{"id":110,"date":"2025-02-13T22:43:53","date_gmt":"2025-02-13T22:43:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-logarithmic-functions\/"},"modified":"2026-03-17T19:05:57","modified_gmt":"2026-03-17T19:05:57","slug":"graphs-of-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-logarithmic-functions\/","title":{"raw":"Graphs of Logarithmic Functions: Learn It 1","rendered":"Graphs of 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>Identify the domain of a logarithmic function.<\/li>\r\n \t<li>Graph logarithmic functions.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Domain of Logarithmic Functions<\/h2>\r\nWhen working with logarithmic functions, understanding their domain is crucial. The domain of a function is the set of all possible input values ([latex]x[\/latex]-values) that the function can accept without causing any mathematical issues. For logarithmic functions, this concept is especially important because it defines where the function is valid and where it isn't.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Why is the Domain Important for Logarithmic Functions?<\/strong>\r\n\r\n<hr \/>\r\n\r\nLogarithmic functions are only defined for positive real numbers. This means you can only take the logarithm of a positive number.\r\n\r\nFor example, [latex]\\mathrm{log}_{b}(x)[\/latex] is only valid if [latex]x \\gt 0[\/latex]. Thus, the <strong>domain<\/strong> is [latex](0, \\infty)[\/latex]. This results in a <strong>vertical asymptote<\/strong> at [latex]x=0[\/latex], which the graph approaches but never touches or crosses. The\u00a0<strong>range<\/strong> of a logarithmic function is always all real numbers, [latex](-\\infty, \\infty)[\/latex].\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">The exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number [latex]x[\/latex] and constant [latex]b&gt;0[\/latex], [latex]b\\ne 1[\/latex], where\r\n<ul>\r\n \t<li>The domain of [latex]y[\/latex]\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]y[\/latex] is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:\r\n<ul>\r\n \t<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]: [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\nPreviously we saw that certain transformations can change the <em>range<\/em> of [latex]y={b}^{x}[\/latex]. Similarly, applying transformations to the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can change the <em>domain<\/em>. Therefore, when finding the domain of a logarithmic function, it is important to remember that the domain consists <em>only of positive real numbers<\/em>. That is, the value you are applying the logarithmic function to, also known as the argument of the logarithmic function, must be greater than zero.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">For example, consider [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex]. This function is defined for any values of [latex]x[\/latex]\u00a0such that the argument, in this case [latex]2x - 3[\/latex], is greater than zero. To find the domain, we set up an inequality and solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x - 3&gt;0\\hfill &amp; \\text{Show the argument greater than zero}.\\hfill \\\\ 2x&gt;3\\hfill &amp; \\text{Add 3}.\\hfill \\\\ x&gt;\\dfrac{3}{2}\\hfill &amp; \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/p>\r\nIn interval notation, the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex] is [latex]\\left(\\dfrac{3}{2},\\infty \\right)[\/latex].\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a logarithmic function, identify the domain\r\n<\/strong>\r\n<ol>\r\n \t<li>Set up an inequality showing the argument greater than zero.<\/li>\r\n \t<li>Solve for [latex]x[\/latex].<\/li>\r\n \t<li>Write the domain in interval notation.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?[reveal-answer q=\"786555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786555\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+3&gt;0\\hfill &amp; \\text{The argument must be positive}.\\hfill \\\\ x&gt;-3\\hfill &amp; \\text{Subtract 3}.\\hfill \\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe logarithmic function is defined only when the argument is positive, so this function is defined when [latex]x+3&gt;0[\/latex].\r\n\r\nThe domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?[reveal-answer q=\"152912\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"152912\"]The logarithmic function is defined only when the argument is positive, so this function is defined when [latex]5 - 2x&gt;0[\/latex]. Solving this inequality,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5 - 2x&gt;0\\hfill &amp; \\text{The argument must be positive}.\\hfill \\\\ -2x&gt;-5\\hfill &amp; \\text{Subtract }5.\\hfill \\\\ x&lt;\\frac{5}{2}\\hfill &amp; \\text{Divide by }-2\\text{ and switch the inequality}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321441[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321442[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321443[\/ohm_question]<\/section><\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Identify the domain of a logarithmic function.<\/li>\n<li>Graph logarithmic functions.<\/li>\n<\/ul>\n<\/section>\n<h2>Domain of Logarithmic Functions<\/h2>\n<p>When working with logarithmic functions, understanding their domain is crucial. The domain of a function is the set of all possible input values ([latex]x[\/latex]-values) that the function can accept without causing any mathematical issues. For logarithmic functions, this concept is especially important because it defines where the function is valid and where it isn&#8217;t.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Why is the Domain Important for Logarithmic Functions?<\/strong><\/p>\n<hr \/>\n<p>Logarithmic functions are only defined for positive real numbers. This means you can only take the logarithm of a positive number.<\/p>\n<p>For example, [latex]\\mathrm{log}_{b}(x)[\/latex] is only valid if [latex]x \\gt 0[\/latex]. Thus, the <strong>domain<\/strong> is [latex](0, \\infty)[\/latex]. This results in a <strong>vertical asymptote<\/strong> at [latex]x=0[\/latex], which the graph approaches but never touches or crosses. The\u00a0<strong>range<\/strong> of a logarithmic function is always all real numbers, [latex](-\\infty, \\infty)[\/latex].<\/p>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number [latex]x[\/latex] and constant [latex]b>0[\/latex], [latex]b\\ne 1[\/latex], where<\/p>\n<ul>\n<li>The domain of [latex]y[\/latex]\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<li>The range of [latex]y[\/latex] is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/section>\n<p>In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:<\/p>\n<ul>\n<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]: [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<p>Previously we saw that certain transformations can change the <em>range<\/em> of [latex]y={b}^{x}[\/latex]. Similarly, applying transformations to the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can change the <em>domain<\/em>. Therefore, when finding the domain of a logarithmic function, it is important to remember that the domain consists <em>only of positive real numbers<\/em>. That is, the value you are applying the logarithmic function to, also known as the argument of the logarithmic function, must be greater than zero.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">For example, consider [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex]. This function is defined for any values of [latex]x[\/latex]\u00a0such that the argument, in this case [latex]2x - 3[\/latex], is greater than zero. To find the domain, we set up an inequality and solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x - 3>0\\hfill & \\text{Show the argument greater than zero}.\\hfill \\\\ 2x>3\\hfill & \\text{Add 3}.\\hfill \\\\ x>\\dfrac{3}{2}\\hfill & \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/p>\n<p>In interval notation, the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{4}\\left(2x - 3\\right)[\/latex] is [latex]\\left(\\dfrac{3}{2},\\infty \\right)[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a logarithmic function, identify the domain<br \/>\n<\/strong><\/p>\n<ol>\n<li>Set up an inequality showing the argument greater than zero.<\/li>\n<li>Solve for [latex]x[\/latex].<\/li>\n<li>Write the domain in interval notation.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q786555\">Show Solution<\/button><\/p>\n<div id=\"q786555\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+3>0\\hfill & \\text{The argument must be positive}.\\hfill \\\\ x>-3\\hfill & \\text{Subtract 3}.\\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The logarithmic function is defined only when the argument is positive, so this function is defined when [latex]x+3>0[\/latex].<\/p>\n<p>The domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q152912\">Show Solution<\/button><\/p>\n<div id=\"q152912\" class=\"hidden-answer\" style=\"display: none\">The logarithmic function is defined only when the argument is positive, so this function is defined when [latex]5 - 2x>0[\/latex]. Solving this inequality,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5 - 2x>0\\hfill & \\text{The argument must be positive}.\\hfill \\\\ -2x>-5\\hfill & \\text{Subtract }5.\\hfill \\\\ x<\\frac{5}{2}\\hfill & \\text{Divide by }-2\\text{ and switch the inequality}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321441\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321441&theme=lumen&iframe_resize_id=ohm321441&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321442\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321442&theme=lumen&iframe_resize_id=ohm321442&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321443\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321443&theme=lumen&iframe_resize_id=ohm321443&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n","protected":false},"author":6,"menu_order":24,"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\/110"}],"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":11,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/110\/revisions"}],"predecessor-version":[{"id":5891,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/110\/revisions\/5891"}],"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\/110\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=110"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=110"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=110"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=110"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}