{"id":235,"date":"2023-11-08T16:10:34","date_gmt":"2023-11-08T16:10:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/introduction-to-logarithmic-properties\/"},"modified":"2025-04-29T16:57:28","modified_gmt":"2025-04-29T16:57:28","slug":"7-7-solving-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/7-7-solving-logarithmic-equations\/","title":{"raw":"7.7 Solving Logarithmic Equations","rendered":"7.7 Solving Logarithmic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve logarithmic equations using the definition of logarithms or 1-1 property.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The One-to-One Property of Logarithms<\/h2>\r\nSome logarithmic equations can be solved in the same manner as the exponential equations in the previous section. Since the logarithm function also has the one-to-one property (its inverse is the exponential function), if two logarithms with common bases are set equal then their arguments must be equal as well.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>ONE-TO-ONE PROPERTY OF LOGARITHMS<\/h3>\r\nLet [latex]b&gt;0[\/latex] and [latex]b \\neq 1.[\/latex] Then [latex]\\log_b(x) = \\log_b(y)[\/latex] implies [latex]x=y.[\/latex]\r\n\r\nThis also applies when the arguments are algebraic expressions.\r\n\r\n<\/div>\r\n<p id=\"eip-625\">For example, if [latex]\\log_2(x-1)=\\log_2(8),\\text{then }x-1=8[\/latex] and [latex]x=9.[\/latex] To check, we can substitute\u00a0\u00a0[latex]x=9[\/latex] into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"37\" height=\"33\" \/>Make sure to always check your answer! The logarithm function has domain [latex](0,\\infty)[\/latex] so any solutions must result in all logarithms having positive arguments. A solution that results in an argument that is zero or negative in a logarithm is an <strong>extraneous solution<\/strong>.\r\n\r\n<\/div>\r\nTry the following example to test your understanding so far.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\log(3x - 2)=\\log(x+4)[\/latex]. Check your solutions.\r\n[reveal-answer q=\"874129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"874129\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log(3x - 2)&amp;=\\log(x+4) \\\\\r\n3x-2&amp;=x+4 &amp;&amp; \\color{blue}{\\textsf{apply the One-to-One Property of Logarithms}} \\\\\r\n2x&amp;=6\\\\\r\nx&amp;=3 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135172191\">To check the solution, substitute [latex]x=3[\/latex] into the original equation. Make sure this does not result in any logarithm arguments that are negative or zero.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log(3x - 2)&amp;=\\log(x+4) \\\\\r\n\\log(3\\cdot 3 - 2)&amp;=\\log(3+4) \\\\\r\n\\log(7)&amp;=\\log(7) \\end{align}[\/latex]<\/p>\r\nThe answer is [latex]x=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another example with more complicated arguments. Note that in the context of these examples, the base of the logarithm does not matter as long as it is the same base on both sides.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\ln(x^2-3)=\\ln(2x)[\/latex].\r\n\r\n[reveal-answer q=\"306816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"306816\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\ln(x^2-3)&amp;=\\ln(2x) \\\\\r\nx^2-3&amp;=2x &amp;&amp; \\color{blue}{\\textsf{apply the One-to-One Property of Logarithms}} \\\\\r\nx^2-2x-3&amp;=0 &amp;&amp; \\color{blue}{\\textsf{prepare to solve by factoring}}\\\\\r\n(x-3)(x+1)&amp;=0\\\\\r\nx-3=0 \\text{ or }&amp;x+1=0\\\\\r\nx=3 \\text{ or }&amp;x=-1 \\end{align}[\/latex]<\/p>\r\nWe now check our solutions. [latex]x=3[\/latex] results in [latex]\\ln(6)=\\ln(6),[\/latex] and\u00a0[latex]x=-1[\/latex] results in [latex]\\ln(-2)=\\ln(-2).[\/latex] Since the argument of the logarithm is negative in the second check, we discard the extraneous solution\u00a0[latex]x=-1.[\/latex]\r\n\r\nThe answer is\u00a0[latex]x=3.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using the Definition of a Logarithm<\/h2>\r\nWe have already seen that every logarithmic equation [latex]\\log_b(x)=y[\/latex] can be written as the exponential equation [latex]b^y=x[\/latex]. We can use this fact to solve logarithmic equations where the argument is an algebraic expression.\r\n<p id=\"fs-id1165134148350\">For example, consider the equation [latex]\\log_2(6x - 10)=3[\/latex]. To solve this equation, we can apply the definition of logarithms which puts the expression containing [latex]x[\/latex] by itself on one side of the equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log_2(6x - 10)&amp;=3 \\\\\r\n2^3&amp;=6x-10 &amp;&amp; \\color{blue}{\\textsf{rewrite using the definition of logarithms}} \\\\\r\n8&amp;=6x-10\\\\\r\n18&amp;=6x\\\\\r\n3&amp;=x \\end{align}[\/latex]<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">Here is another example of the same idea using the natural logarithm. We use a graph to illustrate the point of finding an [latex]x[\/latex] value that makes the two sides equal.\u00a0<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">In this case, we get a logarithmic function and a horizontal line and find that the solution is the point where the two intersect.<\/span><\/div>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\ln(x)=3[\/latex].\r\n[reveal-answer q=\"691419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"691419\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\ln(x)&amp;=3 \\\\\r\ne^{3}&amp;=x &amp;&amp; \\color{blue}{\\textsf{rewrite using the definition of a logarithm}} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137443165\">The graph below\u00a0represents the graphs of each side of the equation. On the graph, the [latex]x[\/latex]-coordinate of the point at which the two graphs intersect is close to [latex]20[\/latex]. In other words, [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\r\n\r\n<figure class=\"small\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201935\/CNX_Precalc_Figure_04_06_0032.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/><\/figure>\r\n<figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\u00a0The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately [latex](20.0855, 3).[\/latex]<\/figcaption><\/figure>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the same way that a story can help you understand a complex concept, graphs can help us understand a wide variety of concepts in mathematics. Visual representations of the parts of an equation can be created by turning each side into a function and plotting the functions on the same set of axes.\r\n\r\n<\/div>\r\n<div>Here is a similar example but only solved algebraically.<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]2\\ln(x+3)=7[\/latex]. Write your final answer both in exact form and approximated to two decimal places.\r\n[reveal-answer q=\"1854\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"1854\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\ln(x+3)&amp;=7 \\\\\r\n\\ln(x+3)&amp;=\\dfrac{7}{2}&amp;&amp;\\color{blue}{\\textsf{isolate the logarithm first}} \\\\\r\ne^{7\/2}&amp;=x+3 &amp;&amp; \\color{blue}{\\textsf{rewrite using the definition of a logarithm}} \\\\\r\ne^{7\/2}-3&amp;=x \\end{align}[\/latex]<\/p>\r\nThe exact answer is [latex]x=e^{7\/2}-3.[\/latex] The answer approximated to two decimal places is [latex]x\\approx 30.12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1165137755280\">\r\n<h3>Summary<\/h3>\r\nThe One-to-One Property of Logarithms can solve logarithmic equations where both sides of the equation have a logarithm with the same base. If there is only a single logarithm, we can use the definition of a logarithm to rewrite the equation as an exponential equation.\r\n\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve logarithmic equations using the definition of logarithms or 1-1 property.<\/li>\n<\/ul>\n<\/div>\n<h2>The One-to-One Property of Logarithms<\/h2>\n<p>Some logarithmic equations can be solved in the same manner as the exponential equations in the previous section. Since the logarithm function also has the one-to-one property (its inverse is the exponential function), if two logarithms with common bases are set equal then their arguments must be equal as well.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>ONE-TO-ONE PROPERTY OF LOGARITHMS<\/h3>\n<p>Let [latex]b>0[\/latex] and [latex]b \\neq 1.[\/latex] Then [latex]\\log_b(x) = \\log_b(y)[\/latex] implies [latex]x=y.[\/latex]<\/p>\n<p>This also applies when the arguments are algebraic expressions.<\/p>\n<\/div>\n<p id=\"eip-625\">For example, if [latex]\\log_2(x-1)=\\log_2(8),\\text{then }x-1=8[\/latex] and [latex]x=9.[\/latex] To check, we can substitute\u00a0\u00a0[latex]x=9[\/latex] into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"37\" height=\"33\" \/>Make sure to always check your answer! The logarithm function has domain [latex](0,\\infty)[\/latex] so any solutions must result in all logarithms having positive arguments. A solution that results in an argument that is zero or negative in a logarithm is an <strong>extraneous solution<\/strong>.<\/p>\n<\/div>\n<p>Try the following example to test your understanding so far.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\log(3x - 2)=\\log(x+4)[\/latex]. Check your solutions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q874129\">Show Solution<\/span><\/p>\n<div id=\"q874129\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log(3x - 2)&=\\log(x+4) \\\\  3x-2&=x+4 && \\color{blue}{\\textsf{apply the One-to-One Property of Logarithms}} \\\\  2x&=6\\\\  x&=3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135172191\">To check the solution, substitute [latex]x=3[\/latex] into the original equation. Make sure this does not result in any logarithm arguments that are negative or zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log(3x - 2)&=\\log(x+4) \\\\  \\log(3\\cdot 3 - 2)&=\\log(3+4) \\\\  \\log(7)&=\\log(7) \\end{align}[\/latex]<\/p>\n<p>The answer is [latex]x=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is another example with more complicated arguments. Note that in the context of these examples, the base of the logarithm does not matter as long as it is the same base on both sides.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\ln(x^2-3)=\\ln(2x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q306816\">Show Solution<\/span><\/p>\n<div id=\"q306816\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\ln(x^2-3)&=\\ln(2x) \\\\  x^2-3&=2x && \\color{blue}{\\textsf{apply the One-to-One Property of Logarithms}} \\\\  x^2-2x-3&=0 && \\color{blue}{\\textsf{prepare to solve by factoring}}\\\\  (x-3)(x+1)&=0\\\\  x-3=0 \\text{ or }&x+1=0\\\\  x=3 \\text{ or }&x=-1 \\end{align}[\/latex]<\/p>\n<p>We now check our solutions. [latex]x=3[\/latex] results in [latex]\\ln(6)=\\ln(6),[\/latex] and\u00a0[latex]x=-1[\/latex] results in [latex]\\ln(-2)=\\ln(-2).[\/latex] Since the argument of the logarithm is negative in the second check, we discard the extraneous solution\u00a0[latex]x=-1.[\/latex]<\/p>\n<p>The answer is\u00a0[latex]x=3.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using the Definition of a Logarithm<\/h2>\n<p>We have already seen that every logarithmic equation [latex]\\log_b(x)=y[\/latex] can be written as the exponential equation [latex]b^y=x[\/latex]. We can use this fact to solve logarithmic equations where the argument is an algebraic expression.<\/p>\n<p id=\"fs-id1165134148350\">For example, consider the equation [latex]\\log_2(6x - 10)=3[\/latex]. To solve this equation, we can apply the definition of logarithms which puts the expression containing [latex]x[\/latex] by itself on one side of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\log_2(6x - 10)&=3 \\\\  2^3&=6x-10 && \\color{blue}{\\textsf{rewrite using the definition of logarithms}} \\\\  8&=6x-10\\\\  18&=6x\\\\  3&=x \\end{align}[\/latex]<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">Here is another example of the same idea using the natural logarithm. We use a graph to illustrate the point of finding an [latex]x[\/latex] value that makes the two sides equal.\u00a0<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">In this case, we get a logarithmic function and a horizontal line and find that the solution is the point where the two intersect.<\/span><\/div>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\ln(x)=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q691419\">Show Solution<\/span><\/p>\n<div id=\"q691419\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\ln(x)&=3 \\\\  e^{3}&=x && \\color{blue}{\\textsf{rewrite using the definition of a logarithm}} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137443165\">The graph below\u00a0represents the graphs of each side of the equation. On the graph, the [latex]x[\/latex]-coordinate of the point at which the two graphs intersect is close to [latex]20[\/latex]. In other words, [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\n<figure class=\"small\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201935\/CNX_Precalc_Figure_04_06_0032.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\u00a0The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately [latex](20.0855, 3).[\/latex]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<p>In the same way that a story can help you understand a complex concept, graphs can help us understand a wide variety of concepts in mathematics. Visual representations of the parts of an equation can be created by turning each side into a function and plotting the functions on the same set of axes.<\/p>\n<\/div>\n<div>Here is a similar example but only solved algebraically.<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]2\\ln(x+3)=7[\/latex]. Write your final answer both in exact form and approximated to two decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1854\">Show Solution<\/span><\/p>\n<div id=\"q1854\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\ln(x+3)&=7 \\\\  \\ln(x+3)&=\\dfrac{7}{2}&&\\color{blue}{\\textsf{isolate the logarithm first}} \\\\  e^{7\/2}&=x+3 && \\color{blue}{\\textsf{rewrite using the definition of a logarithm}} \\\\  e^{7\/2}-3&=x \\end{align}[\/latex]<\/p>\n<p>The exact answer is [latex]x=e^{7\/2}-3.[\/latex] The answer approximated to two decimal places is [latex]x\\approx 30.12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137755280\">\n<h3>Summary<\/h3>\n<p>The One-to-One Property of Logarithms can solve logarithmic equations where both sides of the equation have a logarithm with the same base. If there is only a single logarithm, we can use the definition of a logarithm to rewrite the equation as an exponential equation.<\/p>\n<\/section>\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-235\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/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":395986,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"86fbcdd9-ce63-431d-8a66-5bb6ff2d87ad","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-235","chapter","type-chapter","status-publish","hentry"],"part":225,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/235","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions"}],"predecessor-version":[{"id":2085,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions\/2085"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/225"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/235\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=235"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=235"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=235"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}