{"id":2020,"date":"2016-11-02T23:27:37","date_gmt":"2016-11-02T23:27:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2020"},"modified":"2018-06-25T00:48:32","modified_gmt":"2018-06-25T00:48:32","slug":"introduction-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/chapter\/introduction-logarithmic-functions\/","title":{"raw":"Logarithmic Functions","rendered":"Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\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 with and without a calculator.<\/li>\r\n \t<li>Evaluate logarithms with base 10 and base e.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"mceTemp\"><\/div>\r\nIn 2010, a major earthquake struck Haiti destroying or damaging over 285,000 homes.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote] One year later, another, stronger earthquake devastated Honshu, Japan destroying or damaging over 332,000 buildings[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0like those shown in the picture below. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0whereas the Japanese earthquake registered a 9.0.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013.[\/footnote]\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\/896\/2016\/11\/02232710\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/> Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)[\/caption]\r\n\r\nThe Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.\r\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\r\nIn 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 was 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex] where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?\r\n\r\nWe have not yet learned a method for solving exponential equations algebraically. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3 since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.\r\n\r\n<img class=\"aligncenter size-full wp-image-3089\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195659\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/>\r\n\r\nEstimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.\r\n\r\nWe read a logarithmic expression as, \"The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,\" or, simplified, \"log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.\" We can also say, \"<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,\" because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as \"log base 2 of 32 is 5.\"\r\n\r\nWe can express the relationship between logarithmic form and its corresponding exponential form as follows:\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]\r\n\r\nNote that the base <em>b<\/em>\u00a0is always positive.\r\n\r\n<img class=\"aligncenter size-full wp-image-3090\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/>\r\n\r\nBecause a logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] using parentheses to denote function evaluation just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.\r\n\r\nWe can illustrate the notation of logarithms as follows:\r\n\r\n<img class=\"aligncenter size-full wp-image-3092\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/>\r\n\r\nNotice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Definition of the Logarithmic Function<\/h3>\r\nA <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition:\r\n\r\nFor [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex],\r\n\r\n[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where\r\n<ul>\r\n \t<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\" or the \"log base <em>b<\/em>\u00a0of <em>x<\/em>.\"<\/li>\r\n \t<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\r\n \t<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\r\n<\/ul>\r\nAlso, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,\r\n<ul>\r\n \t<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we take the logarithm of a negative number?<\/strong>\r\n\r\n<em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/h3>\r\n<ol>\r\n \t<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\r\n \t<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting from Logarithmic Form to Exponential Form<\/h3>\r\nWrite the following logarithmic equations in exponential form.\r\n<ol>\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"642511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"642511\"]\r\n\r\nFirst, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].\r\n<ol>\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the following logarithmic equations in exponential form.\r\n<ol style=\"list-style-type: lower-alpha\">\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[reveal-answer q=\"200815\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"200815\"]\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29661&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h3>Converting from Exponential to Logarithmic Form<\/h3>\r\nTo convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting from Exponential Form to Logarithmic Form<\/h3>\r\nWrite the following exponential equations in logarithmic form.\r\n<ol>\r\n \t<li>[latex]{2}^{3}=8[\/latex]<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"583658\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583658\"]\r\n\r\nFirst, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].\r\n<ol>\r\n \t<li>[latex]{2}^{3}=8[\/latex] Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex] Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the following exponential equations in logarithmic form.\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]{3}^{2}=9[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex]<\/li>\r\n \t<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"767260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"767260\"]\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]{3}^{2}=9[\/latex] is equal to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]<\/li>\r\n \t<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equal to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29668&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h2>Evaluating Logarithms<\/h2>\r\nKnowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, \"To what exponent must 2\u00a0be raised in order to get 8?\" Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].\r\n\r\nNow consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.\r\n<ul>\r\n \t<li>We ask, \"To what exponent must 7 be raised in order to get 49?\" We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex].<\/li>\r\n \t<li>We ask, \"To what exponent must 3 be raised in order to get 27?\" We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex].<\/li>\r\n<\/ul>\r\nEven some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.\r\n<ul>\r\n \t<li>We ask, \"To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? \" We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally<\/h3>\r\n<ol>\r\n \t<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\r\n \t<li>Use previous knowledge of powers of <em>b<\/em>\u00a0to identify <em>y<\/em>\u00a0by asking, \"To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?\"<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Logarithms Mentally<\/h3>\r\nSolve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.\r\n\r\n[reveal-answer q=\"879580\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"879580\"]\r\n\r\nFirst we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, \"To what exponent must 4 be raised in order to get 64?\"\r\n\r\nWe know [latex]{4}^{3}=64[\/latex]\r\n\r\nTherefore,\r\n\r\n[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.\r\n\r\n[reveal-answer q=\"143125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"143125\"]\r\n\r\n[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recall that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex] )[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating the Logarithm of a Reciprocal<\/h3>\r\nEvaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.\r\n\r\n[reveal-answer q=\"861965\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"861965\"]\r\n\r\nFirst we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, \"To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]\"?\r\n\r\nWe know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}{3}^{-3}=\\frac{1}{{3}^{3}}=\\frac{1}{27}\\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]y={\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)[\/latex] without using a calculator.\r\n\r\n[reveal-answer q=\"765423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"765423\"][latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15905&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h3>Using Natural Logarithms<\/h3>\r\nThe most frequently used base for logarithms is <em>e<\/em>. Base <em>e<\/em>\u00a0logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base <em>e<\/em>\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].\r\n\r\nMost values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Definition of the Natural Logarithm<\/h3>\r\nA <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition:\r\n\r\nFor [latex]x&gt;0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equal to }{e}^{y}=x[\/latex]\r\nWe read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>\" or \"the natural logarithm of <em>x<\/em>.\"\r\n\r\nThe logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.\r\n\r\nSince the functions [latex]y=e^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for [latex]x&gt;0[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a natural logarithm Of the form [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex], evaluate it using a calculator<\/h3>\r\n<ol>\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter the value given for <em>x<\/em>, followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Natural Logarithm Using a Calculator<\/h3>\r\nEvaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.\r\n\r\n[reveal-answer q=\"810207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"810207\"]\r\n<ul>\r\n \t<li>Press <strong>[LN]<\/strong>.<\/li>\r\n \t<li>Enter 500, followed by <strong>[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong>[ENTER]<\/strong>.<\/li>\r\n<\/ul>\r\nRounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\mathrm{ln}\\left(-500\\right)[\/latex].\r\n\r\n[reveal-answer q=\"342695\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342695\"]It is not possible to take the logarithm of a negative number in the set of real numbers.[\/hidden-answer]\r\n<iframe id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35022&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>Definition of the logarithmic function<\/td>\r\n<td>For [latex]\\text{ } x&gt;0,b&gt;0,b\\ne 1[\/latex], [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{b}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of the common logarithm<\/td>\r\n<td>For [latex]\\text{ }x&gt;0[\/latex], [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{10}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of the natural logarithm<\/td>\r\n<td>For [latex]\\text{ }x&gt;0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{e}^{y}=x[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\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 an equivalent logarithmic form using the definition of a logarithm.<\/li>\r\n \t<li>Logarithmic functions with base <em>b<\/em>\u00a0can be evaluated mentally using previous knowledge of powers of <em>b<\/em>.<\/li>\r\n \t<li>Common logarithms can be evaluated mentally using previous knowledge of powers of 10.<\/li>\r\n \t<li>When common logarithms cannot be evaluated mentally, a calculator can be used.<\/li>\r\n \t<li>Natural logarithms can be evaluated using a calculator.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135397912\" class=\"definition\">\r\n \t<dt><strong>common logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135397918\">the exponent to which 10 must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] is written simply as [latex]\\mathrm{log}\\left(x\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135397926\" class=\"definition\">\r\n \t<dt><strong>logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135397932\">the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>; written [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>natural logarithm<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the exponent to which the number <em>e<\/em>\u00a0must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]<\/dd>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert from logarithmic to exponential form.<\/li>\n<li>Convert from exponential to logarithmic form.<\/li>\n<li>Evaluate logarithms with and without a calculator.<\/li>\n<li>Evaluate logarithms with base 10 and base e.<\/li>\n<\/ul>\n<\/div>\n<div class=\"mceTemp\"><\/div>\n<p>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-2020-1\" href=\"#footnote-2020-1\" 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-2020-2\" href=\"#footnote-2020-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>\u00a0like those shown in the picture below. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013.\" id=\"return-footnote-2020-3\" href=\"#footnote-2020-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0whereas the Japanese earthquake registered a 9.0.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details. Accessed 3\/4\/2013.\" id=\"return-footnote-2020-4\" href=\"#footnote-2020-4\" aria-label=\"Footnote 4\"><sup class=\"footnote\">[4]<\/sup><\/a><\/p>\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\/896\/2016\/11\/02232710\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"488\" height=\"325\" \/><\/p>\n<p class=\"wp-caption-text\">Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)<\/p>\n<\/div>\n<p>The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\n<p>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 was 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex] where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?<\/p>\n<p>We have not yet learned a method for solving exponential equations algebraically. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3 since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3089\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195659\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/><\/p>\n<p>Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\n<p>We read a logarithmic expression as, &#8220;The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,&#8221; or, simplified, &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.&#8221; We can also say, &#8220;<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,&#8221; because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as &#8220;log base 2 of 32 is 5.&#8221;<\/p>\n<p>We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/p>\n<p>Note that the base <em>b<\/em>\u00a0is always positive.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3090\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/><\/p>\n<p>Because a logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] using parentheses to denote function evaluation just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.<\/p>\n<p>We can illustrate the notation of logarithms as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3092\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/><\/p>\n<p>Notice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Definition of the Logarithmic Function<\/h3>\n<p>A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition:<\/p>\n<p>For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n<p>[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where<\/p>\n<ul>\n<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>&#8221; or the &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>.&#8221;<\/li>\n<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\n<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\n<\/ul>\n<p>Also, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\n<ul>\n<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\n<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\n<p><em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/h3>\n<ol>\n<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\n<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Converting from Logarithmic Form to Exponential Form<\/h3>\n<p>Write the following logarithmic equations in exponential form.<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q642511\">Show Solution<\/span><\/p>\n<div id=\"q642511\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the following logarithmic equations in exponential form.<\/p>\n<ol style=\"list-style-type: lower-alpha\">\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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q200815\">Show Solution<\/span><\/p>\n<div id=\"q200815\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29661&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h3>Converting from Exponential to Logarithmic Form<\/h3>\n<p>To convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Converting from Exponential Form to Logarithmic Form<\/h3>\n<p>Write the following exponential equations in logarithmic form.<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex]<\/li>\n<li>[latex]{5}^{2}=25[\/latex]<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q583658\">Show Solution<\/span><\/p>\n<div id=\"q583658\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex] Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\n<li>[latex]{5}^{2}=25[\/latex] Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the following exponential equations in logarithmic form.<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{3}^{2}=9[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex]<\/li>\n<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q767260\">Show Solution<\/span><\/p>\n<div id=\"q767260\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{3}^{2}=9[\/latex] is equal to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]<\/li>\n<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equal to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29668&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h2>Evaluating Logarithms<\/h2>\n<p>Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, &#8220;To what exponent must 2\u00a0be raised in order to get 8?&#8221; Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].<\/p>\n<p>Now consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.<\/p>\n<ul>\n<li>We ask, &#8220;To what exponent must 7 be raised in order to get 49?&#8221; We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex].<\/li>\n<li>We ask, &#8220;To what exponent must 3 be raised in order to get 27?&#8221; We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex].<\/li>\n<\/ul>\n<p>Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.<\/p>\n<ul>\n<li>We ask, &#8220;To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? &#8221; We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally<\/h3>\n<ol>\n<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\n<li>Use previous knowledge of powers of <em>b<\/em>\u00a0to identify <em>y<\/em>\u00a0by asking, &#8220;To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?&#8221;<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Logarithms Mentally<\/h3>\n<p>Solve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q879580\">Show Solution<\/span><\/p>\n<div id=\"q879580\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, &#8220;To what exponent must 4 be raised in order to get 64?&#8221;<\/p>\n<p>We know [latex]{4}^{3}=64[\/latex]<\/p>\n<p>Therefore,<\/p>\n<p>[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]y={\\mathrm{log}}_{121}\\left(11\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q143125\">Show Solution<\/span><\/p>\n<div id=\"q143125\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}[\/latex] (recall that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11[\/latex] )<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating the Logarithm of a Reciprocal<\/h3>\n<p>Evaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q861965\">Show Solution<\/span><\/p>\n<div id=\"q861965\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, &#8220;To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]&#8220;?<\/p>\n<p>We know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}{3}^{-3}=\\frac{1}{{3}^{3}}=\\frac{1}{27}\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]y={\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q765423\">Show Solution<\/span><\/p>\n<div id=\"q765423\" class=\"hidden-answer\" style=\"display: none\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15905&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h3>Using Natural Logarithms<\/h3>\n<p>The most frequently used base for logarithms is <em>e<\/em>. Base <em>e<\/em>\u00a0logarithms are important in calculus and some scientific applications; they are called <strong>natural logarithms<\/strong>. The base <em>e<\/em>\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/p>\n<p>Most values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Definition of the Natural Logarithm<\/h3>\n<p>A <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition:<\/p>\n<p>For [latex]x>0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equal to }{e}^{y}=x[\/latex]<br \/>\nWe read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>&#8221; or &#8220;the natural logarithm of <em>x<\/em>.&#8221;<\/p>\n<p>The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\n<p>Since the functions [latex]y=e^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for [latex]x>0[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a natural logarithm Of the form [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex], evaluate it using a calculator<\/h3>\n<ol>\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter the value given for <em>x<\/em>, followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Natural Logarithm Using a Calculator<\/h3>\n<p>Evaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810207\">Show Solution<\/span><\/p>\n<div id=\"q810207\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>Press <strong>[LN]<\/strong>.<\/li>\n<li>Enter 500, followed by <strong>[ ) ]<\/strong>.<\/li>\n<li>Press <strong>[ENTER]<\/strong>.<\/li>\n<\/ul>\n<p>Rounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\mathrm{ln}\\left(-500\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342695\">Show Solution<\/span><\/p>\n<div id=\"q342695\" class=\"hidden-answer\" style=\"display: none\">It is not possible to take the logarithm of a negative number in the set of real numbers.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35022&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table summary=\"...\">\n<tbody>\n<tr>\n<td>Definition of the logarithmic function<\/td>\n<td>For [latex]\\text{ } x>0,b>0,b\\ne 1[\/latex], [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{b}^{y}=x[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Definition of the common logarithm<\/td>\n<td>For [latex]\\text{ }x>0[\/latex], [latex]y=\\mathrm{log}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{10}^{y}=x[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Definition of the natural logarithm<\/td>\n<td>For [latex]\\text{ }x>0[\/latex], [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] if and only if [latex]\\text{ }{e}^{y}=x[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\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 an equivalent logarithmic form using the definition of a logarithm.<\/li>\n<li>Logarithmic functions with base <em>b<\/em>\u00a0can be evaluated mentally using previous knowledge of powers of <em>b<\/em>.<\/li>\n<li>Common logarithms can be evaluated mentally using previous knowledge of powers of 10.<\/li>\n<li>When common logarithms cannot be evaluated mentally, a calculator can be used.<\/li>\n<li>Natural logarithms can be evaluated using a calculator.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135397912\" class=\"definition\">\n<dt><strong>common logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165135397918\">the exponent to which 10 must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{10}\\left(x\\right)[\/latex] is written simply as [latex]\\mathrm{log}\\left(x\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135397926\" class=\"definition\">\n<dt><strong>logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165135397932\">the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>; written [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>natural logarithm<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">the exponent to which the number <em>e<\/em>\u00a0must be raised to get <em>x<\/em>; [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] is written as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]<\/dd>\n<\/dl>\n\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-2020\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 29668, 29661. <strong>Authored by<\/strong>: McClure, Caren. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 35022. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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-2020-1\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-2020-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-2020-2\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-2020-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-2020-3\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-2020-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><li id=\"footnote-2020-4\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\" rel=\"noopener\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-2020-4\" class=\"return-footnote\" aria-label=\"Return to footnote 4\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 29668, 29661\",\"author\":\"McClure, Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 35022\",\"author\":\"Jim Smart\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2020","chapter","type-chapter","status-publish","hentry"],"part":1964,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2020","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2020\/revisions"}],"predecessor-version":[{"id":4564,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2020\/revisions\/4564"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1964"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2020\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=2020"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2020"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2020"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=2020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}