{"id":597,"date":"2021-02-04T16:34:00","date_gmt":"2021-02-04T16:34:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=597"},"modified":"2022-03-16T22:51:48","modified_gmt":"2022-03-16T22:51:48","slug":"natural-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/natural-logarithms\/","title":{"raw":"Natural Logarithms","rendered":"Natural Logarithms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the definition of the natural logarithm as an integral<\/li>\r\n \t<li>Recognize the derivative of the natural logarithm<\/li>\r\n \t<li>Integrate functions involving the natural logarithmic function<\/li>\r\n \t<li>Define the number \ud835\udc52 through an integral<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Natural Logarithm as an Integral<\/h2>\r\n<p id=\"fs-id1167793958127\">Recall the power rule for integrals:<\/p>\r\n\r\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>\r\nClearly, this does not work when [latex]n=-1,[\/latex] as it would force us to divide by zero. So, what do we do with [latex]\\displaystyle\\int \\frac{1}{x}dx?[\/latex] Recall from the Fundamental Theorem of Calculus that [latex]{\\displaystyle\\int }_{1}^{x}\\dfrac{1}{t}dt[\/latex] is an antiderivative of [latex]\\frac{1}{x}.[\/latex] Therefore, we can make the following definition.\r\n<div id=\"fs-id1167793370244\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793953542\">For [latex]x&gt;0,[\/latex] define the natural logarithm function by<\/p>\r\n\r\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>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793617974\">For [latex]x&gt;1,[\/latex] this is just the area under the curve [latex]y=1\\text{\/}t[\/latex] from 1 to [latex]x.[\/latex] For [latex]x&lt;1,[\/latex] we have [latex]{\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt=\\text{\u2212}{\\displaystyle\\int }_{x}^{1}\\frac{1}{t}dt,[\/latex] so in this case it is the negative of the area under the curve from [latex]x\\text{ to }1[\/latex] (see the following figure).<\/p>\r\n&nbsp;\r\n\r\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]\r\n<p id=\"fs-id1167793932688\">Notice that [latex]\\text{ln}1=0.[\/latex] Furthermore, the function [latex]y=\\frac{1}{t}&gt;0[\/latex] for [latex]x&gt;0.[\/latex] Therefore, by the properties of integrals, it is clear that [latex]\\text{ln}x[\/latex] is increasing for [latex]x&gt;0.[\/latex]<\/p>\r\n\r\n<h2>Properties of the Natural Logarithm<\/h2>\r\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>\r\n\r\n<div id=\"fs-id1167794186831\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Derivative of the Natural Logarithm<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794050615\">For [latex]x&gt;0,[\/latex] the derivative of the natural logarithm is given by<\/p>\r\n\r\n<div id=\"fs-id1167793432137\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\text{ln}x=\\dfrac{1}{x}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793879890\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Corollary to the Derivative of the Natural Logarithm<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794128288\">The function [latex]\\text{ln}x[\/latex] is differentiable; therefore, it is continuous.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794141119\">A graph of [latex]\\text{ln}x[\/latex] is shown in Figure 2. Notice that it is continuous throughout its domain of [latex](0,\\infty ).[\/latex]<\/p>\r\n\r\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]\r\n<div id=\"fs-id1167793967054\" class=\"textbook exercises\">\r\n<h3>Example: Calculating Derivatives of Natural Logarithms<\/h3>\r\n<p id=\"fs-id1167794168088\">Calculate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793966989\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{(\\text{ln}(3x))}^{2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793424342\" class=\"exercise\">[reveal-answer q=\"fs-id1167794171377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794171377\"]\r\n<p id=\"fs-id1167794171377\">We need to apply the chain rule in both cases.<\/p>\r\n\r\n<ol id=\"fs-id1167793926297\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}\\text{ln}(5{x}^{3}-2)=\\frac{15{x}^{2}}{5{x}^{3}-2}[\/latex]<\/li>\r\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>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793398778\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167794334473\">Calculate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793964608\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}\\text{ln}(2{x}^{2}+x)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{(\\text{ln}({x}^{3}))}^{2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793369601\" class=\"exercise\">[reveal-answer q=\"fs-id1167793259644\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793259644\"]\r\n<ol id=\"fs-id1167793259644\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}\\text{ln}(2{x}^{2}+x)=\\frac{4x+1}{2{x}^{2}+x}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{(\\text{ln}({x}^{3}))}^{2}=\\frac{6\\text{ln}({x}^{3})}{x}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167794187630\">Apply the differentiation formula just provided and use the chain rule 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=0&amp;end=135&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\/6.7TryItProblems0to135_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]16155[\/ohm_question]\r\n\r\n<\/div>\r\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](d\\text{\/}(dx))\\text{ln}|x|=1\\text{\/}x.[\/latex] This gives rise to the familiar integration formula.<\/p>\r\n\r\n<div id=\"fs-id1167793432204\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Integral of (1\/[latex]u[\/latex]) <em>du<\/em><\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793832055\">The natural logarithm is the antiderivative of the function [latex]f(u)=1\\text{\/}u\\text{:}[\/latex]<\/p>\r\n\r\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>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793423035\" class=\"textbook exercises\">\r\n<h3>Example: Calculating Integrals Involving Natural Logarithms<\/h3>\r\nCalculate the integral [latex]\\displaystyle\\int \\frac{x}{{x}^{2}+4}dx.[\/latex]\r\n<div id=\"fs-id1167793948151\" class=\"exercise\">[reveal-answer q=\"fs-id1167793887223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793887223\"]\r\n<p id=\"fs-id1167793887223\">Using [latex]u[\/latex]-substitution, let [latex]u={x}^{2}+4.[\/latex] Then [latex]du=2xdx[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1167793971678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{x}{{x}^{2}+4}dx=\\frac{1}{2}\\displaystyle\\int \\frac{1}{u}du\\frac{1}{2}\\text{ln}|u|+C=\\frac{1}{2}\\text{ln}|{x}^{2}+4|+C=\\frac{1}{2}\\text{ln}({x}^{2}+4)+C.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794041523\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nCalculate the integral [latex]\\displaystyle\\int \\frac{{x}^{2}}{{x}^{3}+6}dx.[\/latex]\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793525036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793525036\"]\r\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>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793862633\">Apply the integration formula provided earlier and use [latex]u[\/latex]-substitution 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=137&amp;end=234&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\/6.7TryItProblems137to234_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]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20050[\/ohm_question]\r\n\r\n<\/div>\r\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>\r\n\r\n<div id=\"fs-id1167793366787\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Properties of the Natural Logarithm<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793936147\">If [latex]a,b&gt;0[\/latex] and [latex]r[\/latex] is a rational number, then<\/p>\r\n\r\n<ol id=\"fs-id1167793271023\">\r\n \t<li>[latex]\\text{ln}1=0[\/latex]<\/li>\r\n \t<li>[latex]\\text{ln}(ab)=\\text{ln}a+\\text{ln}b[\/latex]<\/li>\r\n \t<li>[latex]\\text{ln}(\\frac{a}{b})=\\text{ln}a-\\text{ln}b[\/latex]<\/li>\r\n \t<li>[latex]\\text{ln}({a}^{r})=r\\text{ln}a[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167793509941\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<ol id=\"fs-id1167793977078\">\r\n \t<li>By definition, [latex]\\text{ln}1={\\displaystyle\\int }_{1}^{1}\\frac{1}{t}dt=0.[\/latex]<\/li>\r\n \t<li>We have\r\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>\r\nUse [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\r\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>\r\n&nbsp;<\/li>\r\n \t<li>Note that\r\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>\r\nFurthermore,\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(r\\text{ln}x)=\\frac{r}{x}.[\/latex]<\/div>\r\nSince the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have\r\n<div id=\"fs-id1167793563854\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}({x}^{r})=r\\text{ln}x+C[\/latex]<\/div>\r\nfor some constant [latex]C.[\/latex] Taking [latex]x=1,[\/latex] we get\r\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>\r\nThus [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.\r\nPart iii. follows from parts ii. and iv. and the proof is left to you.<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1167794210729\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div class=\"textbook exercises\">\r\n<div id=\"fs-id1167794077125\" class=\"exercise\">\r\n<h3>Example: Using Properties of Logarithms<\/h3>\r\n<p id=\"fs-id1167794003628\">Use properties of logarithms to simplify the following expression into a single logarithm:<\/p>\r\n\r\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;\">\u00a0<\/span><\/div>\r\n[reveal-answer q=\"fs-id1167794337026\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794337026\"]\r\n<p id=\"fs-id1167794337026\">We have<\/p>\r\n\r\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>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793603772\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793607833\">Use properties of logarithms to simplify the following expression into a single logarithm:<\/p>\r\n\r\n<div id=\"fs-id1167793607837\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}8-\\text{ln}2-\\text{ln}(\\frac{1}{4}).[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793949542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793949542\"][latex]4\\text{ln}2[\/latex]\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793949514\">Apply the properties of logarithms.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793514534\" class=\"bc-section section\">\r\n<h2>Defining the Number [latex]e[\/latex]<\/h2>\r\n<p id=\"fs-id1167793877994\">Now that we have the natural logarithm defined, we can use that function to define the number [latex]e.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167793249163\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793249167\">The number [latex]e[\/latex] is defined to be the real number such that<\/p>\r\n\r\n<div id=\"fs-id1167794140148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}e=1[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793871307\">To put it another way, the area under the curve [latex]y=1\\text{\/}t[\/latex] between [latex]t=1[\/latex] and [latex]t=e[\/latex] is 1 (Figure 3). The proof that such a number exists and is unique is left to you. (<em>Hint<\/em>: Use the Intermediate Value Theorem to prove existence and the fact that [latex]\\text{ln}x[\/latex] is increasing to prove uniqueness.)<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213325\/CNX_Calc_Figure_06_07_003.jpg\" alt=\"This figure is a graph. It 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 and to the right at x=e. The area is labeled \u201carea=1\u201d.\" width=\"304\" height=\"316\" \/> Figure 3. The area under the curve from 1 to [latex]e[\/latex] is equal to one.[\/caption]\r\n<p id=\"fs-id1167793939504\">The number [latex]e[\/latex] can be shown to be irrational, although we won\u2019t do so here (see the activity in Taylor and Maclaurin Series in Calculus 2). Its approximate value is given by<\/p>\r\n\r\n<div id=\"fs-id1167793443440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e\\approx 2.71828182846[\/latex]<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the definition of the natural logarithm as an integral<\/li>\n<li>Recognize the derivative of the natural logarithm<\/li>\n<li>Integrate functions involving the natural logarithmic function<\/li>\n<li>Define the number \ud835\udc52 through an integral<\/li>\n<\/ul>\n<\/div>\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>Clearly, this does not work when [latex]n=-1,[\/latex] as it would force us to divide by zero. So, what do we do with [latex]\\displaystyle\\int \\frac{1}{x}dx?[\/latex] Recall from the Fundamental Theorem of Calculus that [latex]{\\displaystyle\\int }_{1}^{x}\\dfrac{1}{t}dt[\/latex] is an antiderivative of [latex]\\frac{1}{x}.[\/latex] Therefore, we can make the following definition.<\/p>\n<div id=\"fs-id1167793370244\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793617974\">For [latex]x>1,[\/latex] this is just the area under the curve [latex]y=1\\text{\/}t[\/latex] from 1 to [latex]x.[\/latex] For [latex]x<1,[\/latex] we have [latex]{\\displaystyle\\int }_{1}^{x}\\frac{1}{t}dt=\\text{\u2212}{\\displaystyle\\int }_{x}^{1}\\frac{1}{t}dt,[\/latex] so in this case it is the negative of the area under the curve from [latex]x\\text{ to }1[\/latex] (see the following figure).<\/p>\n<p>&nbsp;<\/p>\n<div style=\"width: 599px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<p id=\"fs-id1167793932688\">Notice that [latex]\\text{ln}1=0.[\/latex] Furthermore, the function [latex]y=\\frac{1}{t}>0[\/latex] for [latex]x>0.[\/latex] Therefore, by the properties of integrals, it is clear that [latex]\\text{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<div id=\"fs-id1167794186831\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Derivative of the Natural Logarithm<\/h3>\n<hr \/>\n<p id=\"fs-id1167794050615\">For [latex]x>0,[\/latex] the derivative of the natural logarithm is given by<\/p>\n<div id=\"fs-id1167793432137\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\text{ln}x=\\dfrac{1}{x}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1167793879890\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Corollary to the Derivative of the Natural Logarithm<\/h3>\n<hr \/>\n<p id=\"fs-id1167794128288\">The function [latex]\\text{ln}x[\/latex] is differentiable; therefore, it is continuous.<\/p>\n<\/div>\n<p id=\"fs-id1167794141119\">A graph of [latex]\\text{ln}x[\/latex] is shown in Figure 2. Notice that it is continuous throughout its domain of [latex](0,\\infty ).[\/latex]<\/p>\n<div style=\"width: 276px\" 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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The graph of [latex]f(x)=\\text{ln}x[\/latex] shows that it is a continuous function.<\/p>\n<\/div>\n<div id=\"fs-id1167793967054\" class=\"textbook exercises\">\n<h3>Example: Calculating Derivatives of Natural Logarithms<\/h3>\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 id=\"fs-id1167793424342\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794171377\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<div id=\"fs-id1167793398778\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167794334473\">Calculate the following derivatives:<\/p>\n<ol id=\"fs-id1167793964608\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}\\text{ln}(2{x}^{2}+x)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{ln}({x}^{3}))}^{2}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793369601\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793259644\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793259644\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793259644\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}\\text{ln}(2{x}^{2}+x)=\\frac{4x+1}{2{x}^{2}+x}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{ln}({x}^{3}))}^{2}=\\frac{6\\text{ln}({x}^{3})}{x}[\/latex]<\/li>\n<\/ol>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167794187630\">Apply the differentiation formula just provided and use 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=0&amp;end=135&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\/6.7TryItProblems0to135_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=\"ohm16155\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16155&theme=oea&iframe_resize_id=ohm16155&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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](d\\text{\/}(dx))\\text{ln}|x|=1\\text{\/}x.[\/latex] This gives rise to the familiar integration formula.<\/p>\n<div id=\"fs-id1167793432204\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Integral of (1\/[latex]u[\/latex]) <em>du<\/em><\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1167793423035\" class=\"textbook exercises\">\n<h3>Example: Calculating Integrals Involving Natural Logarithms<\/h3>\n<p>Calculate the integral [latex]\\displaystyle\\int \\frac{x}{{x}^{2}+4}dx.[\/latex]<\/p>\n<div id=\"fs-id1167793948151\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793887223\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793887223\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793887223\">Using [latex]u[\/latex]-substitution, let [latex]u={x}^{2}+4.[\/latex] Then [latex]du=2xdx[\/latex] and we have<\/p>\n<div id=\"fs-id1167793971678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{x}{{x}^{2}+4}dx=\\frac{1}{2}\\displaystyle\\int \\frac{1}{u}du\\frac{1}{2}\\text{ln}|u|+C=\\frac{1}{2}\\text{ln}|{x}^{2}+4|+C=\\frac{1}{2}\\text{ln}({x}^{2}+4)+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794041523\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793525036\">Show Solution<\/span><\/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<h4>Hint<\/h4>\n<p id=\"fs-id1167793862633\">Apply the integration formula provided earlier and use [latex]u[\/latex]-substitution 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=137&amp;end=234&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\/6.7TryItProblems137to234_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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20050\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20050&theme=oea&iframe_resize_id=ohm20050&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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<div id=\"fs-id1167793366787\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Properties of the Natural Logarithm<\/h3>\n<hr \/>\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\">\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<\/div>\n<div id=\"fs-id1167793509941\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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;<\/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.<\/li>\n<\/ol>\n<p id=\"fs-id1167794210729\">[latex]_\\blacksquare[\/latex]<\/p>\n<div class=\"textbook exercises\">\n<div id=\"fs-id1167794077125\" class=\"exercise\">\n<h3>Example: Using Properties of Logarithms<\/h3>\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;\">\u00a0<\/span><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794337026\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<div id=\"fs-id1167793603772\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793607833\">Use properties of logarithms to simplify the following expression into a single logarithm:<\/p>\n<div id=\"fs-id1167793607837\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}8-\\text{ln}2-\\text{ln}(\\frac{1}{4}).[\/latex]<\/div>\n<div><\/div>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793949542\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793949542\" class=\"hidden-answer\" style=\"display: none\">[latex]4\\text{ln}2[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793949514\">Apply the properties of logarithms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793514534\" class=\"bc-section section\">\n<h2>Defining the Number [latex]e[\/latex]<\/h2>\n<p id=\"fs-id1167793877994\">Now that we have the natural logarithm defined, we can use that function to define the number [latex]e.[\/latex]<\/p>\n<div id=\"fs-id1167793249163\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167793249167\">The number [latex]e[\/latex] is defined to be the real number such that<\/p>\n<div id=\"fs-id1167794140148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ln}e=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793871307\">To put it another way, the area under the curve [latex]y=1\\text{\/}t[\/latex] between [latex]t=1[\/latex] and [latex]t=e[\/latex] is 1 (Figure 3). The proof that such a number exists and is unique is left to you. (<em>Hint<\/em>: Use the Intermediate Value Theorem to prove existence and the fact that [latex]\\text{ln}x[\/latex] is increasing to prove uniqueness.)<\/p>\n<div style=\"width: 314px\" 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\/11213325\/CNX_Calc_Figure_06_07_003.jpg\" alt=\"This figure is a graph. It 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 and to the right at x=e. The area is labeled \u201carea=1\u201d.\" width=\"304\" height=\"316\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The area under the curve from 1 to [latex]e[\/latex] is equal to one.<\/p>\n<\/div>\n<p id=\"fs-id1167793939504\">The number [latex]e[\/latex] can be shown to be irrational, although we won\u2019t do so here (see the activity in Taylor and Maclaurin Series in Calculus 2). Its approximate value is given by<\/p>\n<div id=\"fs-id1167793443440\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e\\approx 2.71828182846[\/latex]<\/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-597\">\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":27,"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-597","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/597","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":15,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/597\/revisions"}],"predecessor-version":[{"id":4905,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/597\/revisions\/4905"}],"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\/597\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=597"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=597"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=597"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=597"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}