{"id":284,"date":"2023-06-21T13:22:52","date_gmt":"2023-06-21T13:22:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/simplify-logarithmic-expressions\/"},"modified":"2023-07-04T04:23:46","modified_gmt":"2023-07-04T04:23:46","slug":"simplify-logarithmic-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/simplify-logarithmic-expressions\/","title":{"raw":"\u25aa   Change of Base Formula","rendered":"\u25aa   Change of Base Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Rewrite logarithms with a different base using the change of base formula.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using the Change-of-Base Formula for Logarithms<\/h2>\r\nMost calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.\r\n\r\nTo derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.\r\n\r\nGiven any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex], we show\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\r\nLet [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill &amp; ={\\mathrm{log}}_{n}M\\hfill &amp; \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill &amp; ={\\mathrm{log}}_{n}M \\hfill &amp; \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>applying the one-to-one property of logarithms and exponents<\/h3>\r\nIn the demonstration above, deriving the change-of-base formula from the definition of the logarithm, we applied the one-to-one property of logarithms to\r\n<p style=\"text-align: center;\">[latex]b^y=M[\/latex]<\/p>\r\n<p style=\"text-align: left;\">to obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\log_nb^y=\\log_nM[\/latex].<\/p>\r\nThe application of the property is sometimes referred to as a property of equality with regard to\u00a0<em>taking the log base<\/em>\u00a0[latex]n[\/latex]<em>\u00a0on both sides, where<\/em>\u00a0[latex]n[\/latex]<em> is any real number.\u00a0<\/em>\r\n\r\nRecall that the one-to-one property states that [latex]\\log_bM=\\log_bN \\Leftrightarrow M=N[\/latex]. We take the double-headed arrow to mean\u00a0<em>if and only if<\/em> and use it when the equality can be implied in either direction. Therefore, it is just as appropriate to state that\r\n\r\n[latex]M=N \\Leftrightarrow \\log_bM=\\log_bN[\/latex], which is what we did in the derivation above. That is,\r\n\r\n[latex]b^y=M \\Leftrightarrow \\log_nb^y=\\log_nM[\/latex].\r\n\r\nThe same idea applies to the one-to-one property of exponents. Since [latex]a^m=a^n \\Leftrightarrow m=n[\/latex], it is also true that given [latex]m=n[\/latex], we can write [latex]q^m=q^n[\/latex] for [latex]q[\/latex], any real number.\r\n\r\nThis idea leads to important techniques for solving logarithmic and exponential equations. Keep it in mind as you work through the rest of the module.\r\n\r\n<\/div>\r\nFor example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill &amp; =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill &amp; \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill &amp; \\approx 2.2266\\text{ }\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Change-of-Base Formula<\/h3>\r\nThe <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.\r\n\r\nFor any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\r\nIt follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a logarithm Of the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10 and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\r\n \t<li>Rewrite the log as a quotient using the change-of-base formula:\r\n<ul>\r\n \t<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\r\n \t<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\r\nChange [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.\r\n\r\n[reveal-answer q=\"303162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303162\"]\r\n\r\nBecause we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base\u00a0<em>n\u00a0<\/em>= <em>e<\/em>.\r\n\r\nWe rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}M\\hfill &amp; =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}\\hfill \\\\ {\\mathrm{log}}_{5}3\\hfill &amp; =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nEven if your calculator has a logarithm function for bases other than [latex]10[\/latex] or [latex]e[\/latex], you should become familiar with the change-of-base formula. Being able to manipulate formulas by hand is a useful skill in any quantitative or STEM-related field.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nChange [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.\r\n\r\n[reveal-answer q=\"7928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7928\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]86013[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we change common logarithms to natural logarithms?<\/strong>\r\n\r\n<em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.\r\n\r\n[reveal-answer q=\"448676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"448676\"]\r\n\r\nAccording to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.\r\n\r\n[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill &amp; \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.\r\n\r\n[reveal-answer q=\"732930\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"732930\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]35015[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nThe first graphing calculators were programmed to only handle logarithms with base 10. One clever way to create the graph of a logarithm with a different base was to change the base of the logarithm using the principles from this section.\r\n\r\nUse an online graphing tool to plot [latex]f(x)=\\frac{\\log_{10}{x}}{\\log_{10}{2}}[\/latex].\r\n\r\nFollow these steps to see a clever way to graph a logarithmic function with base other than 10 on a graphing tool that only knows base 10.\r\n<ul>\r\n \t<li>Enter the function [latex]g(x) = \\log_{2}{x}[\/latex]<\/li>\r\n \t<li>Can you tell the difference between the graph of this function and the graph of [latex]f(x)[\/latex]? Explain what you think is happening.<\/li>\r\n \t<li>Your challenge is to write two new functions [latex]h(x),\\text{ and }k(x)[\/latex] that include a slider so you can\u00a0change the base of the functions. Remember that there are restrictions on what values the base of a logarithm can take. You can click on the endpoints of the slider to change the input values.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Rewrite logarithms with a different base using the change of base formula.<\/li>\n<\/ul>\n<\/div>\n<h2>Using the Change-of-Base Formula for Logarithms<\/h2>\n<p>Most calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.<\/p>\n<p>To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\n<p>Given any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex], we show<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\n<p>Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill & ={\\mathrm{log}}_{n}M\\hfill & \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill & ={\\mathrm{log}}_{n}M \\hfill & \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>applying the one-to-one property of logarithms and exponents<\/h3>\n<p>In the demonstration above, deriving the change-of-base formula from the definition of the logarithm, we applied the one-to-one property of logarithms to<\/p>\n<p style=\"text-align: center;\">[latex]b^y=M[\/latex]<\/p>\n<p style=\"text-align: left;\">to obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\log_nb^y=\\log_nM[\/latex].<\/p>\n<p>The application of the property is sometimes referred to as a property of equality with regard to\u00a0<em>taking the log base<\/em>\u00a0[latex]n[\/latex]<em>\u00a0on both sides, where<\/em>\u00a0[latex]n[\/latex]<em> is any real number.\u00a0<\/em><\/p>\n<p>Recall that the one-to-one property states that [latex]\\log_bM=\\log_bN \\Leftrightarrow M=N[\/latex]. We take the double-headed arrow to mean\u00a0<em>if and only if<\/em> and use it when the equality can be implied in either direction. Therefore, it is just as appropriate to state that<\/p>\n<p>[latex]M=N \\Leftrightarrow \\log_bM=\\log_bN[\/latex], which is what we did in the derivation above. That is,<\/p>\n<p>[latex]b^y=M \\Leftrightarrow \\log_nb^y=\\log_nM[\/latex].<\/p>\n<p>The same idea applies to the one-to-one property of exponents. Since [latex]a^m=a^n \\Leftrightarrow m=n[\/latex], it is also true that given [latex]m=n[\/latex], we can write [latex]q^m=q^n[\/latex] for [latex]q[\/latex], any real number.<\/p>\n<p>This idea leads to important techniques for solving logarithmic and exponential equations. Keep it in mind as you work through the rest of the module.<\/p>\n<\/div>\n<p>For example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill & =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill & \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill & \\approx 2.2266\\text{ }\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Change-of-Base Formula<\/h3>\n<p>The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\n<p>For any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\n<p>It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a logarithm Of the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex]<\/h3>\n<ol>\n<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10 and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\n<li>Rewrite the log as a quotient using the change-of-base formula:\n<ul>\n<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\n<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\n<p>Change [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303162\">Show Solution<\/span><\/p>\n<div id=\"q303162\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base\u00a0<em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\n<p>We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}M\\hfill & =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}\\hfill \\\\ {\\mathrm{log}}_{5}3\\hfill & =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Even if your calculator has a logarithm function for bases other than [latex]10[\/latex] or [latex]e[\/latex], you should become familiar with the change-of-base formula. Being able to manipulate formulas by hand is a useful skill in any quantitative or STEM-related field.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Change [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7928\">Show Solution<\/span><\/p>\n<div id=\"q7928\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm86013\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86013&theme=oea&iframe_resize_id=ohm86013&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\n<p><em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q448676\">Show Solution<\/span><\/p>\n<div id=\"q448676\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.<\/p>\n<p>[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill & \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q732930\">Show Solution<\/span><\/p>\n<div id=\"q732930\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm35015\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35015&theme=oea&iframe_resize_id=ohm35015&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>The first graphing calculators were programmed to only handle logarithms with base 10. One clever way to create the graph of a logarithm with a different base was to change the base of the logarithm using the principles from this section.<\/p>\n<p>Use an online graphing tool to plot [latex]f(x)=\\frac{\\log_{10}{x}}{\\log_{10}{2}}[\/latex].<\/p>\n<p>Follow these steps to see a clever way to graph a logarithmic function with base other than 10 on a graphing tool that only knows base 10.<\/p>\n<ul>\n<li>Enter the function [latex]g(x) = \\log_{2}{x}[\/latex]<\/li>\n<li>Can you tell the difference between the graph of this function and the graph of [latex]f(x)[\/latex]? Explain what you think is happening.<\/li>\n<li>Your challenge is to write two new functions [latex]h(x),\\text{ and }k(x)[\/latex] that include a slider so you can\u00a0change the base of the functions. Remember that there are restrictions on what values the base of a logarithm can take. You can click on the endpoints of the slider to change the input values.<\/li>\n<\/ul>\n<\/div>\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-284\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Change of Base Graphically Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/umnz24xgl1\">https:\/\/www.desmos.com\/calculator\/umnz24xgl1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Question ID 35015. <strong>Authored by<\/strong>: Smart,Jim. <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>: IMathAS Community License CC-BY + GPL<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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":45,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 35015\",\"author\":\"Smart,Jim\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et 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