{"id":484,"date":"2025-02-13T19:45:22","date_gmt":"2025-02-13T19:45:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-exponential-functions-and-logarithms-learn-it-1\/"},"modified":"2025-02-13T19:45:22","modified_gmt":"2025-02-13T19:45:22","slug":"integrals-exponential-functions-and-logarithms-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-exponential-functions-and-logarithms-learn-it-1\/","title":{"raw":"Integrals, Exponential Functions, and Logarithms: Learn It 1","rendered":"Integrals, Exponential Functions, and Logarithms: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Understand the natural logarithm and the mathematical constant e using integrals<\/li>\n\t<li>Identify how to differentiate the natural logarithm function<\/li>\n\t<li>Perform integrations where the natural logarithm is involved<\/li>\n\t<li>Understand how to find derivatives and integrals of exponential functions<\/li>\n\t<li>Convert logarithmic and exponential expressions to base e forms<\/li>\n<\/ul>\n<\/section>\n<p>We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number [latex]e[\/latex] is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.<\/p>\n<p>For purposes of this section, assume we have not yet defined the natural logarithm, the number [latex]e[\/latex], or any of the integration and differentiation formulas associated with these functions.&nbsp;<\/p>\n<h2>The Natural Logarithm as an Integral<\/h2>\n<p id=\"fs-id1167793958127\">Recall the power rule for integrals:<\/p>\n<div id=\"fs-id1167793299462\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C,n\\ne \\text{\u2212}1.[\/latex]<\/div>\n<p>This rule doesn't work for [latex]n=-1,[\/latex] as it would force us to divide by zero. So, how do we integrate [latex]\\displaystyle\\int \\frac{1}{x}dx?[\/latex] According to the Fundamental Theorem of Calculus, we know that [latex]{\\displaystyle\\int }_{1}^{x}\\dfrac{1}{t}dt[\/latex] is an antiderivative of [latex]\\frac{1}{x}.[\/latex]&nbsp;<\/p>\n<p>Therefore, we define the natural logarithm function as follows:<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>defining the natural logarithm<\/h3>\n<p id=\"fs-id1167793953542\">For [latex]x&gt;0,[\/latex] define the natural logarithm function by<\/p>\n<div id=\"fs-id1167793550098\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{ln}x={\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt[\/latex]<\/div>\n<\/section>\n<p>For [latex]x&gt;1,[\/latex] this represents the area under the curve [latex]y=\\frac{1}{t}[\/latex] from [latex]1[\/latex] to [latex]x.[\/latex] For [latex]x&lt;1,[\/latex] we have,<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt=\\text{\u2212}{\\displaystyle\\int }_{x}^{1}\\frac{1}{t}dt,[\/latex],<\/p>\n<p>so it is the negative of the area under the curve from [latex]x\\text{ to }1[\/latex].<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"589\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213320\/CNX_Calc_Figure_06_07_001.jpg\" alt=\"This figure has two graphs. The first is the curve y=1\/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1. The area is labeled \u201carea=lnx\u201d. The second graph is the same curve y=1\/t. It has shaded area under the curve bounded to the right by x=1. It is labeled \u201carea=-lnx\u201d.\" width=\"589\" height=\"311\"> Figure 1. (a) When [latex]x&gt;1,[\/latex] the natural logarithm is the area under the curve [latex]y=\\frac{1}{t}[\/latex] from [latex]1\\text{ to }x.[\/latex] (b) When [latex]x&lt;1,[\/latex] the natural logarithm is the negative of the area under the curve from [latex]x[\/latex] to 1.[\/caption]\n\n<p id=\"fs-id1167793932688\">Notice that [latex]\\text{ln}1=0.[\/latex] Furthermore, since [latex]\\frac{1}{t} &gt; 0 [\/latex] for [latex]t &gt; 0[\/latex], [latex]\\ln x [\/latex] is increasing for [latex]x &gt; 0[\/latex].<\/p>\n<h2>Properties of the Natural Logarithm<\/h2>\n<p id=\"fs-id1167794139011\">Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>derivative of the natural logarithm<\/h3>\n<p id=\"fs-id1167794050615\">For [latex]x&gt;0,[\/latex] the derivative of the natural logarithm is given by<\/p>\n<center>[latex]\\frac{d}{dx}\\text{ln}x=\\dfrac{1}{x}.[\/latex]<\/center><\/section>\n<p>The function [latex]\\text{ln}x[\/latex] is differentiable; therefore, it is continuous.<\/p>\n<p id=\"fs-id1167794141119\">A graph of [latex]\\text{ln}x[\/latex] is shown below. Notice that it is continuous throughout its domain of [latex](0,\\infty ).[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"266\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213322\/CNX_Calc_Figure_06_07_002.jpg\" alt=\"This figure is a graph. It is an increasing curve labeled f(x)=lnx. The curve is increasing with the y-axis as an asymptote. The curve intersects the x-axis at x=1.\" width=\"266\" height=\"347\"> Figure 2. The graph of [latex]f(x)=\\text{ln}x[\/latex] shows that it is a continuous function.[\/caption]\n\n<section class=\"textbox example\">\n<p id=\"fs-id1167794168088\">Calculate the following derivatives:<\/p>\n<ol id=\"fs-id1167793966989\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}{(\\text{ln}(3x))}^{2}[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167794171377\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794171377\"]\n\n<p id=\"fs-id1167794171377\">We need to apply the chain rule in both cases.<\/p>\n<ol id=\"fs-id1167793926297\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)=\\frac{15{x}^{2}}{5{x}^{3}-2}[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}{(\\text{ln}(3x))}^{2}=\\frac{2(\\text{ln}(3x))\u00b73}{3x}=\\frac{2(\\text{ln}(3x))}{x}[\/latex]<\/li>\n<\/ol>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<p id=\"fs-id1167793609503\">Note that if we use the absolute value function and create a new function [latex]\\text{ln}|x|,[\/latex] we can extend the domain of the natural logarithm to include [latex]x&lt;0.[\/latex] Then [latex](\\frac{d}{dx})\\text{ln}|x|=\\frac{1}{x}.[\/latex] This gives rise to the familiar integration formula.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>integral of (1\/[latex]u[\/latex]) du<\/h3>\n<p id=\"fs-id1167793832055\">The natural logarithm is the antiderivative of the function [latex]f(u)=1\\text{\/}u\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793973017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{1}{u}du=\\text{ln}|u|+C[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Calculate the integral [latex]\\displaystyle\\int \\frac{{x}^{2}}{{x}^{3}+6}dx.[\/latex]<\/p>\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793525036\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793525036\"]\n\n<p id=\"fs-id1167793525036\">[latex]\\displaystyle\\int \\frac{{x}^{2}}{{x}^{3}+6}dx=\\frac{1}{3}\\text{ln}|{x}^{3}+6|+C[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6_9o-2mK1tk?controls=0&amp;start=137&amp;end=234&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\n<p>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\/6.7TryItProblems137to234_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]<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288449[\/ohm_question]<\/p>\n<\/section>\n<p id=\"fs-id1167794062403\">Although we have called our function a \u201clogarithm,\u201d we have not actually proved that any of the properties of logarithms hold for this function. We do so here.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>properties of the natural logarithm<\/h3>\n<p id=\"fs-id1167793936147\">If [latex]a,b&gt;0[\/latex] and [latex]r[\/latex] is a rational number, then<\/p>\n<ol id=\"fs-id1167793271023\" style=\"list-style-type: lower-roman;\">\n\t<li>[latex]\\text{ln}1=0[\/latex]<\/li>\n\t<li>[latex]\\text{ln}(ab)=\\text{ln}a+\\text{ln}b[\/latex]<\/li>\n\t<li>[latex]\\text{ln}(\\frac{a}{b})=\\text{ln}a-\\text{ln}b[\/latex]<\/li>\n\t<li>[latex]\\text{ln}({a}^{r})=r\\text{ln}a[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr>\n<ol id=\"fs-id1167793977078\">\n\t<li>By definition, [latex]\\text{ln}1={\\displaystyle\\int }_{1}^{1}\\frac{1}{t}dt=0.[\/latex]<\/li>\n\t<li>We have<br>\n<div id=\"fs-id1167793964775\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}(ab)={\\displaystyle\\int }_{1}^{ab}\\frac{1}{t}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{a}^{ab}\\frac{1}{t}dt.[\/latex]<\/div>\n<p>Use [latex]u\\text{-substitution}[\/latex] on the last integral in this expression. Let [latex]u=t\\text{\/}a.[\/latex] Then [latex]du=(1\\text{\/}a)dt.[\/latex] Furthermore, when [latex]t=a,u=1,[\/latex] and when [latex]t=ab,u=b.[\/latex] So we get<\/p>\n<div id=\"fs-id1167793490878\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}(ab)={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{a}^{ab}\\frac{1}{t}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{1}^{ab}\\frac{a}{t}\u00b7\\frac{1}{a}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{1}^{b}\\frac{1}{u}du=\\text{ln}a+\\text{ln}b.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/li>\n\t<li>Note that<br>\n<div id=\"fs-id1167793951598\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\text{ln}({x}^{r})=\\frac{r{x}^{r-1}}{{x}^{r}}=\\frac{r}{x}.[\/latex]<\/div>\n<p>Furthermore,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(r\\text{ln}x)=\\frac{r}{x}.[\/latex]<\/div>\n<p>Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have<\/p>\n<div id=\"fs-id1167793563854\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}({x}^{r})=r\\text{ln}x+C[\/latex]<\/div>\n<p>for some constant [latex]C.[\/latex] Taking [latex]x=1,[\/latex] we get<\/p>\n<div id=\"fs-id1167793510762\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}({1}^{r})&amp; =\\hfill &amp; r\\text{ln}(1)+C\\hfill \\\\ \\hfill 0&amp; =\\hfill &amp; r(0)+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; 0.\\hfill \\end{array}[\/latex]<\/div>\n<p>Thus [latex]\\text{ln}({x}^{r})=r\\text{ln}x[\/latex] and the proof is complete. Note that we can extend this property to irrational values of [latex]r[\/latex] later in this section.<br>\nPart iii. follows from parts ii. and iv. and the proof is left to you.<\/p>\n<\/li>\n<\/ol>\n<p id=\"fs-id1167794210729\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794003628\">Use properties of logarithms to simplify the following expression into a single logarithm:<\/p>\n<div id=\"fs-id1167793590440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}9-2\\text{ln}3+\\text{ln}(\\frac{1}{3})[\/latex]<span style=\"background-color: initial; font-size: 0.9em;\">&nbsp;<\/span><\/div>\n<p>[reveal-answer q=\"fs-id1167794337026\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794337026\"]<\/p>\n<p id=\"fs-id1167794337026\">We have<\/p>\n<div id=\"fs-id1167794337029\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}9-2\\text{ln}3+\\text{ln}(\\frac{1}{3})=\\text{ln}({3}^{2})-2\\text{ln}3+\\text{ln}({3}^{-1})=2\\text{ln}3-2\\text{ln}3-\\text{ln}3=\\text{\u2212}\\text{ln}3.[\/latex][\/hidden-answer]<\/div>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand the natural logarithm and the mathematical constant e using integrals<\/li>\n<li>Identify how to differentiate the natural logarithm function<\/li>\n<li>Perform integrations where the natural logarithm is involved<\/li>\n<li>Understand how to find derivatives and integrals of exponential functions<\/li>\n<li>Convert logarithmic and exponential expressions to base e forms<\/li>\n<\/ul>\n<\/section>\n<p>We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number [latex]e[\/latex] is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.<\/p>\n<p>For purposes of this section, assume we have not yet defined the natural logarithm, the number [latex]e[\/latex], or any of the integration and differentiation formulas associated with these functions.&nbsp;<\/p>\n<h2>The Natural Logarithm as an Integral<\/h2>\n<p id=\"fs-id1167793958127\">Recall the power rule for integrals:<\/p>\n<div id=\"fs-id1167793299462\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C,n\\ne \\text{\u2212}1.[\/latex]<\/div>\n<p>This rule doesn&#8217;t work for [latex]n=-1,[\/latex] as it would force us to divide by zero. So, how do we integrate [latex]\\displaystyle\\int \\frac{1}{x}dx?[\/latex] According to the Fundamental Theorem of Calculus, we know that [latex]{\\displaystyle\\int }_{1}^{x}\\dfrac{1}{t}dt[\/latex] is an antiderivative of [latex]\\frac{1}{x}.[\/latex]&nbsp;<\/p>\n<p>Therefore, we define the natural logarithm function as follows:<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>defining the natural logarithm<\/h3>\n<p id=\"fs-id1167793953542\">For [latex]x>0,[\/latex] define the natural logarithm function by<\/p>\n<div id=\"fs-id1167793550098\" class=\"equation\" style=\"text-align: center;\">[latex]\\text{ln}x={\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt[\/latex]<\/div>\n<\/section>\n<p>For [latex]x>1,[\/latex] this represents the area under the curve [latex]y=\\frac{1}{t}[\/latex] from [latex]1[\/latex] to [latex]x.[\/latex] For [latex]x<1,[\/latex] we have,<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt=\\text{\u2212}{\\displaystyle\\int }_{x}^{1}\\frac{1}{t}dt,[\/latex],<\/p>\n<p>so it is the negative of the area under the curve from [latex]x\\text{ to }1[\/latex].<\/p>\n<figure style=\"width: 589px\" 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\/11213320\/CNX_Calc_Figure_06_07_001.jpg\" alt=\"This figure has two graphs. The first is the curve y=1\/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1. The area is labeled \u201carea=lnx\u201d. The second graph is the same curve y=1\/t. It has shaded area under the curve bounded to the right by x=1. It is labeled \u201carea=-lnx\u201d.\" width=\"589\" height=\"311\" \/><figcaption class=\"wp-caption-text\">Figure 1. (a) When [latex]x&gt;1,[\/latex] the natural logarithm is the area under the curve [latex]y=\\frac{1}{t}[\/latex] from [latex]1\\text{ to }x.[\/latex] (b) When [latex]x&lt;1,[\/latex] the natural logarithm is the negative of the area under the curve from [latex]x[\/latex] to 1.<\/figcaption><\/figure>\n<p id=\"fs-id1167793932688\">Notice that [latex]\\text{ln}1=0.[\/latex] Furthermore, since [latex]\\frac{1}{t} > 0[\/latex] for [latex]t > 0[\/latex], [latex]\\ln x[\/latex] is increasing for [latex]x > 0[\/latex].<\/p>\n<h2>Properties of the Natural Logarithm<\/h2>\n<p id=\"fs-id1167794139011\">Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>derivative of the natural logarithm<\/h3>\n<p id=\"fs-id1167794050615\">For [latex]x>0,[\/latex] the derivative of the natural logarithm is given by<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{d}{dx}\\text{ln}x=\\dfrac{1}{x}.[\/latex]<\/div>\n<\/section>\n<p>The function [latex]\\text{ln}x[\/latex] is differentiable; therefore, it is continuous.<\/p>\n<p id=\"fs-id1167794141119\">A graph of [latex]\\text{ln}x[\/latex] is shown below. Notice that it is continuous throughout its domain of [latex](0,\\infty ).[\/latex]<\/p>\n<figure style=\"width: 266px\" 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\/11213322\/CNX_Calc_Figure_06_07_002.jpg\" alt=\"This figure is a graph. It is an increasing curve labeled f(x)=lnx. The curve is increasing with the y-axis as an asymptote. The curve intersects the x-axis at x=1.\" width=\"266\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 2. The graph of [latex]f(x)=\\text{ln}x[\/latex] shows that it is a continuous function.<\/figcaption><\/figure>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794168088\">Calculate the following derivatives:<\/p>\n<ol id=\"fs-id1167793966989\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{ln}(3x))}^{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794171377\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794171377\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794171377\">We need to apply the chain rule in both cases.<\/p>\n<ol id=\"fs-id1167793926297\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)=\\frac{15{x}^{2}}{5{x}^{3}-2}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{ln}(3x))}^{2}=\\frac{2(\\text{ln}(3x))\u00b73}{3x}=\\frac{2(\\text{ln}(3x))}{x}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1167793609503\">Note that if we use the absolute value function and create a new function [latex]\\text{ln}|x|,[\/latex] we can extend the domain of the natural logarithm to include [latex]x<0.[\/latex] Then [latex](\\frac{d}{dx})\\text{ln}|x|=\\frac{1}{x}.[\/latex] This gives rise to the familiar integration formula.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>integral of (1\/[latex]u[\/latex]) du<\/h3>\n<p id=\"fs-id1167793832055\">The natural logarithm is the antiderivative of the function [latex]f(u)=1\\text{\/}u\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167793973017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{1}{u}du=\\text{ln}|u|+C[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Calculate the integral [latex]\\displaystyle\\int \\frac{{x}^{2}}{{x}^{3}+6}dx.[\/latex]<\/p>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793525036\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793525036\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793525036\">[latex]\\displaystyle\\int \\frac{{x}^{2}}{{x}^{3}+6}dx=\\frac{1}{3}\\text{ln}|{x}^{3}+6|+C[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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=137&amp;end=234&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>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\/6.7TryItProblems137to234_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>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288449\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288449&theme=lumen&iframe_resize_id=ohm288449&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<p id=\"fs-id1167794062403\">Although we have called our function a \u201clogarithm,\u201d we have not actually proved that any of the properties of logarithms hold for this function. We do so here.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>properties of the natural logarithm<\/h3>\n<p id=\"fs-id1167793936147\">If [latex]a,b>0[\/latex] and [latex]r[\/latex] is a rational number, then<\/p>\n<ol id=\"fs-id1167793271023\" style=\"list-style-type: lower-roman;\">\n<li>[latex]\\text{ln}1=0[\/latex]<\/li>\n<li>[latex]\\text{ln}(ab)=\\text{ln}a+\\text{ln}b[\/latex]<\/li>\n<li>[latex]\\text{ln}(\\frac{a}{b})=\\text{ln}a-\\text{ln}b[\/latex]<\/li>\n<li>[latex]\\text{ln}({a}^{r})=r\\text{ln}a[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<ol id=\"fs-id1167793977078\">\n<li>By definition, [latex]\\text{ln}1={\\displaystyle\\int }_{1}^{1}\\frac{1}{t}dt=0.[\/latex]<\/li>\n<li>We have\n<div id=\"fs-id1167793964775\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}(ab)={\\displaystyle\\int }_{1}^{ab}\\frac{1}{t}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{a}^{ab}\\frac{1}{t}dt.[\/latex]<\/div>\n<p>Use [latex]u\\text{-substitution}[\/latex] on the last integral in this expression. Let [latex]u=t\\text{\/}a.[\/latex] Then [latex]du=(1\\text{\/}a)dt.[\/latex] Furthermore, when [latex]t=a,u=1,[\/latex] and when [latex]t=ab,u=b.[\/latex] So we get<\/p>\n<div id=\"fs-id1167793490878\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}(ab)={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{a}^{ab}\\frac{1}{t}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{1}^{ab}\\frac{a}{t}\u00b7\\frac{1}{a}dt={\\displaystyle\\int }_{1}^{a}\\frac{1}{t}dt+{\\displaystyle\\int }_{1}^{b}\\frac{1}{u}du=\\text{ln}a+\\text{ln}b.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/li>\n<li>Note that\n<div id=\"fs-id1167793951598\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\text{ln}({x}^{r})=\\frac{r{x}^{r-1}}{{x}^{r}}=\\frac{r}{x}.[\/latex]<\/div>\n<p>Furthermore,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(r\\text{ln}x)=\\frac{r}{x}.[\/latex]<\/div>\n<p>Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have<\/p>\n<div id=\"fs-id1167793563854\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}({x}^{r})=r\\text{ln}x+C[\/latex]<\/div>\n<p>for some constant [latex]C.[\/latex] Taking [latex]x=1,[\/latex] we get<\/p>\n<div id=\"fs-id1167793510762\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}({1}^{r})& =\\hfill & r\\text{ln}(1)+C\\hfill \\\\ \\hfill 0& =\\hfill & r(0)+C\\hfill \\\\ \\hfill C& =\\hfill & 0.\\hfill \\end{array}[\/latex]<\/div>\n<p>Thus [latex]\\text{ln}({x}^{r})=r\\text{ln}x[\/latex] and the proof is complete. Note that we can extend this property to irrational values of [latex]r[\/latex] later in this section.<br \/>\nPart iii. follows from parts ii. and iv. and the proof is left to you.<\/p>\n<\/li>\n<\/ol>\n<p id=\"fs-id1167794210729\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794003628\">Use properties of logarithms to simplify the following expression into a single logarithm:<\/p>\n<div id=\"fs-id1167793590440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}9-2\\text{ln}3+\\text{ln}(\\frac{1}{3})[\/latex]<span style=\"background-color: initial; font-size: 0.9em;\">&nbsp;<\/span><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794337026\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794337026\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794337026\">We have<\/p>\n<div id=\"fs-id1167794337029\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}9-2\\text{ln}3+\\text{ln}(\\frac{1}{3})=\\text{ln}({3}^{2})-2\\text{ln}3+\\text{ln}({3}^{-1})=2\\text{ln}3-2\\text{ln}3-\\text{ln}3=\\text{\u2212}\\text{ln}3.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":5,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":479,"module-header":"","content_attributions":[{"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":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/484"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/484\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/479"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/484\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=484"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=484"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=484"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}