this section, and we will not consider them here.<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137534455\" class=\"precalculus tryit\">\n<h3>Try it #10<\/h3>\n<div id=\"ti_04_05_02\">\n<div id=\"fs-id1165137534464\">\n<p id=\"fs-id1165137534466\">Expand [latex]\\mathrm{log}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x-1\\right)\\left(x-2\\right)}\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137939600\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137939600\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137939600\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137939602\">[latex]\\mathrm{log}\\left(x+3\\right)-\\mathrm{log}\\left(x-1\\right)-\\mathrm{log}\\left(x-2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137627625\" class=\"bc-section section\">\n<h4>Power Rule for Logarithms<\/h4>\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}?[\/latex]\u00a0 One method is as follows:<\/p>\n<div id=\"eip-271\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ll}{\\mathrm{log}}_{b}\\left({x}^{2}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(x\\right)\\hfill \\\\ \\hfill & =2{\\mathrm{log}}_{b}\\left(x\\right)\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137534037\">Notice that we used the <span class=\"no-emphasis\">product rule for logarithms<\/span> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\n<div id=\"eip-702\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{lll}100={10}^{2}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill & \\sqrt{3}={3}^{\\frac{1}{2}}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\hfill & \\frac{1}{e}={e}^{-1}\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\n<div id=\"fs-id1165137676322\">\n<p id=\"fs-id1165137676330\">The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\n<div style=\"text-align: center\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}\\left(M\\right)[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1165137627625\" class=\"bc-section section\">\n<div id=\"fs-id1165137639704\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137639709\"><strong>Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/strong><\/p>\n<ol id=\"fs-id1165137761651\" type=\"1\">\n<li>Express the argument as a power, if needed.<\/li>\n<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_03\" class=\"textbox examples\">\n<div id=\"fs-id1165135593557\">\n<div id=\"fs-id1165135593559\">\n<h3>example 11:\u00a0 Expanding a Logarithm with Powers<\/h3>\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137767301\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137767301\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137767301\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, [latex]x,[\/latex] and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<div id=\"eip-id1165132032525\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135508373\" class=\"precalculus tryit\">\n<h3>Try it #11<\/h3>\n<div id=\"ti_04_05_03\">\n<div id=\"fs-id1165135508382\">\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}\\left({x}^{2}\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135180118\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135180118\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135180118\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135180121\">[latex]2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_04\" class=\"textbox examples\">\n<div id=\"fs-id1165134163985\">\n<div id=\"fs-id1165134163988\">\n<h3>example 12:\u00a0 Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\n<\/div>\n<div id=\"fs-id1165137827656\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137827656\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137827656\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right).[\/latex]<\/p>\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<div id=\"eip-id1165133276220\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137643493\" class=\"precalculus tryit\">\n<h3>Try it #12<\/h3>\n<div id=\"ti_04_05_04\">\n<div id=\"fs-id1165137643502\">\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137705119\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137705119\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137705119\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137705121\">[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137451079\">Access this online resource for additional instruction and practice with logarithms.<\/p>\n<ul id=\"fs-id1165137732886\">\n<li><a href=\"http:\/\/openstax.org\/l\/intrologarithms\">Introduction to Logarithms<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137870892\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id1983134\" summary=\"...\">\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 126.5px\">Definition of the logarithmic function<\/td>\n<td class=\"border\" style=\"width: 584.5px\">For [latex]x>0,b>0,b\\ne 1,[\/latex]<\/p>\n<div><\/div>\n<p>[latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]{b}^{y}=x.[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 126.5px\">Definition of the common logarithm<\/td>\n<td class=\"border\" style=\"width: 584.5px\">For [latex]x>0,[\/latex] [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]{10}^{y}=x.[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 126.5px\">Definition of the natural logarithm<\/td>\n<td class=\"border\" style=\"width: 584.5px\">For [latex]x>0,[\/latex] [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]{e}^{y}=x.[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 126.5px\">Properties of\u00a0 logarithms<\/td>\n<td class=\"border\" style=\"width: 584.5px\">[latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex]<\/p>\n<p>[latex]{b}^{{\\mathrm{log}}_{b}x}=x,\\text{ }x>0[\/latex]<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex] for [latex]b>0[\/latex]<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}\\left(M\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165135699130\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137574258\">\n<li>The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.<\/li>\n<li>Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm.<\/li>\n<li>Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm.<\/li>\n<li>Logarithmic functions with base [latex]b[\/latex] can be evaluated mentally using previous knowledge of powers of [latex]b.[\/latex]<\/li>\n<li>Common logarithms can be evaluated mentally using previous knowledge of powers of [latex]10.[\/latex]<\/li>\n<li>When common logarithms cannot be evaluated mentally, a calculator can be used.<\/li>\n<li>Real-world exponential problems with base [latex]10[\/latex] can be rewritten as a common logarithm and then evaluated using a calculator.<\/li>\n<li>Natural logarithms can be evaluated using a calculator<strong>.<\/strong><\/li>\n<li>Properties of logarithms can be used to simplify expressions and expand them into sums and differences.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135160066\">\n<dt>common logarithm<\/dt>\n<dd id=\"fs-id1165137571387\">the exponent to which 10 must be raised to get [latex]x;[\/latex] [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\">\n<dt>logarithm<\/dt>\n<dd id=\"fs-id1165137849198\">the exponent to which [latex]b[\/latex] must be raised to get [latex]x;[\/latex] written [latex]y={\\mathrm{log}}_{b}\\left(x\\right).[\/latex]<\/dd>\n<\/dl>\n<dl>\n<dt>natural logarithm<\/dt>\n<dd>the exponent to which the number [latex]e[\/latex] must be raised to get [latex]x;[\/latex] [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right).[\/latex]<\/dd>\n<\/dl>\n<\/div>\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-303\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Logarithmic Functions. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:dGtL5139@7\/Logarithmic-Functions\">https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:dGtL5139@7\/Logarithmic-Functions<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 1. <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-303-1\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary. Accessed 3\/4\/2013. <a href=\"#return-footnote-303-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-303-2\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary. Accessed 3\/4\/2013. <a href=\"#return-footnote-303-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-303-3\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013. <a href=\"#return-footnote-303-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><li id=\"footnote-303-4\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details. Accessed 3\/4\/2013. <a href=\"#return-footnote-303-4\" class=\"return-footnote\" aria-label=\"Return to footnote 4\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":311,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Logarithmic Functions\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:dGtL5139@7\/Logarithmic-Functions\",\"project\":\"Essential Precalcus, Part 1\",\"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-303","chapter","type-chapter","status-publish","hentry"],"part":223,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/303","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":25,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/303\/revisions"}],"predecessor-version":[{"id":2690,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/303\/revisions\/2690"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/223"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/303\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=303"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=303"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=303"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=303"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}