{"id":109,"date":"2025-02-13T22:43:52","date_gmt":"2025-02-13T22:43:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/logarithmic-properties\/"},"modified":"2026-03-17T18:08:52","modified_gmt":"2026-03-17T18:08:52","slug":"logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/logarithmic-properties\/","title":{"raw":"Logarithmic Functions: Learn It 1","rendered":"Logarithmic Functions: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Convert between logarithmic to exponential form.<\/li>\r\n \t<li>Evaluate logarithms.<\/li>\r\n \t<li>Use common and natural logarithms<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\r\nHave you ever wondered how we can work backward from an exponential equation? That\u2019s where logarithms come in! They are super handy tools that help us solve problems involving exponential functions.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Let\u2019s break it down:\r\n[latex]\\\\[\/latex]\r\nIf you know that [latex]2^3 = 8[\/latex], you can say that the logarithm base [latex]2[\/latex] of [latex]8[\/latex] is [latex]3[\/latex].\r\n\r\nIn other words:\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_2(8) = 3[\/latex]<\/p>\r\nA logarithm tells us the power we need to raise a base to get a certain number.\r\n\r\nSo, when we see [latex]{\\mathrm{log}}_2(8) = 3[\/latex], it means \"to get [latex]8[\/latex], what power do we raise [latex]2[\/latex] to?\u201d The answer is [latex]3[\/latex]!\r\n\r\n<\/section>We can say that the equations [latex]y = {\\mathrm{log}}_a(x)[\/latex] and [latex]x = a^y[\/latex] are equivalent. This means that we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth, let\u2019s take a close look at the equations. Notice the positions of the exponent and base.\r\n\r\n<img class=\"aligncenter wp-image-5012\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-300x60.png\" alt=\"A diagram showing that y equals log base a of x and x equals a to the y represent the same relationship, with arrows highlighting how the base and exponent correspond in each form.\" width=\"525\" height=\"105\" \/>\r\n\r\nIf we remember the logarithm is the exponent, it makes the conversion easier. You may want to repeat, \u201cbase to the exponent gives us the number.\u201d\r\n\r\n<img class=\"aligncenter wp-image-5013 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03182012\/7.3.L.1.Diagram2.png\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"289\" height=\"143\" \/>\r\n\r\nNotice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>logarithmic function<\/h3>\r\nA <strong>logarithm<\/strong> base [latex]b[\/latex]\u00a0of a positive number [latex]x[\/latex]\u00a0satisfies the following definition:\r\n\r\nFor [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex],\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex]<\/p>\r\n<p style=\"text-align: left;\">where<\/p>\r\n\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]\u00a0of [latex]x[\/latex]\" or the \"log base [latex]b[\/latex]\u00a0of [latex]x[\/latex].\"<\/li>\r\n \t<li>the logarithm [latex]y[\/latex]\u00a0is the exponent to which [latex]b[\/latex]\u00a0must 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\nAlso, since the logarithmic and exponential functions switch the [latex]x[\/latex]\u00a0and [latex]y[\/latex]\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,\r\n<ul>\r\n \t<li>the <strong>domain of the logarithm function<\/strong> with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>the <strong>range of the logarithm function<\/strong> with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Can we take the logarithm of a negative number?<\/strong>\r\n\r\n<hr \/>\r\n\r\nNo. 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.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/strong>\r\n<ol>\r\n \t<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex].<\/li>\r\n \t<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Write the following logarithmic equations in exponential form.\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<\/ol>\r\n[reveal-answer q=\"642511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"642511\"]\r\n\r\nFirst, 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].\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 = 3, y = 2, \\text{ and } x = 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<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321425[\/ohm_question]<\/section><section aria-label=\"Try It\">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].<\/section><section class=\"textbox example\" aria-label=\"Example\">Write the following exponential equations in logarithmic form.\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[reveal-answer q=\"583658\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583658\"]\r\n\r\nFirst, 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].\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].\r\n[latex]\\\\[\/latex]\r\nTherefore, 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].\r\n[latex]\\\\[\/latex]\r\nTherefore, 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].\r\n[latex]\\\\[\/latex]\r\nTherefore, 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[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321426[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321427[\/ohm_question]<\/section><\/div>\r\n<section id=\"fs-id1165135699130\" class=\"key-concepts\">\r\n<dl id=\"fs-id1165137507853\" class=\"definition\">\r\n \t<dd id=\"fs-id1165134037589\"><\/dd>\r\n<\/dl>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Convert between logarithmic to exponential form.<\/li>\n<li>Evaluate logarithms.<\/li>\n<li>Use common and natural logarithms<\/li>\n<\/ul>\n<\/section>\n<h2>Converting Between Logarithmic And Exponential Form<\/h2>\n<p>Have you ever wondered how we can work backward from an exponential equation? That\u2019s where logarithms come in! They are super handy tools that help us solve problems involving exponential functions.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Let\u2019s break it down:<br \/>\n[latex]\\\\[\/latex]<br \/>\nIf you know that [latex]2^3 = 8[\/latex], you can say that the logarithm base [latex]2[\/latex] of [latex]8[\/latex] is [latex]3[\/latex].<\/p>\n<p>In other words:<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_2(8) = 3[\/latex]<\/p>\n<p>A logarithm tells us the power we need to raise a base to get a certain number.<\/p>\n<p>So, when we see [latex]{\\mathrm{log}}_2(8) = 3[\/latex], it means &#8220;to get [latex]8[\/latex], what power do we raise [latex]2[\/latex] to?\u201d The answer is [latex]3[\/latex]!<\/p>\n<\/section>\n<p>We can say that the equations [latex]y = {\\mathrm{log}}_a(x)[\/latex] and [latex]x = a^y[\/latex] are equivalent. This means that we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth, let\u2019s take a close look at the equations. Notice the positions of the exponent and base.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5012\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-300x60.png\" alt=\"A diagram showing that y equals log base a of x and x equals a to the y represent the same relationship, with arrows highlighting how the base and exponent correspond in each form.\" width=\"525\" height=\"105\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-300x60.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-65x13.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-225x45.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1-350x70.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03181918\/7.3.L.1.Diagram1.png 532w\" sizes=\"(max-width: 525px) 100vw, 525px\" \/><\/p>\n<p>If we remember the logarithm is the exponent, it makes the conversion easier. You may want to repeat, \u201cbase to the exponent gives us the number.\u201d<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5013 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03182012\/7.3.L.1.Diagram2.png\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"289\" height=\"143\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03182012\/7.3.L.1.Diagram2.png 289w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03182012\/7.3.L.1.Diagram2-65x32.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/02\/03182012\/7.3.L.1.Diagram2-225x111.png 225w\" sizes=\"(max-width: 289px) 100vw, 289px\" \/><\/p>\n<p>Notice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>logarithmic function<\/h3>\n<p>A <strong>logarithm<\/strong> base [latex]b[\/latex]\u00a0of a positive number [latex]x[\/latex]\u00a0satisfies the following definition:<\/p>\n<p>For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex]<\/p>\n<p style=\"text-align: left;\">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]\u00a0of [latex]x[\/latex]&#8221; or the &#8220;log base [latex]b[\/latex]\u00a0of [latex]x[\/latex].&#8221;<\/li>\n<li>the logarithm [latex]y[\/latex]\u00a0is the exponent to which [latex]b[\/latex]\u00a0must 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<p>Also, since the logarithmic and exponential functions switch the [latex]x[\/latex]\u00a0and [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 <strong>domain of the logarithm function<\/strong> with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\n<li>the <strong>range of the logarithm function<\/strong> with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Can we take the logarithm of a negative number?<\/strong><\/p>\n<hr \/>\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\" aria-label=\"Question Help\"><strong>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/strong><\/p>\n<ol>\n<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify [latex]b[\/latex], [latex]y[\/latex], and [latex]x[\/latex].<\/li>\n<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<\/ol>\n<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\">\n<p>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 = 3, y = 2, \\text{ and } x = 9[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321425\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321425&theme=lumen&iframe_resize_id=ohm321425&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">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].<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<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\">\n<p>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].<br \/>\n[latex]\\\\[\/latex]<br \/>\nTherefore, 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].<br \/>\n[latex]\\\\[\/latex]<br \/>\nTherefore, 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].<br \/>\n[latex]\\\\[\/latex]<br \/>\nTherefore, 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\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321426\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321426&theme=lumen&iframe_resize_id=ohm321426&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321427\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321427&theme=lumen&iframe_resize_id=ohm321427&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<section id=\"fs-id1165135699130\" class=\"key-concepts\">\n<dl id=\"fs-id1165137507853\" class=\"definition\">\n<dd id=\"fs-id1165134037589\"><\/dd>\n<\/dl>\n<\/section>\n","protected":false},"author":6,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":105,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/109"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/109\/revisions"}],"predecessor-version":[{"id":5885,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/109\/revisions\/5885"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/105"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/109\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=109"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=109"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=109"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}