{"id":2757,"date":"2021-04-07T20:43:48","date_gmt":"2021-04-07T20:43:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2757"},"modified":"2022-03-16T22:52:12","modified_gmt":"2022-03-16T22:52:12","slug":"exponential-and-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/exponential-and-logarithmic-functions\/","title":{"raw":"Exponential and Logarithmic Functions","rendered":"Exponential and Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize the derivative and integral of the exponential function.<\/li>\r\n \t<li>Prove properties of logarithms and exponential functions using integrals.<\/li>\r\n \t<li>Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Exponential Function<\/h2>\r\n<p id=\"fs-id1167794051417\">We now turn our attention to the function [latex]{e}^{x}.[\/latex] Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by [latex]\\text{exp}x.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1167793514717\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{exp}(\\text{ln}x)=x\\text{ for }x&gt;0\\text{ and }\\text{ln}(\\text{exp}x)=x\\text{ for all }x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793932178\">The following figure shows the graphs of [latex]\\text{exp}x[\/latex] and [latex]\\text{ln}x.[\/latex]<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"421\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213328\/CNX_Calc_Figure_06_07_004.jpg\" alt=\"This figure is a graph. It has three curves. The first curve is labeled exp x. It is an increasing curve with the x-axis as a horizontal asymptote. It intersects the y-axis at y=1. The second curve is a diagonal line through the origin. The third curve is labeled lnx. It is an increasing curve with the y-axis as an vertical axis. It intersects the x-axis at x=1.\" width=\"421\" height=\"422\" \/> Figure 4. The graphs of [latex]\\text{ln}x[\/latex] and [latex]\\text{exp}x.[\/latex][\/caption]\r\n<p id=\"fs-id1167793905821\">We hypothesize that [latex]\\text{exp}x={e}^{x}.[\/latex] For rational values of [latex]x,[\/latex] this is easy to show. If [latex]x[\/latex] is rational, then we have [latex]\\text{ln}({e}^{x})=x\\text{ln}e=x.[\/latex] Thus, when [latex]x[\/latex] is rational, [latex]{e}^{x}=\\text{exp}x.[\/latex] For irrational values of [latex]x,[\/latex] we simply define [latex]{e}^{x}[\/latex] as the inverse function of [latex]\\text{ln}x.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793579578\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793367959\">For any real number [latex]x,[\/latex] define [latex]y={e}^{x}[\/latex] to be the number for which<\/p>\r\n\r\n<div id=\"fs-id1167793447883\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{ln}y=\\text{ln}({e}^{x})=x[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793489536\">Then we have [latex]{e}^{x}=\\text{exp}(x)[\/latex] for all [latex]x,[\/latex] and thus<\/p>\r\n\r\n<div id=\"fs-id1167793564127\" class=\"equation\" style=\"text-align: center;\">[latex]{e}^{\\text{ln}x}=x\\text{ for }x&gt;0\\text{ and }\\text{ln}({e}^{x})=x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794036848\">for all [latex]x.[\/latex]<\/p>\r\n\r\n<h2>Properties of the Exponential Function<\/h2>\r\n<p id=\"fs-id1167793516179\">Since the exponential function was defined in terms of an inverse function, and not in terms of a power of [latex]e,[\/latex] we must verify that the usual laws of exponents hold for the function [latex]{e}^{x}.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167794176164\" class=\"textbox shaded\">\r\n<h3>Properties of the Exponential Function<\/h3>\r\n<p id=\"fs-id1167794210866\">If [latex]p[\/latex] and [latex]q[\/latex] are any real numbers and [latex]r[\/latex] is a rational number, then<\/p>\r\n\r\n<ol id=\"fs-id1167793361174\">\r\n \t<li>[latex]{e}^{p}{e}^{q}={e}^{p+q}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{{e}^{p}}{{e}^{q}}={e}^{p-q}[\/latex]<\/li>\r\n \t<li>[latex]{({e}^{p})}^{r}={e}^{pr}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167794064034\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1167793420459\">Note that if [latex]p[\/latex] and [latex]q[\/latex] are rational, the properties hold. However, if [latex]p[\/latex] or [latex]q[\/latex] are irrational, we must apply the inverse function definition of [latex]{e}^{x}[\/latex] and verify the properties. Only the first property is verified here; the other two are left to you. We have<\/p>\r\n\r\n<div id=\"fs-id1167793301084\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}({e}^{p}{e}^{q})=\\text{ln}({e}^{p})+\\text{ln}({e}^{q})=p+q=\\text{ln}({e}^{p+q}).[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793937142\">Since [latex]\\text{ln}x[\/latex] is one-to-one, then<\/p>\r\n\r\n<div id=\"fs-id1167794329135\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{e}^{p}{e}^{q}={e}^{p+q}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793503220\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1167793443585\">As with part iv. of the logarithm properties, we can extend property iii. to irrational values of [latex]r,[\/latex] and we do so by the end of the section.<\/p>\r\n<p id=\"fs-id1167793952464\">We also want to verify the differentiation formula for the function [latex]y={e}^{x}.[\/latex] To do this, we need to use implicit differentiation. Let [latex]y={e}^{x}.[\/latex] Then<\/p>\r\n\r\n<div id=\"fs-id1167794095942\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}y&amp; =\\hfill &amp; x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{ln}y&amp; =\\hfill &amp; \\frac{d}{dx}x\\hfill \\\\ \\hfill \\frac{1}{y}\\frac{dy}{dx}&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill \\frac{dy}{dx}&amp; =\\hfill &amp; y.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794147337\">Thus, we see<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{e}^{x}={e}^{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793370445\">as desired, which leads immediately to the integration formula<\/p>\r\n\r\n<div id=\"fs-id1167793924216\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}dx={e}^{x}+C[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793272155\">We apply these formulas in the following examples.<\/p>\r\n\r\n<div id=\"fs-id1167793272158\" class=\"textbook exercises\">\r\n<h3>Example: Using Properties of Exponential Functions<\/h3>\r\n<p id=\"fs-id1167793393570\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793393574\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{e}^{3{x}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167794293256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794293256\"]\r\n<p id=\"fs-id1167794293256\">We apply the chain rule as necessary.<\/p>\r\n\r\n<ol id=\"fs-id1167794293259\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793372329\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793374990\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793374993\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\frac{{e}^{{x}^{2}}}{{e}^{5x}})[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dt}{({e}^{2t})}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793442905\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1167793872351\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793872351\"]\r\n<ol id=\"fs-id1167793872351\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\frac{{e}^{{x}^{2}}}{{e}^{5x}})={e}^{{x}^{2}-5x}(2x-5)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dt}{({e}^{2t})}^{3}=6{e}^{6t}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793568428\">Use the properties of exponential functions and the chain rule as necessary.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=238&amp;end=356&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\/6.7TryItProblems238to356_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.7 Try It Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793944672\" class=\"textbook exercises\">\r\n<h3>Example: Using Properties of Exponential Functions<\/h3>\r\nEvaluate the following integral: [latex]\\displaystyle\\int 2x{e}^{\\text{\u2212}{x}^{2}}dx.[\/latex]\r\n<div id=\"fs-id1167794031059\" class=\"exercise\">[reveal-answer q=\"fs-id1167793501968\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793501968\"]\r\n<p id=\"fs-id1167793501968\">Using [latex]u[\/latex]-substitution, let [latex]u=\\text{\u2212}{x}^{2}.[\/latex] Then [latex]du=-2xdx,[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1167793956578\" class=\"equation unnumbered\">[latex]\\displaystyle\\int 2x{e}^{\\text{\u2212}{x}^{2}}dx=\\text{\u2212}\\displaystyle\\int {e}^{u}du=\\text{\u2212}{e}^{u}+C=\\text{\u2212}{e}^{\\text{\u2212}{x}^{2}}+C.[\/latex][\/hidden-answer]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793478795\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate the following integral: [latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793510888\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793510888\"]\r\n<p id=\"fs-id1167793510888\">[latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx=-\\frac{4}{3}{e}^{-3x}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793943946\">Use the properties of exponential functions and [latex]u\\text{-substitution}[\/latex] as necessary.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=359&amp;end=460&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\/6.7TryItProblems359to460_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.7 Try It Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20047[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>General Logarithmic and Exponential Functions<\/h2>\r\n<p id=\"fs-id1167793400845\">We close this section by looking at exponential functions and logarithms with bases other than [latex]e.[\/latex] Exponential functions are functions of the form [latex]f(x)={a}^{x}.[\/latex] Note that unless [latex]a=e,[\/latex] we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let\u2019s rectify that here by defining the function [latex]f(x)={a}^{x}[\/latex] in terms of the exponential function [latex]{e}^{x}.[\/latex] We then examine logarithms with bases other than [latex]e[\/latex] as inverse functions of exponential functions.<\/p>\r\n\r\n<div id=\"fs-id1167793285142\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nFor any [latex]a&gt;0,[\/latex] and for any real number [latex]x,[\/latex] define [latex]y={a}^{x}[\/latex] as follows:\r\n<div id=\"fs-id1167793420642\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={a}^{x}={e}^{x\\text{ln}a}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793455076\">Now [latex]{a}^{x}[\/latex] is defined rigorously for all values of [latex]x[\/latex]. This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of [latex]r.[\/latex] It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.<\/p>\r\n<p id=\"fs-id1167793384514\">Let\u2019s now apply this definition to calculate a differentiation formula for [latex]{a}^{x}.[\/latex] We have<\/p>\r\n\r\n<div id=\"fs-id1167793559160\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{a}^{x}=\\frac{d}{dx}{e}^{x\\text{ln}a}={e}^{x\\text{ln}a}\\text{ln}a={a}^{x}\\text{ln}a.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794075642\">The corresponding integration formula follows immediately.<\/p>\r\n\r\n<div id=\"fs-id1167794075645\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Derivatives and Integrals Involving General Exponential Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793293670\">Let [latex]a&gt;0.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1167793271586\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{a}^{x}={a}^{x}\\text{ln}a[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793562027\">and<\/p>\r\n\r\n<div id=\"fs-id1167793562030\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {a}^{x}dx=\\frac{1}{\\text{ln}a}{a}^{x}+C[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793267617\">If [latex]a\\ne 1,[\/latex] then the function [latex]{a}^{x}[\/latex] is one-to-one and has a well-defined inverse. Its inverse is denoted by [latex]{\\text{log}}_{a}x.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1167793776857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={\\text{log}}_{a}x\\text{if and only if}x={a}^{y}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793929151\">Note that general logarithm functions can be written in terms of the natural logarithm. Let [latex]y={\\text{log}}_{a}x.[\/latex] Then, [latex]x={a}^{y}.[\/latex] Taking the natural logarithm of both sides of this second equation, we get<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}x&amp; =\\hfill &amp; \\text{ln}({a}^{y})\\hfill \\\\ \\hfill \\text{ln}x&amp; =\\hfill &amp; y\\text{ln}a\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{\\text{ln}x}{\\text{ln}a}\\hfill \\\\ \\hfill {\\text{log}}_{}x&amp; =\\hfill &amp; \\frac{\\text{ln}x}{\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793450615\">Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base [latex]a.[\/latex] Again, let [latex]y={\\text{log}}_{a}x.[\/latex] Then,<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{dy}{dx}&amp; =\\frac{d}{dx}({\\text{log}}_{a}x)\\hfill \\\\ &amp; =\\frac{d}{dx}(\\frac{\\text{ln}x}{\\text{ln}a})\\hfill \\\\ &amp; =(\\frac{1}{\\text{ln}a})\\frac{d}{dx}(\\text{ln}x)\\hfill \\\\ &amp; =\\frac{1}{\\text{ln}a}\u00b7\\frac{1}{x}\\hfill \\\\ &amp; =\\frac{1}{x\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167794324570\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Derivatives of General Logarithm Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793432748\">Let [latex]a&gt;0.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1167794139845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{\\text{log}}_{a}x=\\frac{1}{x\\text{ln}a}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbook exercises\">\r\n<div id=\"fs-id1167793948186\" class=\"exercise\">\r\n<h3>Example: Calculating Derivatives of General Exponential and Logarithm Functions<\/h3>\r\n<p id=\"fs-id1167793640049\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793640052\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167793298214\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793298214\"]\r\n<p id=\"fs-id1167793298214\">We need to apply the chain rule as necessary.<\/p>\r\n\r\n<ol id=\"fs-id1167793829825\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t+{t}^{2}})={2}^{2t+{t}^{2}}\\text{ln}(2)(2+2t)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)=\\frac{1}{(7{x}^{2}+4)(\\text{ln}8)}(14x)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793455288\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793543534\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}{4}^{{t}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{\\text{log}}_{3}(\\sqrt{{x}^{2}+1})[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793455291\" class=\"exercise\">[reveal-answer q=\"fs-id1167793978416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793978416\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dt}{4}^{{t}^{4}}={4}^{{t}^{4}}(\\text{ln}4)(4{t}^{3})[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{\\text{log}}_{3}(\\sqrt{{x}^{2}+1})=\\frac{x}{(\\text{ln}3)({x}^{2}+1)}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Hint<\/h4>\r\nUse the formulas and apply the chain rule as necessary.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=461&amp;end=662&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]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\/6.7TryItProblems461to662_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.7 Try It Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793499138\" class=\"textbook exercises\">\r\n<h3>Example: Integrating General Exponential Functions<\/h3>\r\nEvaluate the following integral: [latex]\\displaystyle\\int \\frac{3}{{2}^{3x}}dx.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793956537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793956537\"]\r\n<p id=\"fs-id1167793956537\">Use [latex]u\\text{-substitution}[\/latex] and let [latex]u=-3x.[\/latex] Then [latex]du=-3dx[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1167793929418\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{3}{{2}^{3x}}dx=\\displaystyle\\int 3\u00b7{2}^{-3x}dx=\\text{\u2212}\\displaystyle\\int {2}^{u}du=-\\frac{1}{\\text{ln}2}{2}^{u}+C=-\\frac{1}{\\text{ln}2}{2}^{-3x}+C.[\/latex][\/hidden-answer]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793789604\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate the following integral: [latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx.[\/latex]\r\n<div id=\"fs-id1167793789608\" class=\"exercise\">[reveal-answer q=\"fs-id1167793979123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793979123\"]\r\n<p id=\"fs-id1167793979123\">[latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx=\\frac{1}{3\\text{ln}2}{2}^{{x}^{3}}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793423554\">Use the properties of exponential functions and [latex]u\\text{-substitution}[\/latex] as necessary.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=664&amp;end=789&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266836\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266836\"]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\/6.7TryItProblems664to789_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.7 Try It Problems\" here (opens in new window)<\/a>.[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize the derivative and integral of the exponential function.<\/li>\n<li>Prove properties of logarithms and exponential functions using integrals.<\/li>\n<li>Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.<\/li>\n<\/ul>\n<\/div>\n<h2>The Exponential Function<\/h2>\n<p id=\"fs-id1167794051417\">We now turn our attention to the function [latex]{e}^{x}.[\/latex] Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by [latex]\\text{exp}x.[\/latex] Then,<\/p>\n<div id=\"fs-id1167793514717\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{exp}(\\text{ln}x)=x\\text{ for }x>0\\text{ and }\\text{ln}(\\text{exp}x)=x\\text{ for all }x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793932178\">The following figure shows the graphs of [latex]\\text{exp}x[\/latex] and [latex]\\text{ln}x.[\/latex]<\/p>\n<div style=\"width: 431px\" 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\/11213328\/CNX_Calc_Figure_06_07_004.jpg\" alt=\"This figure is a graph. It has three curves. The first curve is labeled exp x. It is an increasing curve with the x-axis as a horizontal asymptote. It intersects the y-axis at y=1. The second curve is a diagonal line through the origin. The third curve is labeled lnx. It is an increasing curve with the y-axis as an vertical axis. It intersects the x-axis at x=1.\" width=\"421\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The graphs of [latex]\\text{ln}x[\/latex] and [latex]\\text{exp}x.[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167793905821\">We hypothesize that [latex]\\text{exp}x={e}^{x}.[\/latex] For rational values of [latex]x,[\/latex] this is easy to show. If [latex]x[\/latex] is rational, then we have [latex]\\text{ln}({e}^{x})=x\\text{ln}e=x.[\/latex] Thus, when [latex]x[\/latex] is rational, [latex]{e}^{x}=\\text{exp}x.[\/latex] For irrational values of [latex]x,[\/latex] we simply define [latex]{e}^{x}[\/latex] as the inverse function of [latex]\\text{ln}x.[\/latex]<\/p>\n<div id=\"fs-id1167793579578\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167793367959\">For any real number [latex]x,[\/latex] define [latex]y={e}^{x}[\/latex] to be the number for which<\/p>\n<div id=\"fs-id1167793447883\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{ln}y=\\text{ln}({e}^{x})=x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793489536\">Then we have [latex]{e}^{x}=\\text{exp}(x)[\/latex] for all [latex]x,[\/latex] and thus<\/p>\n<div id=\"fs-id1167793564127\" class=\"equation\" style=\"text-align: center;\">[latex]{e}^{\\text{ln}x}=x\\text{ for }x>0\\text{ and }\\text{ln}({e}^{x})=x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794036848\">for all [latex]x.[\/latex]<\/p>\n<h2>Properties of the Exponential Function<\/h2>\n<p id=\"fs-id1167793516179\">Since the exponential function was defined in terms of an inverse function, and not in terms of a power of [latex]e,[\/latex] we must verify that the usual laws of exponents hold for the function [latex]{e}^{x}.[\/latex]<\/p>\n<div id=\"fs-id1167794176164\" class=\"textbox shaded\">\n<h3>Properties of the Exponential Function<\/h3>\n<p id=\"fs-id1167794210866\">If [latex]p[\/latex] and [latex]q[\/latex] are any real numbers and [latex]r[\/latex] is a rational number, then<\/p>\n<ol id=\"fs-id1167793361174\">\n<li>[latex]{e}^{p}{e}^{q}={e}^{p+q}[\/latex]<\/li>\n<li>[latex]\\frac{{e}^{p}}{{e}^{q}}={e}^{p-q}[\/latex]<\/li>\n<li>[latex]{({e}^{p})}^{r}={e}^{pr}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167794064034\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1167793420459\">Note that if [latex]p[\/latex] and [latex]q[\/latex] are rational, the properties hold. However, if [latex]p[\/latex] or [latex]q[\/latex] are irrational, we must apply the inverse function definition of [latex]{e}^{x}[\/latex] and verify the properties. Only the first property is verified here; the other two are left to you. We have<\/p>\n<div id=\"fs-id1167793301084\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}({e}^{p}{e}^{q})=\\text{ln}({e}^{p})+\\text{ln}({e}^{q})=p+q=\\text{ln}({e}^{p+q}).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793937142\">Since [latex]\\text{ln}x[\/latex] is one-to-one, then<\/p>\n<div id=\"fs-id1167794329135\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{e}^{p}{e}^{q}={e}^{p+q}.[\/latex]<\/div>\n<p id=\"fs-id1167793503220\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1167793443585\">As with part iv. of the logarithm properties, we can extend property iii. to irrational values of [latex]r,[\/latex] and we do so by the end of the section.<\/p>\n<p id=\"fs-id1167793952464\">We also want to verify the differentiation formula for the function [latex]y={e}^{x}.[\/latex] To do this, we need to use implicit differentiation. Let [latex]y={e}^{x}.[\/latex] Then<\/p>\n<div id=\"fs-id1167794095942\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}y& =\\hfill & x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{ln}y& =\\hfill & \\frac{d}{dx}x\\hfill \\\\ \\hfill \\frac{1}{y}\\frac{dy}{dx}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{dy}{dx}& =\\hfill & y.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794147337\">Thus, we see<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{e}^{x}={e}^{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793370445\">as desired, which leads immediately to the integration formula<\/p>\n<div id=\"fs-id1167793924216\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}dx={e}^{x}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793272155\">We apply these formulas in the following examples.<\/p>\n<div id=\"fs-id1167793272158\" class=\"textbook exercises\">\n<h3>Example: Using Properties of Exponential Functions<\/h3>\n<p id=\"fs-id1167793393570\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793393574\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{e}^{3{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794293256\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794293256\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794293256\">We apply the chain rule as necessary.<\/p>\n<ol id=\"fs-id1167794293259\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793372329\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793374990\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793374993\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\frac{{e}^{{x}^{2}}}{{e}^{5x}})[\/latex]<\/li>\n<li>[latex]\\frac{d}{dt}{({e}^{2t})}^{3}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793442905\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793872351\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793872351\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793872351\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\frac{{e}^{{x}^{2}}}{{e}^{5x}})={e}^{{x}^{2}-5x}(2x-5)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dt}{({e}^{2t})}^{3}=6{e}^{6t}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793568428\">Use the properties of exponential functions and the chain rule as necessary.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=238&amp;end=356&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\/6.7TryItProblems238to356_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.7 Try It Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793944672\" class=\"textbook exercises\">\n<h3>Example: Using Properties of Exponential Functions<\/h3>\n<p>Evaluate the following integral: [latex]\\displaystyle\\int 2x{e}^{\\text{\u2212}{x}^{2}}dx.[\/latex]<\/p>\n<div id=\"fs-id1167794031059\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793501968\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793501968\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793501968\">Using [latex]u[\/latex]-substitution, let [latex]u=\\text{\u2212}{x}^{2}.[\/latex] Then [latex]du=-2xdx,[\/latex] and we have<\/p>\n<div id=\"fs-id1167793956578\" class=\"equation unnumbered\">[latex]\\displaystyle\\int 2x{e}^{\\text{\u2212}{x}^{2}}dx=\\text{\u2212}\\displaystyle\\int {e}^{u}du=\\text{\u2212}{e}^{u}+C=\\text{\u2212}{e}^{\\text{\u2212}{x}^{2}}+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793478795\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate the following integral: [latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793510888\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793510888\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793510888\">[latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx=-\\frac{4}{3}{e}^{-3x}+C[\/latex]<\/p>\n<\/div>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793943946\">Use the properties of exponential functions and [latex]u\\text{-substitution}[\/latex] as necessary.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=359&amp;end=460&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\/6.7TryItProblems359to460_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.7 Try It Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20047\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20047&theme=oea&iframe_resize_id=ohm20047&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>General Logarithmic and Exponential Functions<\/h2>\n<p id=\"fs-id1167793400845\">We close this section by looking at exponential functions and logarithms with bases other than [latex]e.[\/latex] Exponential functions are functions of the form [latex]f(x)={a}^{x}.[\/latex] Note that unless [latex]a=e,[\/latex] we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let\u2019s rectify that here by defining the function [latex]f(x)={a}^{x}[\/latex] in terms of the exponential function [latex]{e}^{x}.[\/latex] We then examine logarithms with bases other than [latex]e[\/latex] as inverse functions of exponential functions.<\/p>\n<div id=\"fs-id1167793285142\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>For any [latex]a>0,[\/latex] and for any real number [latex]x,[\/latex] define [latex]y={a}^{x}[\/latex] as follows:<\/p>\n<div id=\"fs-id1167793420642\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={a}^{x}={e}^{x\\text{ln}a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793455076\">Now [latex]{a}^{x}[\/latex] is defined rigorously for all values of [latex]x[\/latex]. This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of [latex]r.[\/latex] It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.<\/p>\n<p id=\"fs-id1167793384514\">Let\u2019s now apply this definition to calculate a differentiation formula for [latex]{a}^{x}.[\/latex] We have<\/p>\n<div id=\"fs-id1167793559160\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{a}^{x}=\\frac{d}{dx}{e}^{x\\text{ln}a}={e}^{x\\text{ln}a}\\text{ln}a={a}^{x}\\text{ln}a.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794075642\">The corresponding integration formula follows immediately.<\/p>\n<div id=\"fs-id1167794075645\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Derivatives and Integrals Involving General Exponential Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1167793293670\">Let [latex]a>0.[\/latex] Then,<\/p>\n<div id=\"fs-id1167793271586\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{a}^{x}={a}^{x}\\text{ln}a[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793562027\">and<\/p>\n<div id=\"fs-id1167793562030\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {a}^{x}dx=\\frac{1}{\\text{ln}a}{a}^{x}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793267617\">If [latex]a\\ne 1,[\/latex] then the function [latex]{a}^{x}[\/latex] is one-to-one and has a well-defined inverse. Its inverse is denoted by [latex]{\\text{log}}_{a}x.[\/latex] Then,<\/p>\n<div id=\"fs-id1167793776857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={\\text{log}}_{a}x\\text{if and only if}x={a}^{y}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793929151\">Note that general logarithm functions can be written in terms of the natural logarithm. Let [latex]y={\\text{log}}_{a}x.[\/latex] Then, [latex]x={a}^{y}.[\/latex] Taking the natural logarithm of both sides of this second equation, we get<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}x& =\\hfill & \\text{ln}({a}^{y})\\hfill \\\\ \\hfill \\text{ln}x& =\\hfill & y\\text{ln}a\\hfill \\\\ \\hfill y& =\\hfill & \\frac{\\text{ln}x}{\\text{ln}a}\\hfill \\\\ \\hfill {\\text{log}}_{}x& =\\hfill & \\frac{\\text{ln}x}{\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793450615\">Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base [latex]a.[\/latex] Again, let [latex]y={\\text{log}}_{a}x.[\/latex] Then,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{dy}{dx}& =\\frac{d}{dx}({\\text{log}}_{a}x)\\hfill \\\\ & =\\frac{d}{dx}(\\frac{\\text{ln}x}{\\text{ln}a})\\hfill \\\\ & =(\\frac{1}{\\text{ln}a})\\frac{d}{dx}(\\text{ln}x)\\hfill \\\\ & =\\frac{1}{\\text{ln}a}\u00b7\\frac{1}{x}\\hfill \\\\ & =\\frac{1}{x\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167794324570\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Derivatives of General Logarithm Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1167793432748\">Let [latex]a>0.[\/latex] Then,<\/p>\n<div id=\"fs-id1167794139845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}{\\text{log}}_{a}x=\\frac{1}{x\\text{ln}a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbook exercises\">\n<div id=\"fs-id1167793948186\" class=\"exercise\">\n<h3>Example: Calculating Derivatives of General Exponential and Logarithm Functions<\/h3>\n<p id=\"fs-id1167793640049\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793640052\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793298214\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793298214\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793298214\">We need to apply the chain rule as necessary.<\/p>\n<ol id=\"fs-id1167793829825\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t+{t}^{2}})={2}^{2t+{t}^{2}}\\text{ln}(2)(2+2t)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)=\\frac{1}{(7{x}^{2}+4)(\\text{ln}8)}(14x)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793455288\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793543534\">Evaluate the following derivatives:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}{4}^{{t}^{4}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{\\text{log}}_{3}(\\sqrt{{x}^{2}+1})[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793455291\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793978416\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793978416\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dt}{4}^{{t}^{4}}={4}^{{t}^{4}}(\\text{ln}4)(4{t}^{3})[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{\\text{log}}_{3}(\\sqrt{{x}^{2}+1})=\\frac{x}{(\\text{ln}3)({x}^{2}+1)}[\/latex]<\/li>\n<\/ol>\n<h4>Hint<\/h4>\n<p>Use the formulas and apply the chain rule as necessary.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=461&amp;end=662&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=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" 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\/6.7TryItProblems461to662_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.7 Try It Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793499138\" class=\"textbook exercises\">\n<h3>Example: Integrating General Exponential Functions<\/h3>\n<p>Evaluate the following integral: [latex]\\displaystyle\\int \\frac{3}{{2}^{3x}}dx.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793956537\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793956537\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793956537\">Use [latex]u\\text{-substitution}[\/latex] and let [latex]u=-3x.[\/latex] Then [latex]du=-3dx[\/latex] and we have<\/p>\n<div id=\"fs-id1167793929418\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{3}{{2}^{3x}}dx=\\displaystyle\\int 3\u00b7{2}^{-3x}dx=\\text{\u2212}\\displaystyle\\int {2}^{u}du=-\\frac{1}{\\text{ln}2}{2}^{u}+C=-\\frac{1}{\\text{ln}2}{2}^{-3x}+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793789604\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate the following integral: [latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx.[\/latex]<\/p>\n<div id=\"fs-id1167793789608\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793979123\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793979123\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793979123\">[latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx=\\frac{1}{3\\text{ln}2}{2}^{{x}^{3}}+C[\/latex]<\/p>\n<\/div>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793423554\">Use the properties of exponential functions and [latex]u\\text{-substitution}[\/latex] as necessary.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=664&amp;end=789&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=\"q266836\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266836\" 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\/6.7TryItProblems664to789_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.7 Try It Problems&#8221; here (opens in new window)<\/a>.<\/div>\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-2757\">\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>6.7 Try It Problems. <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":28,"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 free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"6.7 Try It Problems\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2757","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2757","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2757\/revisions"}],"predecessor-version":[{"id":4907,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2757\/revisions\/4907"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/65"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2757\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2757"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2757"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2757"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}