{"id":166,"date":"2021-02-03T22:19:10","date_gmt":"2021-02-03T22:19:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=166"},"modified":"2022-03-11T21:46:10","modified_gmt":"2022-03-11T21:46:10","slug":"logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/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>Identify the form of a logarithmic function<\/li>\r\n \t<li>Explain the relationship between exponential and logarithmic functions<\/li>\r\n \t<li>Describe how to calculate a logarithm to a different base<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Recall the definition of an inverse function.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall:\u00a0Inverse Function<\/h3>\r\nFor any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].\r\n\r\nThe notation [latex]{f}^{-1}[\/latex] is read \"[latex]f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]\".\r\n\r\n<\/div>\r\nLogarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.\r\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <em>logarithmic function with base<\/em> [latex]b[\/latex]. For any [latex]b&gt;0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\r\n\r\n<div id=\"fs-id1170572551548\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex]<\/div>\r\n<p id=\"fs-id1170572545100\">For example,<\/p>\r\n\r\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b&gt;0.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nFurthermore, since [latex]y=\\log_b (x)[\/latex] and [latex]y=b^x[\/latex] are inverse functions,\r\n<div id=\"fs-id1170572169121\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_b (b^x)=x \\, \\text{and} \\, b^{\\log_b (x)}=x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex]. For example,<\/p>\r\n\r\n<div id=\"fs-id1170572243723\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (e)=\\log_e (e)=1, \\, \\ln(e^3)=\\log_e (e^3)=3, \\, \\ln(1)=\\log_e (1)=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572243836\">Since the functions [latex]f(x)=e^x[\/latex] and [latex]g(x)=\\ln(x)[\/latex] are inverses of each other,<\/p>\r\n\r\n<div id=\"fs-id1170572243887\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln(e^x)=x \\, \\text{ and } \\, e^{\\ln x}=x[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572229068\">and their graphs are symmetric about the line [latex]y=x[\/latex] (Figure 4).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202652\/CNX_Calc_Figure_01_05_004.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -3 to 4. The graph is of two functions. The first function is \u201cf(x) = e to power of x\u201d, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1) and there is no x intercept. The second function is \u201cf(x) = ln(x)\u201d, an increasing curved function. The x intercept is at the point (1, 0) and there is no y intercept. A dotted line with label \u201cy = x\u201d is also plotted on the graph, to show that the functions are mirror images over this line.\" width=\"325\" height=\"312\" \/> Figure 4: The functions [latex]y=e^x[\/latex] and [latex]y=\\ln(x)[\/latex] are inverses of each other, so their graphs are symmetric about the line [latex]y=x[\/latex].[\/caption]\r\n<div id=\"fs-id1170572229152\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1170572229156\"><a href=\"https:\/\/www.gcaudio.com\/tips-tricks\/the-relationship-of-voltage-loudness-power-and-decibels\/\" target=\"_blank\" rel=\"noopener\">At this site you can see an example of a base-10 logarithmic scale.<\/a><\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572229166\">In general, for any base [latex]b&gt;0, \\, b\\ne 1[\/latex], the function [latex]g(x)=\\log_b (x)[\/latex] is symmetric about the line [latex]y=x[\/latex] with the function [latex]f(x)=b^x[\/latex]. Using this fact and the graphs of the exponential functions, we graph functions [latex]\\log_b (x)[\/latex] for several values of [latex]b&gt;1[\/latex] (Figure 5).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202654\/CNX_Calc_Figure_01_05_005.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of three functions. All three functions a log functions that are increasing curved functions that start slightly to the right of the y axis and have an x intercept at (1, 0). The first function is \u201cy = log base 10 (x)\u201d, the second function is \u201cf(x) = ln(x)\u201d, and the third function is \u201cy = log base 2 (x)\u201d. The third function increases the most rapidly, the second function increases next most rapidly, and the third function increases the slowest.\" width=\"325\" height=\"312\" \/> Figure 5: Graphs of [latex]y=\\log_b (x)[\/latex] are depicted for [latex]b=2, \\, e, \\, 10[\/latex].[\/caption]\r\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Properties of Logarithms<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572482707\">If [latex]a,b,c&gt;0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill &amp; &amp; &amp; \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572551970\" class=\"textbook exercises\">\r\n<h3>Example: Solving Equations Involving Exponential Functions<\/h3>\r\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]5^x=2[\/latex]<\/li>\r\n \t<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572550555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572550555\"]\r\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Applying the natural logarithm function to both sides of the equation, we have\r\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\r\nUsing the power property of logarithms,\r\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\r\nTherefore, [latex]x=\\ln 2 \/ \\ln 5[\/latex].<\/li>\r\n \t<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation\r\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\r\nRewriting this equation as\r\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\r\nwe can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:\r\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\r\nNow we can solve the quadratic equation. Factoring this equation, we obtain\r\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\r\nTherefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions\r\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Solving Equations Involving Exponential Functions[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.5 Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572232044\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572232053\">Solve [latex]\\dfrac{e^{2x}}{(3+e^{2x})}=\\dfrac{1}{2}[\/latex].<\/p>\r\n[reveal-answer q=\"708533\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"708533\"]\r\n<p id=\"fs-id1165042579095\">First solve the equation for [latex]e^{2x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572174654\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572174654\"]\r\n<p id=\"fs-id1170572174654\">[latex]x=\\dfrac{\\ln 3}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572174674\" class=\"textbook exercises\">\r\n<h3>Example: Solving Equations Involving Logarithmic Functions<\/h3>\r\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\r\n \t<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\r\n \t<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572174799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572174799\"]\r\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>By the definition of the natural logarithm function,\r\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\r\nTherefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/li>\r\n \t<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as\r\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\r\nTherefore, the equation can be rewritten as\r\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\r\nThe solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/li>\r\n \t<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].\r\nUsing the quotient property, this becomes\r\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\r\nTherefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Solving Equations Involving Logarithmic Functions[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=827&amp;end=1126&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.5ExponentialAndLogarithmicFunctions827to1126_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.5 Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572552646\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572552654\">Solve [latex]\\ln(x^3)-4 \\ln (x)=1[\/latex].<\/p>\r\n[reveal-answer q=\"662277\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"662277\"]\r\n<p id=\"fs-id1165043161250\">First use the power property, then use the product property of logarithms.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572552698\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572552698\"]\r\n<p id=\"fs-id1170572552698\">[latex]x=\\frac{1}{e}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]14391[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572552715\">When evaluating a logarithmic function with a calculator, you may have noticed that the only options are [latex]\\log_{10}[\/latex] or log, called the <span class=\"no-emphasis\"><em>common logarithm<\/em><\/span>, or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base [latex]b[\/latex]. If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]217547[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Change-of-Base Formulas<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572267920\">Let [latex]a&gt;0, \\, b&gt;0[\/latex], and [latex]a\\ne 1, \\, b\\ne 1[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572267962\">\r\n \t<li>[latex]a^x=b^{x \\log_b a}[\/latex] for any real number [latex]x[\/latex].\r\nIf [latex]b=e[\/latex], this equation reduces to [latex]a^x=e^{x \\log_e a}=e^{x \\ln a}[\/latex].<\/li>\r\n \t<li>[latex]\\log_a x=\\frac{\\log_b x}{\\log_b a}[\/latex] for any real number [latex]x&gt;0[\/latex].\r\nIf [latex]b=e[\/latex], this equation reduces to [latex]\\log_a x=\\frac{\\ln x}{\\ln a}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170572219413\">For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base [latex]b&gt;0, \\, b\\ne 1, \\, \\log_b (a^x)=x \\log_b a[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1170572219472\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{\\log_b(a^x)}=b^{x \\log_b a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572219523\">In addition, we know that [latex]b^x[\/latex] and [latex]\\log_b (x)[\/latex] are inverse functions. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1170572219556\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{\\log_b (a^x)}=a^x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572309868\">Combining these last two equalities, we conclude that [latex]a^x=b^{x \\log_b a}[\/latex].<\/p>\r\n<p id=\"fs-id1170572309898\">To prove the second property, we show that<\/p>\r\n\r\n<div id=\"fs-id1170572309902\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](\\log_b a)\u00b7(\\log_a x)=\\log_b x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572309957\">Let [latex]u=\\log_b a, \\, v=\\log_a x[\/latex], and [latex]w=\\log_b x[\/latex]. We will show that [latex]u\u00b7v=w[\/latex]. By the definition of logarithmic functions, we know that [latex]b^u=a, \\, a^v=x[\/latex], and [latex]b^w=x[\/latex]. From the previous equations, we see that<\/p>\r\n\r\n<div id=\"fs-id1170572434870\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{uv}=(b^u)^v=a^v=x=b^w[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572434927\">Therefore, [latex]b^{uv}=b^w[\/latex]. Since exponential functions are one-to-one, we can conclude that [latex]u\u00b7v=w[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<div id=\"fs-id1170572434970\" class=\"textbook exercises\">\r\n<h3>Example: Changing Bases<\/h3>\r\n<p id=\"fs-id1170572434979\">Use a calculating utility to evaluate [latex]\\log_3 7[\/latex] with the change-of-base formula presented earlier.<\/p>\r\n[reveal-answer q=\"fs-id1170572435001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572435001\"]\r\n<p id=\"fs-id1170572435001\">Use the second equation with [latex]a=3[\/latex] and [latex]e=3[\/latex]:<\/p>\r\n<p id=\"fs-id1170572435025\">[latex]\\log_3 7=\\frac{\\ln 7}{\\ln 3} \\approx 1.77124[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572435059\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572435068\">Use the change-of-base formula and a calculating utility to evaluate [latex]\\log_4 6[\/latex].<\/p>\r\n[reveal-answer q=\"3762229\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3762229\"]\r\n<p id=\"fs-id1165042853660\">Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572128662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572128662\"]\r\n<p id=\"fs-id1170572128662\">1.29248<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572478059\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572478067\">Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.<\/p>\r\n<p id=\"fs-id1170572478135\">[reveal-answer q=\"3287755\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3287755\"]<\/p>\r\n[latex]R_1-R_2=\\log_{10}(A_1 \/ A_2)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170448623\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170448623\"]\r\n\r\nThe magnitude 8.4 earthquake is roughly 10 times as severe as the magnitude 7.4 earthquake.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]217554[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify the form of a logarithmic function<\/li>\n<li>Explain the relationship between exponential and logarithmic functions<\/li>\n<li>Describe how to calculate a logarithm to a different base<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Recall the definition of an inverse function.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall:\u00a0Inverse Function<\/h3>\n<p>For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].<\/p>\n<p>The notation [latex]{f}^{-1}[\/latex] is read &#8220;[latex]f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]&#8220;.<\/p>\n<\/div>\n<p>Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.<\/p>\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <em>logarithmic function with base<\/em> [latex]b[\/latex]. For any [latex]b>0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\n<div id=\"fs-id1170572551548\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex]<\/div>\n<p id=\"fs-id1170572545100\">For example,<\/p>\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b>0.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Furthermore, since [latex]y=\\log_b (x)[\/latex] and [latex]y=b^x[\/latex] are inverse functions,<\/p>\n<div id=\"fs-id1170572169121\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_b (b^x)=x \\, \\text{and} \\, b^{\\log_b (x)}=x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex]. For example,<\/p>\n<div id=\"fs-id1170572243723\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln (e)=\\log_e (e)=1, \\, \\ln(e^3)=\\log_e (e^3)=3, \\, \\ln(1)=\\log_e (1)=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572243836\">Since the functions [latex]f(x)=e^x[\/latex] and [latex]g(x)=\\ln(x)[\/latex] are inverses of each other,<\/p>\n<div id=\"fs-id1170572243887\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln(e^x)=x \\, \\text{ and } \\, e^{\\ln x}=x[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572229068\">and their graphs are symmetric about the line [latex]y=x[\/latex] (Figure 4).<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202652\/CNX_Calc_Figure_01_05_004.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -3 to 4. The graph is of two functions. The first function is \u201cf(x) = e to power of x\u201d, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1) and there is no x intercept. The second function is \u201cf(x) = ln(x)\u201d, an increasing curved function. The x intercept is at the point (1, 0) and there is no y intercept. A dotted line with label \u201cy = x\u201d is also plotted on the graph, to show that the functions are mirror images over this line.\" width=\"325\" height=\"312\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4: The functions [latex]y=e^x[\/latex] and [latex]y=\\ln(x)[\/latex] are inverses of each other, so their graphs are symmetric about the line [latex]y=x[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170572229152\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1170572229156\"><a href=\"https:\/\/www.gcaudio.com\/tips-tricks\/the-relationship-of-voltage-loudness-power-and-decibels\/\" target=\"_blank\" rel=\"noopener\">At this site you can see an example of a base-10 logarithmic scale.<\/a><\/p>\n<\/div>\n<p id=\"fs-id1170572229166\">In general, for any base [latex]b>0, \\, b\\ne 1[\/latex], the function [latex]g(x)=\\log_b (x)[\/latex] is symmetric about the line [latex]y=x[\/latex] with the function [latex]f(x)=b^x[\/latex]. Using this fact and the graphs of the exponential functions, we graph functions [latex]\\log_b (x)[\/latex] for several values of [latex]b>1[\/latex] (Figure 5).<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202654\/CNX_Calc_Figure_01_05_005.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of three functions. All three functions a log functions that are increasing curved functions that start slightly to the right of the y axis and have an x intercept at (1, 0). The first function is \u201cy = log base 10 (x)\u201d, the second function is \u201cf(x) = ln(x)\u201d, and the third function is \u201cy = log base 2 (x)\u201d. The third function increases the most rapidly, the second function increases next most rapidly, and the third function increases the slowest.\" width=\"325\" height=\"312\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5: Graphs of [latex]y=\\log_b (x)[\/latex] are depicted for [latex]b=2, \\, e, \\, 10[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Properties of Logarithms<\/h3>\n<hr \/>\n<p id=\"fs-id1170572482707\">If [latex]a,b,c>0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill & & & \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill & & & \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill & & & \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<div id=\"fs-id1170572551970\" class=\"textbook exercises\">\n<h3>Example: Solving Equations Involving Exponential Functions<\/h3>\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]5^x=2[\/latex]<\/li>\n<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572550555\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572550555\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\n<li>Applying the natural logarithm function to both sides of the equation, we have\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\n<p>Using the power property of logarithms,<\/p>\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\n<p>Therefore, [latex]x=\\ln 2 \/ \\ln 5[\/latex].<\/li>\n<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\n<p>Rewriting this equation as<\/p>\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\n<p>we can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:<\/p>\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\n<p>Now we can solve the quadratic equation. Factoring this equation, we obtain<\/p>\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\n<p>Therefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions<\/p>\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Solving Equations Involving Exponential Functions<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.5 Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572232044\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572232053\">Solve [latex]\\dfrac{e^{2x}}{(3+e^{2x})}=\\dfrac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q708533\">Hint<\/span><\/p>\n<div id=\"q708533\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042579095\">First solve the equation for [latex]e^{2x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572174654\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572174654\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572174654\">[latex]x=\\dfrac{\\ln 3}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572174674\" class=\"textbook exercises\">\n<h3>Example: Solving Equations Involving Logarithmic Functions<\/h3>\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\n<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\n<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572174799\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572174799\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\n<li>By the definition of the natural logarithm function,\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\n<p>Therefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/li>\n<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\n<p>Therefore, the equation can be rewritten as<\/p>\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\n<p>The solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/li>\n<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].<br \/>\nUsing the quotient property, this becomes<\/p>\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\n<p>Therefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Solving Equations Involving Logarithmic Functions<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=827&amp;end=1126&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.5ExponentialAndLogarithmicFunctions827to1126_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.5 Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572552646\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572552654\">Solve [latex]\\ln(x^3)-4 \\ln (x)=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q662277\">Hint<\/span><\/p>\n<div id=\"q662277\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043161250\">First use the power property, then use the product property of logarithms.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572552698\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572552698\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572552698\">[latex]x=\\frac{1}{e}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14391\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14391&theme=oea&iframe_resize_id=ohm14391&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1170572552715\">When evaluating a logarithmic function with a calculator, you may have noticed that the only options are [latex]\\log_{10}[\/latex] or log, called the <span class=\"no-emphasis\"><em>common logarithm<\/em><\/span>, or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base [latex]b[\/latex]. If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm217547\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=217547&theme=oea&iframe_resize_id=ohm217547&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Change-of-Base Formulas<\/h3>\n<hr \/>\n<p id=\"fs-id1170572267920\">Let [latex]a>0, \\, b>0[\/latex], and [latex]a\\ne 1, \\, b\\ne 1[\/latex].<\/p>\n<ol id=\"fs-id1170572267962\">\n<li>[latex]a^x=b^{x \\log_b a}[\/latex] for any real number [latex]x[\/latex].<br \/>\nIf [latex]b=e[\/latex], this equation reduces to [latex]a^x=e^{x \\log_e a}=e^{x \\ln a}[\/latex].<\/li>\n<li>[latex]\\log_a x=\\frac{\\log_b x}{\\log_b a}[\/latex] for any real number [latex]x>0[\/latex].<br \/>\nIf [latex]b=e[\/latex], this equation reduces to [latex]\\log_a x=\\frac{\\ln x}{\\ln a}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<h3>Proof<\/h3>\n<p id=\"fs-id1170572219413\">For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base [latex]b>0, \\, b\\ne 1, \\, \\log_b (a^x)=x \\log_b a[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1170572219472\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{\\log_b(a^x)}=b^{x \\log_b a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572219523\">In addition, we know that [latex]b^x[\/latex] and [latex]\\log_b (x)[\/latex] are inverse functions. Therefore,<\/p>\n<div id=\"fs-id1170572219556\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{\\log_b (a^x)}=a^x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572309868\">Combining these last two equalities, we conclude that [latex]a^x=b^{x \\log_b a}[\/latex].<\/p>\n<p id=\"fs-id1170572309898\">To prove the second property, we show that<\/p>\n<div id=\"fs-id1170572309902\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](\\log_b a)\u00b7(\\log_a x)=\\log_b x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572309957\">Let [latex]u=\\log_b a, \\, v=\\log_a x[\/latex], and [latex]w=\\log_b x[\/latex]. We will show that [latex]u\u00b7v=w[\/latex]. By the definition of logarithmic functions, we know that [latex]b^u=a, \\, a^v=x[\/latex], and [latex]b^w=x[\/latex]. From the previous equations, we see that<\/p>\n<div id=\"fs-id1170572434870\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]b^{uv}=(b^u)^v=a^v=x=b^w[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572434927\">Therefore, [latex]b^{uv}=b^w[\/latex]. Since exponential functions are one-to-one, we can conclude that [latex]u\u00b7v=w[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1170572434970\" class=\"textbook exercises\">\n<h3>Example: Changing Bases<\/h3>\n<p id=\"fs-id1170572434979\">Use a calculating utility to evaluate [latex]\\log_3 7[\/latex] with the change-of-base formula presented earlier.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572435001\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572435001\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572435001\">Use the second equation with [latex]a=3[\/latex] and [latex]e=3[\/latex]:<\/p>\n<p id=\"fs-id1170572435025\">[latex]\\log_3 7=\\frac{\\ln 7}{\\ln 3} \\approx 1.77124[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572435059\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572435068\">Use the change-of-base formula and a calculating utility to evaluate [latex]\\log_4 6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3762229\">Hint<\/span><\/p>\n<div id=\"q3762229\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042853660\">Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572128662\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572128662\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572128662\">1.29248<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572478059\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572478067\">Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.<\/p>\n<p id=\"fs-id1170572478135\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3287755\">Hint<\/span><\/p>\n<div id=\"q3287755\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]R_1-R_2=\\log_{10}(A_1 \/ A_2)[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170448623\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170448623\" class=\"hidden-answer\" style=\"display: none\">\n<p>The magnitude 8.4 earthquake is roughly 10 times as severe as the magnitude 7.4 earthquake.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm217554\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=217554&theme=oea&iframe_resize_id=ohm217554&show_question_numbers\" width=\"100%\" height=\"150\"><\/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-166\">\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>1.5 Exponential and Logarithmic Functions. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":23,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for 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