{"id":101,"date":"2021-06-04T18:08:58","date_gmt":"2021-06-04T18:08:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/logarithmic-properties-2\/"},"modified":"2021-06-14T23:52:41","modified_gmt":"2021-06-14T23:52:41","slug":"logarithmic-properties-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/logarithmic-properties-2\/","title":{"raw":"5.5 Logarithmic Properties","rendered":"5.5 Logarithmic Properties"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the product rule for logarithms.<\/li>\r\n \t<li>Use the quotient rule for logarithms.<\/li>\r\n \t<li>Use the power rule for logarithms.<\/li>\r\n \t<li>Expand logarithmic expressions.<\/li>\r\n \t<li>Condense logarithmic expressions.<\/li>\r\n \t<li>Use the change-of-base formula for logarithms.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<figure id=\"CNX_Precalc_Figure_04_05_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"244\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_05_001F2.jpg\" alt=\"Testing of the pH of hydrochloric acid.\" width=\"244\" height=\"382\" \/> <b>Figure 1.<\/b> The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137759741\">In chemistry, <strong>pH<\/strong> is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\r\n\r\n<ul id=\"fs-id1165135253210\">\r\n \t<li>Battery acid: 0.8<\/li>\r\n \t<li>Stomach acid: 2.7<\/li>\r\n \t<li>Orange juice: 3.3<\/li>\r\n \t<li>Pure water: 7 (at 25\u00b0 C)<\/li>\r\n \t<li>Human blood: 7.35<\/li>\r\n \t<li>Fresh coconut: 7.8<\/li>\r\n \t<li>Sodium hydroxide (lye): 14<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where <em>a<\/em>\u00a0is the concentration of hydrogen ion in the solution<\/p>\r\n\r\n<div id=\"eip-396\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\text{pH}&amp;=-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right) \\\\ &amp;=\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right) \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137472164\">The equivalence of [latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex] and [latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] is one of the logarithm properties we will examine in this section.<\/p>\r\n\r\n<h2>Use the product rule for logarithms<\/h2>\r\n<p id=\"fs-id1165137405402\">Recall that the logarithmic and exponential functions \"undo\" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\r\n\r\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}1=0\\\\ &amp;{\\mathrm{log}}_{b}b=1\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137628765\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\r\n<p id=\"fs-id1165137772010\">Next, we have the inverse property.<\/p>\r\n\r\n<div id=\"eip-896\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&amp;=x\\hfill \\\\ {b}^{{\\mathrm{log}}_{b}x}&amp;=x,x&gt;0 \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137696455\">For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\r\n<p id=\"fs-id1165134297163\">To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\r\n<p id=\"fs-id1165137592421\">Finally, we have the <strong>one-to-one<\/strong> property.<\/p>\r\n\r\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\r\n<p id=\"fs-id1165137723139\">We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for <em>x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-448\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}3x&amp;=2x+5 &amp;&amp; \\text{Set the arguments equal.} \\\\ x&amp;=5 &amp;&amp; \\text{Subtract 2}x. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135194712\">But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\r\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\r\n<p id=\"fs-id1165137645446\">Given any real number <em>x<\/em>\u00a0and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\r\n\r\n<div id=\"eip-214\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right) &amp;&amp; \\text{Substitute for }M\\text{ and }N. \\\\ &amp; ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right) &amp;&amp; \\text{Apply the product rule for exponents}. \\\\ &amp; =m+n &amp;&amp; \\text{Apply the inverse property of logs}. \\\\ &amp; ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right) &amp;&amp; \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\r\n\r\n<div id=\"eip-502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\r\n<div id=\"fs-id1165137891324\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b&gt;0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137541378\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165134223340\">How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\r\n<ol id=\"fs-id1165137748303\">\r\n \t<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\r\n \t<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_01\" class=\"example\">\r\n<div id=\"fs-id1165135458651\" class=\"exercise\">\r\n<div id=\"fs-id1165135458654\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using the Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"868590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"868590\"]\r\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137871801\">Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"989002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989002\"]\r\n\r\n[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/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\n[ohm_question hide_question_numbers=1]174314[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use the quotient and power rules for logarithms<\/h2>\r\n<section id=\"fs-id1165135151295\">\r\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{b}}={x}^{a-b}[\/latex]. The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\r\n<p id=\"fs-id1165137431410\">Given any real number <em>x\u00a0<\/em>and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\r\n\r\n<div id=\"eip-589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-303\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right) &amp;&amp; \\text{Substitute for }M\\text{ and }N. \\\\ &amp; ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right) &amp;&amp; \\text{Apply the quotient rule for exponents}. \\\\ &amp; =m-n &amp;&amp; \\text{Apply the inverse property of logs}. \\\\ &amp; ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right) &amp;&amp; \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\r\n\r\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) &amp; =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right) &amp;&amp; \\text{Factor the numerator and denominator}. \\\\ &amp; =\\mathrm{log}\\left(\\frac{2x}{3}\\right) &amp;&amp; \\text{Cancel the common factors}. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\r\n\r\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&amp;=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right) \\\\ &amp;=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right) \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137733855\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Quotient Rule for Logarithms<\/h3>\r\n<p id=\"eip-id1165135390834\">The <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137834642\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137749807\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137749813\">How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\r\n<ol id=\"fs-id1165137749817\">\r\n \t<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\r\n \t<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\r\n \t<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_02\" class=\"example\">\r\n<div id=\"fs-id1165135185904\" class=\"exercise\">\r\n<div id=\"fs-id1165135185906\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using the Quotient Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"47414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"47414\"]\r\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\[1mm] &amp;= \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right] \\\\[1mm] &amp;={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right) \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137629471\">There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and <em>x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that <em>x\u00a0<\/em>&gt; 0, <em>x\u00a0<\/em>&gt; 1, [latex]x&gt;-\\frac{4}{3}[\/latex], and <em>x\u00a0<\/em>&lt; 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137534466\">Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"51755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"51755\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/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\n[ohm_question hide_question_numbers=1]174320[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Using the Power Rule for Logarithms<\/span>\r\n<div id=\"fs-id1165137939600\" class=\"solution\"><section id=\"fs-id1165137627625\">\r\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[\/latex]? One method is as follows:<\/p>\r\n\r\n<div id=\"eip-271\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({x}^{2}\\right) &amp; ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right) \\\\ &amp; ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x \\\\ &amp; =2{\\mathrm{log}}_{b}x \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137534037\">Notice that we used the <strong>product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\r\n\r\n<div id=\"eip-702\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;100={10}^{2} &amp;&amp; \\sqrt{3}={3}^{\\frac{1}{2}} &amp;&amp; \\frac{1}{e}={e}^{-1} \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137676322\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Power Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165137676330\">The <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137639704\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137639709\">How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\r\n<ol id=\"fs-id1165137761651\">\r\n \t<li>Express the argument as a power, if needed.<\/li>\r\n \t<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_03\" class=\"example\">\r\n<div id=\"fs-id1165135593557\" class=\"exercise\">\r\n<div id=\"fs-id1165135593559\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Expanding a Logarithm with Powers<\/h3>\r\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\r\n[reveal-answer q=\"476325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"476325\"]\r\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\r\n[reveal-answer q=\"229979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229979\"]\r\n\r\n[latex]2\\mathrm{ln}x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_04\" class=\"example\">\r\n<div id=\"fs-id1165134163985\" class=\"exercise\">\r\n<div id=\"fs-id1165134163988\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\r\n[reveal-answer q=\"400323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"400323\"]\r\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"153337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153337\"]\r\n\r\n[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_05\" class=\"example\">\r\n<div id=\"fs-id1165134435871\" class=\"exercise\">\r\n<div id=\"fs-id1165134435874\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Using the Power Rule in Reverse<\/h3>\r\n<p id=\"fs-id1165137714060\">Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\n[reveal-answer q=\"851055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"851055\"]\r\n<p id=\"fs-id1165137416106\">Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\r\n<p style=\"text-align: center;\">[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137502321\">Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\n[reveal-answer q=\"294897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"294897\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}16[\/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\n[ohm_question hide_question_numbers=1]45617[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]25581[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Expand logarithmic expressions<\/span>\r\n\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165137657409\">\r\n<p id=\"fs-id1165137558543\">Taken together, the product rule, quotient rule, and power rule are often called \"laws of logs.\" Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\r\n\r\n<div id=\"eip-423\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)&amp; ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y \\\\ &amp;={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135545872\">We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:<\/p>\r\n\r\n<div id=\"eip-622\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right) &amp; ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right) \\\\ &amp; ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right) \\\\ &amp; ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C \\\\ &amp; ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135153099\">We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\r\n<p id=\"fs-id1165135153103\">With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\r\n\r\n<div id=\"Example_04_05_06\" class=\"example\">\r\n<div id=\"fs-id1165135173497\" class=\"exercise\">\r\n<div id=\"fs-id1165135173499\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Expanding Logarithms Using Product, Quotient, and Power Rules<\/h3>\r\n<p id=\"fs-id1165135173504\">Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\r\n[reveal-answer q=\"737111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737111\"]\r\n<p id=\"fs-id1165135253768\">First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137854981\">Then seeing the product in the first term, we use the product rule:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134154611\">Finally, we use the power rule on the first term:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137635162\">Expand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"765910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"765910\"]\r\n\r\n[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_07\" class=\"example\">\r\n<div id=\"fs-id1165137811250\" class=\"exercise\">\r\n<div id=\"fs-id1165137811252\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\r\n<p id=\"fs-id1165135637419\">Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"304354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"304354\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\sqrt{x}\\right) &amp; =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)} \\\\ &amp; =\\frac{1}{2}\\mathrm{log}x \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137804498\">Expand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"410566\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410566\"]\r\n\r\n[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135173426\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1165134197968\"><strong>Can we expand<\/strong> [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/p>\r\n<p id=\"fs-id1165135440437\"><em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_08\" class=\"example\">\r\n<div id=\"fs-id1165135440448\" class=\"exercise\">\r\n<div id=\"fs-id1165135440450\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Expanding Complex Logarithmic Expressions<\/h3>\r\n<p id=\"fs-id1165135150641\">Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"840332\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"840332\"]\r\n<p id=\"fs-id1165134085801\">We can expand by applying the Product and Quotient Rules.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right) &amp; ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; \\text{Apply the Quotient Rule}. \\\\ &amp; ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; {\\text{Simplify by writing 64 as 2}}^{6}. \\\\ &amp; =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) &amp;&amp; \\text{Apply the Power Rule}.\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135635247\">Expand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"303340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303340\"]\r\n\r\n[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/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\n[ohm_question hide_question_numbers=1]35034[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174326[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Condense logarithmic expressions<\/span>\r\n\r\n<\/section>\r\n<p id=\"fs-id1165135190860\">We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\r\n\r\n<div id=\"fs-id1165135190866\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135190871\">How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\r\n<ol id=\"fs-id1165137833816\">\r\n \t<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\r\n \t<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\r\n \t<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_05_09\" class=\"example\">\r\n<div id=\"fs-id1165137833837\" class=\"exercise\">\r\n<div id=\"fs-id1165137833839\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\r\n<p id=\"fs-id1165135484124\">Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"118143\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"118143\"]\r\n<p id=\"fs-id1165135527077\">Using the product and quotient rules<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135400169\">This reduces our original expression to<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137846453\">Then, using the quotient rule<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134047712\">Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\r\n[reveal-answer q=\"144049\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"144049\"]\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_10\" class=\"example\">\r\n<div id=\"fs-id1165134435881\" class=\"exercise\">\r\n<div id=\"fs-id1165134435883\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Condensing Complex Logarithmic Expressions<\/h3>\r\n<p id=\"fs-id1165134435888\">Condense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"788800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788800\"]\r\n<p id=\"fs-id1165135344097\">We apply the power rule first:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135531546\">Next we apply the product rule to the sum:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134280852\">Finally, we apply the quotient rule to the difference:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_05_11\" class=\"example\">\r\n<div id=\"fs-id1165135519254\" class=\"exercise\">\r\n<div id=\"fs-id1165135519256\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Rewriting as a Single Logarithm<\/h3>\r\n<p id=\"fs-id1165134156051\">Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"302963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"302963\"]\r\n<p id=\"fs-id1165135177652\">We apply the power rule first:<\/p>\r\n<p style=\"text-align: center;\">[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135195405\">Next we apply the product rule to the sum:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134042304\">Finally, we apply the quotient rule to the difference:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135511375\">Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\r\n[reveal-answer q=\"235871\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"235871\"]\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135627841\">Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"628183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"628183\"]\r\n\r\n[latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/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\n[ohm_question hide_question_numbers=1]174327[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_12\" class=\"example\">\r\n<div id=\"fs-id1165137400159\" class=\"exercise\">\r\n<div id=\"fs-id1165137400162\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Applying of the Laws of Logs<\/h3>\r\n<p id=\"fs-id1165135530298\">Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\r\n[reveal-answer q=\"197365\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197365\"]\r\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\r\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135251378\">How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\r\n[reveal-answer q=\"371569\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371569\"]\r\n\r\nThe pH increases by about 0.301.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Use the change-of-base formula for logarithms<\/h2>\r\n<section id=\"fs-id1165137675210\">\r\n<p id=\"fs-id1165137675216\">Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the <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>\r\n<p id=\"fs-id1165137855374\">To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\r\n<p id=\"fs-id1165137855378\">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>\r\n\r\n<div id=\"eip-643\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/div>\r\n<p id=\"fs-id1165137932683\">Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. By taking the log base [latex]n[\/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[\/latex]. It follows that<\/p>\r\n\r\n<div id=\"eip-226\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{n}\\left({b}^{y}\\right) &amp; ={\\mathrm{log}}_{n}M &amp;&amp; \\text{Apply the one-to-one property}. \\\\ y{\\mathrm{log}}_{n}b\\hfill &amp; ={\\mathrm{log}}_{n}M &amp;&amp; \\text{Apply the power rule for logarithms}. \\\\ y &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} &amp;&amp; \\text{Isolate }y. \\\\ {\\mathrm{log}}_{b}M &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} &amp;&amp; \\text{Substitute for }y. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135207389\">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>\r\n\r\n<div id=\"eip-428\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{5}36 &amp; =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)} &amp;&amp; \\text{Apply the change of base formula using base 10.} \\\\ &amp; \\approx 2.2266 &amp;&amp; \\text{Use a calculator to evaluate to 4 decimal places.} \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165134381722\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Change-of-Base Formula<\/h3>\r\n<p id=\"fs-id1165135342066\">The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\r\n<p id=\"fs-id1165135342073\">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>\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\n<p id=\"fs-id1165134042184\">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>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\r\n<p id=\"fs-id1165137935512\">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 id=\"fs-id1165137893333\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137893339\">How To: Given a logarithm with 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 id=\"fs-id1165134172563\">\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 id=\"fs-id1165134039298\">\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 id=\"Example_04_05_13\" class=\"example\">\r\n<div id=\"fs-id1165134196189\" class=\"exercise\">\r\n<div id=\"fs-id1165134196191\" class=\"problem textbox shaded\">\r\n<h3>Example 13: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\r\n<p id=\"fs-id1165134196197\">Change [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.<\/p>\r\n[reveal-answer q=\"164120\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164120\"]\r\n<p id=\"fs-id1165135444049\">Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base, <em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\r\n<p id=\"fs-id1165135690112\">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>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}M &amp; =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b} \\\\[1mm] {\\mathrm{log}}_{5}3 &amp; =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134223311\">Change [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.<\/p>\r\n[reveal-answer q=\"562111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"562111\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135181811\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1172294599410\"><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\r\n<p id=\"fs-id1165135193274\"><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>\r\n\r\n<\/div>\r\n<div id=\"Example_04_05_14\" class=\"example\">\r\n<div id=\"fs-id1165134084328\" class=\"exercise\">\r\n<div id=\"fs-id1165134084330\" class=\"problem textbox shaded\">\r\n<h3>Example 14: Using the Change-of-Base Formula with a Calculator<\/h3>\r\n<p id=\"fs-id1165134084335\">Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\r\n[reveal-answer q=\"264040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264040\"]\r\n<p id=\"fs-id1165135353038\">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>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{2}10&amp;=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2} &amp;&amp; \\text{Apply the change of base formula using base }e. \\\\ &amp;\\approx 3.3219 &amp;&amp; \\text{Use a calculator to evaluate to 4 decimal places}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 21<\/h3>\r\n<p id=\"fs-id1165135358918\">Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\r\n[reveal-answer q=\"97931\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"97931\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 22<\/h3>\r\n[ohm_question hide_question_numbers=1]13570[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #1d1d1d; font-size: 1.5em; font-weight: bold;\">Key Equations<\/span>\r\n\r\n<section id=\"fs-id1165135570408\" class=\"key-equations\">\r\n<table id=\"fs-id2922999\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>The Product Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Quotient Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Power Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Change-of-Base Formula<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n&gt;0,n\\ne 1,b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165134049414\" class=\"key-concepts\">\r\n<h1>Key Concepts<\/h1>\r\n<ul id=\"fs-id1165134049421\">\r\n \t<li>We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.<\/li>\r\n \t<li>We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.<\/li>\r\n \t<li>We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base.<\/li>\r\n \t<li>We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.<\/li>\r\n \t<li>The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.<\/li>\r\n \t<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\r\n \t<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and <i>e<\/i>\u00a0as the quotient of natural or common logs. That way a calculator can be used to evaluate.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137890644\" class=\"definition\">\r\n \t<dt><strong>change-of-base formula<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890649\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890654\" class=\"definition\">\r\n \t<dt><strong>power rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890659\">a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890664\" class=\"definition\">\r\n \t<dt><strong>product rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890670\">a rule of logarithms that states that the log of a product is equal to a sum of logarithms<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137890674\" class=\"definition\">\r\n \t<dt><strong>quotient rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137890679\">a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms<\/dd>\r\n<\/dl>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the product rule for logarithms.<\/li>\n<li>Use the quotient rule for logarithms.<\/li>\n<li>Use the power rule for logarithms.<\/li>\n<li>Expand logarithmic expressions.<\/li>\n<li>Condense logarithmic expressions.<\/li>\n<li>Use the change-of-base formula for logarithms.<\/li>\n<\/ul>\n<\/div>\n<figure id=\"CNX_Precalc_Figure_04_05_001\" class=\"small\">\n<div style=\"width: 254px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_05_001F2.jpg\" alt=\"Testing of the pH of hydrochloric acid.\" width=\"244\" height=\"382\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137759741\">In chemistry, <strong>pH<\/strong> is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\n<ul id=\"fs-id1165135253210\">\n<li>Battery acid: 0.8<\/li>\n<li>Stomach acid: 2.7<\/li>\n<li>Orange juice: 3.3<\/li>\n<li>Pure water: 7 (at 25\u00b0 C)<\/li>\n<li>Human blood: 7.35<\/li>\n<li>Fresh coconut: 7.8<\/li>\n<li>Sodium hydroxide (lye): 14<\/li>\n<\/ul>\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where <em>a<\/em>\u00a0is the concentration of hydrogen ion in the solution<\/p>\n<div id=\"eip-396\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\text{pH}&=-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right) \\\\ &=\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right) \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137472164\">The equivalence of [latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex] and [latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] is one of the logarithm properties we will examine in this section.<\/p>\n<h2>Use the product rule for logarithms<\/h2>\n<p id=\"fs-id1165137405402\">Recall that the logarithmic and exponential functions &#8220;undo&#8221; each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}1=0\\\\ &{\\mathrm{log}}_{b}b=1\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137628765\">For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\n<p id=\"fs-id1165137772010\">Next, we have the inverse property.<\/p>\n<div id=\"eip-896\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({b}^{x}\\right)&=x\\hfill \\\\ {b}^{{\\mathrm{log}}_{b}x}&=x,x>0 \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137696455\">For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\n<p id=\"fs-id1165134297163\">To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\n<p id=\"fs-id1165137592421\">Finally, we have the <strong>one-to-one<\/strong> property.<\/p>\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\n<p id=\"fs-id1165137723139\">We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for <em>x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for <em>x<\/em>:<\/p>\n<div id=\"eip-448\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}3x&=2x+5 && \\text{Set the arguments equal.} \\\\ x&=5 && \\text{Subtract 2}x. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135194712\">But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\n<p id=\"fs-id1165137645446\">Given any real number <em>x<\/em>\u00a0and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\n<div id=\"eip-214\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n<p id=\"fs-id1165135160334\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\n<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right) && \\text{Substitute for }M\\text{ and }N. \\\\ & ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right) && \\text{Apply the product rule for exponents}. \\\\ & =m+n && \\text{Apply the inverse property of logs}. \\\\ & ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right) && \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137749030\">Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\n<div id=\"eip-502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\n<div id=\"fs-id1165137891324\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137541378\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165134223340\">How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\n<ol id=\"fs-id1165137748303\">\n<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\n<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_01\" class=\"example\">\n<div id=\"fs-id1165135458651\" class=\"exercise\">\n<div id=\"fs-id1165135458654\" class=\"problem textbox shaded\">\n<h3>Example 1: Using the Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165137585196\">Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q868590\">Show Solution<\/span><\/p>\n<div id=\"q868590\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137871801\">Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989002\">Show Solution<\/span><\/p>\n<div id=\"q989002\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174314\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174314&theme=oea&iframe_resize_id=ohm174314\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use the quotient and power rules for logarithms<\/h2>\n<section id=\"fs-id1165135151295\">\n<p id=\"fs-id1165135151301\">For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{b}}={x}^{a-b}[\/latex]. The <strong>quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\n<p id=\"fs-id1165137431410\">Given any real number <em>x\u00a0<\/em>and positive real numbers <em>M<\/em>, <em>N<\/em>, and <em>b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\n<div id=\"eip-589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n<p id=\"fs-id1165137733602\">Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\n<div id=\"eip-303\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right) && \\text{Substitute for }M\\text{ and }N. \\\\ & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right) && \\text{Apply the quotient rule for exponents}. \\\\ & =m-n && \\text{Apply the inverse property of logs}. \\\\ & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right) && \\text{Substitute for }m\\text{ and }n. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137474733\">For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) & =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right) && \\text{Factor the numerator and denominator}. \\\\ & =\\mathrm{log}\\left(\\frac{2x}{3}\\right) && \\text{Cancel the common factors}. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\frac{2x}{3}\\right)&=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right) \\\\ &=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right) \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137733855\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Quotient Rule for Logarithms<\/h3>\n<p id=\"eip-id1165135390834\">The <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\n<div id=\"fs-id1165137834642\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137749807\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137749813\">How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\n<ol id=\"fs-id1165137749817\">\n<li>Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\n<li>Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\n<li>Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_02\" class=\"example\">\n<div id=\"fs-id1165135185904\" class=\"exercise\">\n<div id=\"fs-id1165135185906\" class=\"problem textbox shaded\">\n<h3>Example 2: Using the Quotient Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135696743\">Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q47414\">Show Solution<\/span><\/p>\n<div id=\"q47414\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137534102\">First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137634442\">Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\[1mm] &= \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right] \\\\[1mm] &={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right) \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137629471\">There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and <em>x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that <em>x\u00a0<\/em>&gt; 0, <em>x\u00a0<\/em>&gt; 1, [latex]x>-\\frac{4}{3}[\/latex], and <em>x\u00a0<\/em>&lt; 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137534466\">Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q51755\">Show Solution<\/span><\/p>\n<div id=\"q51755\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174320\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174320&theme=oea&iframe_resize_id=ohm174320\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Using the Power Rule for Logarithms<\/span><\/p>\n<div id=\"fs-id1165137939600\" class=\"solution\">\n<section id=\"fs-id1165137627625\">\n<p id=\"fs-id1165137732439\">We\u2019ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[\/latex]? One method is as follows:<\/p>\n<div id=\"eip-271\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left({x}^{2}\\right) & ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right) \\\\ & ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x \\\\ & =2{\\mathrm{log}}_{b}x \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137534037\">Notice that we used the <strong>product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the <strong>power rule for logarithms<\/strong>, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\n<div id=\"eip-702\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&100={10}^{2} && \\sqrt{3}={3}^{\\frac{1}{2}} && \\frac{1}{e}={e}^{-1} \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137676322\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Power Rule for Logarithms<\/h3>\n<p id=\"fs-id1165137676330\">The <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137639704\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137639709\">How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\n<ol id=\"fs-id1165137761651\">\n<li>Express the argument as a power, if needed.<\/li>\n<li>Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_03\" class=\"example\">\n<div id=\"fs-id1165135593557\" class=\"exercise\">\n<div id=\"fs-id1165135593559\" class=\"problem textbox shaded\">\n<h3>Example 3: Expanding a Logarithm with Powers<\/h3>\n<p id=\"fs-id1165135593564\">Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476325\">Show Solution<\/span><\/p>\n<div id=\"q476325\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137843823\">The argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135508384\">Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229979\">Show Solution<\/span><\/p>\n<div id=\"q229979\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{ln}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_04\" class=\"example\">\n<div id=\"fs-id1165134163985\" class=\"exercise\">\n<div id=\"fs-id1165134163988\" class=\"problem textbox shaded\">\n<h3>Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n<p id=\"fs-id1165135181650\">Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q400323\">Show Solution<\/span><\/p>\n<div id=\"q400323\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827658\">Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137834568\">Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137643504\">Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q153337\">Show Solution<\/span><\/p>\n<div id=\"q153337\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_05\" class=\"example\">\n<div id=\"fs-id1165134435871\" class=\"exercise\">\n<div id=\"fs-id1165134435874\" class=\"problem textbox shaded\">\n<h3>Example 5: Using the Power Rule in Reverse<\/h3>\n<p id=\"fs-id1165137714060\">Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q851055\">Show Solution<\/span><\/p>\n<div id=\"q851055\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137416106\">Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\n<p style=\"text-align: center;\">[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137502321\">Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q294897\">Show Solution<\/span><\/p>\n<div id=\"q294897\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}16[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm45617\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=45617&theme=oea&iframe_resize_id=ohm45617\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm25581\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25581&theme=oea&iframe_resize_id=ohm25581\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Expand logarithmic expressions<\/span><\/p>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137657409\">\n<p id=\"fs-id1165137558543\">Taken together, the product rule, quotient rule, and power rule are often called &#8220;laws of logs.&#8221; Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\n<div id=\"eip-423\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)& ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y \\\\ &={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135545872\">We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:<\/p>\n<div id=\"eip-622\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right) & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right) \\\\ & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right) \\\\ & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C \\\\ & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135153099\">We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n<p id=\"fs-id1165135153103\">With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\n<div id=\"Example_04_05_06\" class=\"example\">\n<div id=\"fs-id1165135173497\" class=\"exercise\">\n<div id=\"fs-id1165135173499\" class=\"problem textbox shaded\">\n<h3>Example 6: Expanding Logarithms Using Product, Quotient, and Power Rules<\/h3>\n<p id=\"fs-id1165135173504\">Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737111\">Show Solution<\/span><\/p>\n<div id=\"q737111\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135253768\">First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137854981\">Then seeing the product in the first term, we use the product rule:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134154611\">Finally, we use the power rule on the first term:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137635162\">Expand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q765910\">Show Solution<\/span><\/p>\n<div id=\"q765910\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_07\" class=\"example\">\n<div id=\"fs-id1165137811250\" class=\"exercise\">\n<div id=\"fs-id1165137811252\" class=\"problem textbox shaded\">\n<h3>Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\n<p id=\"fs-id1165135637419\">Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q304354\">Show Solution<\/span><\/p>\n<div id=\"q304354\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(\\sqrt{x}\\right) & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)} \\\\ & =\\frac{1}{2}\\mathrm{log}x \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137804498\">Expand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410566\">Show Solution<\/span><\/p>\n<div id=\"q410566\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135173426\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1165134197968\"><strong>Can we expand<\/strong> [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/p>\n<p id=\"fs-id1165135440437\"><em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em><\/p>\n<\/div>\n<div id=\"Example_04_05_08\" class=\"example\">\n<div id=\"fs-id1165135440448\" class=\"exercise\">\n<div id=\"fs-id1165135440450\" class=\"problem textbox shaded\">\n<h3>Example 8: Expanding Complex Logarithmic Expressions<\/h3>\n<p id=\"fs-id1165135150641\">Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q840332\">Show Solution<\/span><\/p>\n<div id=\"q840332\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134085801\">We can expand by applying the Product and Quotient Rules.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right) & ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && \\text{Apply the Quotient Rule}. \\\\ & ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && {\\text{Simplify by writing 64 as 2}}^{6}. \\\\ & =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right) && \\text{Apply the Power Rule}.\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135635247\">Expand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303340\">Show Solution<\/span><\/p>\n<div id=\"q303340\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35034\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35034&theme=oea&iframe_resize_id=ohm35034\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174326\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174326&theme=oea&iframe_resize_id=ohm174326\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Condense logarithmic expressions<\/span><\/p>\n<\/section>\n<p id=\"fs-id1165135190860\">We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\n<div id=\"fs-id1165135190866\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135190871\">How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\n<ol id=\"fs-id1165137833816\">\n<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\n<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\n<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_05_09\" class=\"example\">\n<div id=\"fs-id1165137833837\" class=\"exercise\">\n<div id=\"fs-id1165137833839\" class=\"problem textbox shaded\">\n<h3>Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\n<p id=\"fs-id1165135484124\">Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q118143\">Show Solution<\/span><\/p>\n<div id=\"q118143\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135527077\">Using the product and quotient rules<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135400169\">This reduces our original expression to<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137846453\">Then, using the quotient rule<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134047712\">Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q144049\">Show Solution<\/span><\/p>\n<div id=\"q144049\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_10\" class=\"example\">\n<div id=\"fs-id1165134435881\" class=\"exercise\">\n<div id=\"fs-id1165134435883\" class=\"problem textbox shaded\">\n<h3>Example 10: Condensing Complex Logarithmic Expressions<\/h3>\n<p id=\"fs-id1165134435888\">Condense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788800\">Show Solution<\/span><\/p>\n<div id=\"q788800\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135344097\">We apply the power rule first:<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135531546\">Next we apply the product rule to the sum:<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134280852\">Finally, we apply the quotient rule to the difference:<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_05_11\" class=\"example\">\n<div id=\"fs-id1165135519254\" class=\"exercise\">\n<div id=\"fs-id1165135519256\" class=\"problem textbox shaded\">\n<h3>Example 11: Rewriting as a Single Logarithm<\/h3>\n<p id=\"fs-id1165134156051\">Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q302963\">Show Solution<\/span><\/p>\n<div id=\"q302963\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135177652\">We apply the power rule first:<\/p>\n<p style=\"text-align: center;\">[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135195405\">Next we apply the product rule to the sum:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134042304\">Finally, we apply the quotient rule to the difference:<\/p>\n<p style=\"text-align: center;\">[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135511375\">Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q235871\">Show Solution<\/span><\/p>\n<div id=\"q235871\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135627841\">Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q628183\">Show Solution<\/span><\/p>\n<div id=\"q628183\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174327\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174327&theme=oea&iframe_resize_id=ohm174327\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_04_05_12\" class=\"example\">\n<div id=\"fs-id1165137400159\" class=\"exercise\">\n<div id=\"fs-id1165137400162\" class=\"problem textbox shaded\">\n<h3>Example 12: Applying of the Laws of Logs<\/h3>\n<p id=\"fs-id1165135530298\">Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197365\">Show Solution<\/span><\/p>\n<div id=\"q197365\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135251378\">How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371569\">Show Solution<\/span><\/p>\n<div id=\"q371569\" class=\"hidden-answer\" style=\"display: none\">\n<p>The pH increases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Use the change-of-base formula for logarithms<\/h2>\n<section id=\"fs-id1165137675210\">\n<p id=\"fs-id1165137675216\">Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the <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 id=\"fs-id1165137855374\">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 id=\"fs-id1165137855378\">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<div id=\"eip-643\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/div>\n<p id=\"fs-id1165137932683\">Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. By taking the log base [latex]n[\/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[\/latex]. It follows that<\/p>\n<div id=\"eip-226\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{n}\\left({b}^{y}\\right) & ={\\mathrm{log}}_{n}M && \\text{Apply the one-to-one property}. \\\\ y{\\mathrm{log}}_{n}b\\hfill & ={\\mathrm{log}}_{n}M && \\text{Apply the power rule for logarithms}. \\\\ y & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} && \\text{Isolate }y. \\\\ {\\mathrm{log}}_{b}M & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b} && \\text{Substitute for }y. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135207389\">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<div id=\"eip-428\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{5}36 & =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)} && \\text{Apply the change of base formula using base 10.} \\\\ & \\approx 2.2266 && \\text{Use a calculator to evaluate to 4 decimal places.} \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165134381722\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Change-of-Base Formula<\/h3>\n<p id=\"fs-id1165135342066\">The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\n<p id=\"fs-id1165135342073\">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 id=\"fs-id1165134042184\">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 id=\"fs-id1165137935512\">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 id=\"fs-id1165137893333\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137893339\">How To: Given a logarithm with 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 id=\"fs-id1165134172563\">\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 id=\"fs-id1165134039298\">\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 id=\"Example_04_05_13\" class=\"example\">\n<div id=\"fs-id1165134196189\" class=\"exercise\">\n<div id=\"fs-id1165134196191\" class=\"problem textbox shaded\">\n<h3>Example 13: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\n<p id=\"fs-id1165134196197\">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=\"q164120\">Show Solution<\/span><\/p>\n<div id=\"q164120\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135444049\">Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base, <em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\n<p id=\"fs-id1165135690112\">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{align}{\\mathrm{log}}_{b}M & =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b} \\\\[1mm] {\\mathrm{log}}_{5}3 & =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134223311\">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=\"q562111\">Show Solution<\/span><\/p>\n<div id=\"q562111\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135181811\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1172294599410\"><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\n<p id=\"fs-id1165135193274\"><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 id=\"Example_04_05_14\" class=\"example\">\n<div id=\"fs-id1165134084328\" class=\"exercise\">\n<div id=\"fs-id1165134084330\" class=\"problem textbox shaded\">\n<h3>Example 14: Using the Change-of-Base Formula with a Calculator<\/h3>\n<p id=\"fs-id1165134084335\">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=\"q264040\">Show Solution<\/span><\/p>\n<div id=\"q264040\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135353038\">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 style=\"text-align: center;\">[latex]\\begin{align}{\\mathrm{log}}_{2}10&=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2} && \\text{Apply the change of base formula using base }e. \\\\ &\\approx 3.3219 && \\text{Use a calculator to evaluate to 4 decimal places}. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 21<\/h3>\n<p id=\"fs-id1165135358918\">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=\"q97931\">Show Solution<\/span><\/p>\n<div id=\"q97931\" 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>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>try it 22<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm13570\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=13570&theme=oea&iframe_resize_id=ohm13570\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #1d1d1d; font-size: 1.5em; font-weight: bold;\">Key Equations<\/span><\/p>\n<section id=\"fs-id1165135570408\" class=\"key-equations\">\n<table id=\"fs-id2922999\" summary=\"...\">\n<tbody>\n<tr>\n<td>The Product Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Quotient Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Power Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Change-of-Base Formula<\/td>\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n>0,n\\ne 1,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165134049414\" class=\"key-concepts\">\n<h1>Key Concepts<\/h1>\n<ul id=\"fs-id1165134049421\">\n<li>We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.<\/li>\n<li>We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.<\/li>\n<li>We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base.<\/li>\n<li>We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.<\/li>\n<li>The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.<\/li>\n<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\n<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and <i>e<\/i>\u00a0as the quotient of natural or common logs. That way a calculator can be used to evaluate.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137890644\" class=\"definition\">\n<dt><strong>change-of-base formula<\/strong><\/dt>\n<dd id=\"fs-id1165137890649\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137890654\" class=\"definition\">\n<dt><strong>power rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137890659\">a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137890664\" class=\"definition\">\n<dt><strong>product rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137890670\">a rule of logarithms that states that the log of a product is equal to a sum of logarithms<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137890674\" class=\"definition\">\n<dt><strong>quotient rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137890679\">a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms<\/dd>\n<\/dl>\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-101\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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":23485,"menu_order":7,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-101","chapter","type-chapter","status-publish","hentry"],"part":96,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/101","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/101\/revisions"}],"predecessor-version":[{"id":678,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/101\/revisions\/678"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/parts\/96"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/101\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/media?parent=101"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=101"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/contributor?post=101"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/license?post=101"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}