{"id":1708,"date":"2024-04-24T17:05:57","date_gmt":"2024-04-24T17:05:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1708"},"modified":"2025-08-17T23:46:56","modified_gmt":"2025-08-17T23:46:56","slug":"contextual-applications-of-derivatives-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/contextual-applications-of-derivatives-background-youll-need-1\/","title":{"raw":"Contextual Applications of Derivatives: Background You'll Need 1","rendered":"Contextual Applications of Derivatives: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Change logarithmic equations into exponential equations&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:769,&quot;3&quot;:{&quot;1&quot;:0},&quot;11&quot;:4,&quot;12&quot;:0}\">Change logarithmic equations into exponential equations<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Convert from Logarithmic to Exponential Form<\/h2>\r\n<p>Understanding the relationship between logarithmic and exponential forms is fundamental. This conversion can be succinctly represented as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]<\/p>\r\n<p>Here, [latex]b[\/latex] is always a positive number and cannot be equal to [latex]1[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_3090\" align=\"aligncenter\" width=\"487\"]<img class=\"wp-image-3090 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/> Relationship between logarithmic and exponential forms[\/caption]\r\n\r\n<section class=\"textbox proTip\">\r\n<p>The logarithm function [latex]\\log_{b}(x)[\/latex] is conventionally written with parentheses to clearly denote the function's input, similar to [latex]f(x)[\/latex]. However, when dealing with a single variable or a simple expression, parentheses might be omitted, leading to the notation [latex]\\log_{b}x[\/latex]. It\u2019s important to note that many calculators might still require parentheses around the input [latex]x[\/latex].<\/p>\r\n<\/section>\r\n<p>The notation [latex]\\log_{b}(c)=a[\/latex] can be interpreted as [latex]b^a=c[\/latex]. This implies that the base [latex]b[\/latex] raised to the power [latex]a[\/latex] equals [latex]c[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_3092\" align=\"aligncenter\" width=\"487\"]<img class=\"wp-image-3092 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/> Diagram showing that a logarithm expresses an exponent.[\/caption]\r\n\r\n<p>It\u2019s common to see [latex]ln[\/latex] representing the natural logarithm, which uses [latex]e[\/latex] (approximately [latex]2.718[\/latex]) as its base. The notation [latex]ln[\/latex] corresponds to [latex]\\log_e[\/latex], emphasizing the natural logarithm\u2019s specific base.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>logarithmic functions<\/h3>\r\n<p>A <strong>logarithm<\/strong> base [latex]b[\/latex]\u00a0of a positive number [latex]x[\/latex]\u00a0satisfies the following definition: For [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex], [latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where<\/p>\r\n<ul>\r\n\t<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \"the logarithm with base [latex]b[\/latex] of [latex]x[\/latex]\" or the \"log base [latex]b[\/latex] of [latex]x[\/latex].\"<\/li>\r\n\t<li>the logarithm <em>y<\/em> is the exponent to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/li>\r\n\t<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>Also, since the logarithmic and exponential functions switch the [latex]x[\/latex] and [latex]y[\/latex]\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\r\n<ul>\r\n\t<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n\t<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<section class=\"textbox proTip\">\r\n<p><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\r\n<p>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Convert a Logarithmic Equation to Exponential Form<\/strong><\/p>\r\n<p>Given a logarithmic equation in the format [latex]\\log_{b}(x)=y[\/latex]:<\/p>\r\n<ol>\r\n\t<li><strong>Identify the Components<\/strong>: Recognize [latex]b[\/latex] as the base, [latex]y[\/latex] as the logarithmic result, and [latex]x[\/latex] as the argument of the logarithm.<\/li>\r\n\t<li><strong>Convert to Exponential Form<\/strong>: Rewrite the equation from logarithmic to exponential form by setting the base [latex]b[\/latex] raised to the power [latex]y[\/latex] equal to [latex]x[\/latex]. This translates to [latex]b^y=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Write the following logarithmic equations in exponential form.<\/p>\r\n<ol>\r\n\t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"642511\"]Show Solution[\/reveal-answer] [hidden-answer a=\"642511\"] First, identify the values of [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex]. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\r\n<ol>\r\n\t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, [latex]b\u00a0= 3[\/latex], [latex]y\u00a0= 2[\/latex], and [latex]x\u00a0= 9[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\r\n\t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288397[\/ohm_question]<\/p>\r\n<\/section>\r\n<h2>Convert from Exponential to Logarithmic Form<\/h2>\r\n<p>To convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base [latex]b[\/latex], exponent [latex]x[\/latex], and output [latex]y[\/latex]. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n<section class=\"textbox example\">\r\n<p>Write the following exponential equations in logarithmic form.<\/p>\r\n<ol>\r\n\t<li>[latex]{2}^{3}=8[\/latex]<\/li>\r\n\t<li>[latex]{5}^{2}=25[\/latex]<\/li>\r\n\t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"583658\"]Show Solution[\/reveal-answer] [hidden-answer a=\"583658\"] First, identify the values of [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex]. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n<ol>\r\n\t<li>[latex]{2}^{3}=8[\/latex] Here, [latex]b\u00a0= 2[\/latex], [latex]x\u00a0= 3[\/latex], and [latex]y\u00a0= 8[\/latex]. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\r\n\t<li>[latex]{5}^{2}=25[\/latex] Here, [latex]b\u00a0= 5[\/latex], [latex]x\u00a0= 2[\/latex], and [latex]y\u00a0= 25[\/latex]. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\r\n\t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, [latex]b\u00a0= 10[\/latex], [latex]x\u00a0= \u20134,[\/latex] and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288398[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Change logarithmic equations into exponential equations&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:769,&quot;3&quot;:{&quot;1&quot;:0},&quot;11&quot;:4,&quot;12&quot;:0}\">Change logarithmic equations into exponential equations<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Convert from Logarithmic to Exponential Form<\/h2>\n<p>Understanding the relationship between logarithmic and exponential forms is fundamental. This conversion can be succinctly represented as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/p>\n<p>Here, [latex]b[\/latex] is always a positive number and cannot be equal to [latex]1[\/latex].<\/p>\n<figure id=\"attachment_3090\" aria-describedby=\"caption-attachment-3090\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3090 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/><figcaption id=\"caption-attachment-3090\" class=\"wp-caption-text\">Relationship between logarithmic and exponential forms<\/figcaption><\/figure>\n<section class=\"textbox proTip\">\n<p>The logarithm function [latex]\\log_{b}(x)[\/latex] is conventionally written with parentheses to clearly denote the function&#8217;s input, similar to [latex]f(x)[\/latex]. However, when dealing with a single variable or a simple expression, parentheses might be omitted, leading to the notation [latex]\\log_{b}x[\/latex]. It\u2019s important to note that many calculators might still require parentheses around the input [latex]x[\/latex].<\/p>\n<\/section>\n<p>The notation [latex]\\log_{b}(c)=a[\/latex] can be interpreted as [latex]b^a=c[\/latex]. This implies that the base [latex]b[\/latex] raised to the power [latex]a[\/latex] equals [latex]c[\/latex].<\/p>\n<figure id=\"attachment_3092\" aria-describedby=\"caption-attachment-3092\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3092 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/><figcaption id=\"caption-attachment-3092\" class=\"wp-caption-text\">Diagram showing that a logarithm expresses an exponent.<\/figcaption><\/figure>\n<p>It\u2019s common to see [latex]ln[\/latex] representing the natural logarithm, which uses [latex]e[\/latex] (approximately [latex]2.718[\/latex]) as its base. The notation [latex]ln[\/latex] corresponds to [latex]\\log_e[\/latex], emphasizing the natural logarithm\u2019s specific base.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>logarithmic functions<\/h3>\n<p>A <strong>logarithm<\/strong> base [latex]b[\/latex]\u00a0of a positive number [latex]x[\/latex]\u00a0satisfies the following definition: For [latex]x>0,b>0,b\\ne 1[\/latex], [latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where<\/p>\n<ul>\n<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base [latex]b[\/latex] of [latex]x[\/latex]&#8221; or the &#8220;log base [latex]b[\/latex] of [latex]x[\/latex].&#8221;<\/li>\n<li>the logarithm <em>y<\/em> is the exponent to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/li>\n<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\n<\/ul>\n<\/section>\n<p>Also, since the logarithmic and exponential functions switch the [latex]x[\/latex] and [latex]y[\/latex]\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\n<ul>\n<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\n<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<section class=\"textbox proTip\">\n<p><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\n<p>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Convert a Logarithmic Equation to Exponential Form<\/strong><\/p>\n<p>Given a logarithmic equation in the format [latex]\\log_{b}(x)=y[\/latex]:<\/p>\n<ol>\n<li><strong>Identify the Components<\/strong>: Recognize [latex]b[\/latex] as the base, [latex]y[\/latex] as the logarithmic result, and [latex]x[\/latex] as the argument of the logarithm.<\/li>\n<li><strong>Convert to Exponential Form<\/strong>: Rewrite the equation from logarithmic to exponential form by setting the base [latex]b[\/latex] raised to the power [latex]y[\/latex] equal to [latex]x[\/latex]. This translates to [latex]b^y=x[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Write the following logarithmic equations in exponential form.<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q642511\">Show Solution<\/button> <\/p>\n<div id=\"q642511\" class=\"hidden-answer\" style=\"display: none\"> First, identify the values of [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex]. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, [latex]b\u00a0= 3[\/latex], [latex]y\u00a0= 2[\/latex], and [latex]x\u00a0= 9[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288397\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288397&theme=lumen&iframe_resize_id=ohm288397&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h2>Convert from Exponential to Logarithmic Form<\/h2>\n<p>To convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base [latex]b[\/latex], exponent [latex]x[\/latex], and output [latex]y[\/latex]. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<section class=\"textbox example\">\n<p>Write the following exponential equations in logarithmic form.<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex]<\/li>\n<li>[latex]{5}^{2}=25[\/latex]<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q583658\">Show Solution<\/button> <\/p>\n<div id=\"q583658\" class=\"hidden-answer\" style=\"display: none\"> First, identify the values of [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex]. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex] Here, [latex]b\u00a0= 2[\/latex], [latex]x\u00a0= 3[\/latex], and [latex]y\u00a0= 8[\/latex]. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\n<li>[latex]{5}^{2}=25[\/latex] Here, [latex]b\u00a0= 5[\/latex], [latex]x\u00a0= 2[\/latex], and [latex]y\u00a0= 25[\/latex]. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, [latex]b\u00a0= 10[\/latex], [latex]x\u00a0= \u20134,[\/latex] and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288398\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288398&theme=lumen&iframe_resize_id=ohm288398&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"background_you_need","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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1708"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":17,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1708\/revisions"}],"predecessor-version":[{"id":4779,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1708\/revisions\/4779"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1708\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1708"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1708"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1708"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}