{"id":5339,"date":"2021-10-13T18:24:24","date_gmt":"2021-10-13T18:24:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-logarithmic-functions\/"},"modified":"2021-11-23T03:19:30","modified_gmt":"2021-11-23T03:19:30","slug":"introduction-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-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 between logarithmic and exponential forms.<\/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 \t<li>Determine the domain and range of a logarithmic function.<\/li>\r\n \t<li>Determine the x-intercept and vertical asymptote of a logarithmic function.<\/li>\r\n \t<li>Identify the features of a logarithmic function that make it an inverse of an exponential function.<\/li>\r\n \t<li>Graph logarithmic functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"mceTemp\"><\/div>\r\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\r\nObserve that graphs of exponential functions pass 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\u00a0<strong>logarithm,\u00a0<\/strong>base <em>b,\u00a0<\/em>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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert between logarithmic and exponential forms.<\/li>\n<li>Evaluate logarithms with and without a calculator.<\/li>\n<li>Evaluate logarithms with base 10 and base e.<\/li>\n<li>Determine the domain and range of a logarithmic function.<\/li>\n<li>Determine the x-intercept and vertical asymptote of a logarithmic function.<\/li>\n<li>Identify the features of a logarithmic function that make it an inverse of an exponential function.<\/li>\n<li>Graph logarithmic functions<\/li>\n<\/ul>\n<\/div>\n<div class=\"mceTemp\"><\/div>\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\n<p>Observe that graphs of exponential functions pass 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\u00a0<strong>logarithm,\u00a0<\/strong>base <em>b,\u00a0<\/em>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\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-5339\">\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>","protected":false},"author":167848,"menu_order":5,"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-5339","chapter","type-chapter","status-publish","hentry"],"part":5352,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5339","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5339\/revisions"}],"predecessor-version":[{"id":5749,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5339\/revisions\/5749"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5352"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5339\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5339"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5339"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5339"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5339"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}