{"id":2070,"date":"2025-08-29T14:19:28","date_gmt":"2025-08-29T14:19:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=2070"},"modified":"2025-08-29T14:19:28","modified_gmt":"2025-08-29T14:19:28","slug":"introduction-to-differential-equations-background-youll-need-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-differential-equations-background-youll-need-4\/","title":{"raw":"Introduction to Differential Equations: Background You'll Need 4","rendered":"Introduction to Differential Equations: Background You&#8217;ll Need 4"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Use logarithms to solve exponential equations<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Logarithmic Functions<\/h2>\r\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.<\/p>\r\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <strong>logarithmic function<\/strong><em> with base<\/em> [latex]b[\/latex].<\/p>\r\nFor any [latex]b&gt;0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies\r\n<p style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>logarithmic functions<\/h3>\r\nA logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[\/latex]. For a given base [latex]b[\/latex], it tells us the power to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].\r\n\r\n<\/section><section class=\"textbox example\">\r\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b&gt;0.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/section>\r\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex].<\/p>\r\n\r\n<section class=\"textbox example\"><center>[latex]\\begin{array}{l}\\ln (e)=\\log_e (e)=1 \\\\ \\ln(e^3)=\\log_e (e^3)=3 \\\\ \\ln(1)=\\log_e (1)=0\\end{array}[\/latex]<\/center><\/section>\r\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>properties of logarithms<\/h3>\r\n<p id=\"fs-id1170572482707\">If [latex]a,b,c&gt;0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill &amp; &amp; &amp; \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\r\n \t<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\r\n \t<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572174799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572174799\"]\r\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>By the definition of the natural logarithm function,\r\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\r\nTherefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/li>\r\n \t<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as\r\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\r\nTherefore, the equation can be rewritten as\r\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\r\nThe solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/li>\r\n \t<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].\r\nUsing the quotient property, this becomes\r\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\r\nTherefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]5^x=2[\/latex]<\/li>\r\n \t<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572550555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572550555\"]\r\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Applying the natural logarithm function to both sides of the equation, we have\r\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\r\nUsing the power property of logarithms,\r\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\r\nTherefore, [latex]x=\\frac{\\ln 2 }{\\ln 5}[\/latex].<\/li>\r\n \t<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation\r\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\r\nRewriting this equation as\r\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\r\nwe can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:\r\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\r\nNow we can solve the quadratic equation. Factoring this equation, we obtain\r\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\r\nTherefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions\r\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div><\/li>\r\n<\/ol>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>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.You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]217547[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Use logarithms to solve exponential equations<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Logarithmic Functions<\/h2>\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.<\/p>\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <strong>logarithmic function<\/strong><em> with base<\/em> [latex]b[\/latex].<\/p>\n<p>For any [latex]b>0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>logarithmic functions<\/h3>\n<p>A logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[\/latex]. For a given base [latex]b[\/latex], it tells us the power to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b>0.\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex].<\/p>\n<section class=\"textbox example\">\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\ln (e)=\\log_e (e)=1 \\\\ \\ln(e^3)=\\log_e (e^3)=3 \\\\ \\ln(1)=\\log_e (1)=0\\end{array}[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>properties of logarithms<\/h3>\n<p id=\"fs-id1170572482707\">If [latex]a,b,c>0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill & & & \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill & & & \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill & & & \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\n<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\n<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572174799\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572174799\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\n<li>By the definition of the natural logarithm function,\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\n<p>Therefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/li>\n<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\n<p>Therefore, the equation can be rewritten as<\/p>\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\n<p>The solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/li>\n<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].<br \/>\nUsing the quotient property, this becomes<\/p>\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\n<p>Therefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]5^x=2[\/latex]<\/li>\n<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572550555\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572550555\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\n<li>Applying the natural logarithm function to both sides of the equation, we have\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\n<p>Using the power property of logarithms,<\/p>\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\n<p>Therefore, [latex]x=\\frac{\\ln 2 }{\\ln 5}[\/latex].<\/li>\n<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\n<p>Rewriting this equation as<\/p>\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\n<p>we can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:<\/p>\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\n<p>Now we can solve the quadratic equation. Factoring this equation, we obtain<\/p>\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\n<p>Therefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions<\/p>\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div>\n<\/li>\n<\/ol>\n<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\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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.You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm217547\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=217547&theme=lumen&iframe_resize_id=ohm217547&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"module-header":"- Select Header -","content_attributions":[],"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\/2070"}],"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\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/2070\/revisions"}],"predecessor-version":[{"id":2072,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/2070\/revisions\/2072"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/2070\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=2070"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=2070"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=2070"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=2070"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}