{"id":2076,"date":"2016-11-03T17:08:24","date_gmt":"2016-11-03T17:08:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2076"},"modified":"2020-11-19T23:46:17","modified_gmt":"2020-11-19T23:46:17","slug":"introduction-logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/chapter\/introduction-logarithmic-properties\/","title":{"raw":"Properties of Logarithms","rendered":"Properties of Logarithms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Rewrite a logarithmic expression using the power rule, product rule, or quotient rule.<\/li>\r\n \t<li>Expand logarithmic expressions using a combination of logarithm rules.<\/li>\r\n \t<li>Condense logarithmic expressions using logarithm rules.<\/li>\r\n \t<li>Expand a logarithm\u00a0using a combination of logarithm rules.<\/li>\r\n \t<li>Condense a logarithmic expression into one logarithm.<\/li>\r\n \t<li>Rewrite logarithms with a different base using the change of base formula.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"244\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03170558\/CNX_Precalc_Figure_04_05_001F2.jpg\" alt=\"Testing of the pH of hydrochloric acid.\" width=\"244\" height=\"382\" \/> The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)[\/caption]\r\n\r\nIn 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:\r\n<ul>\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\nTo 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 [latex]a[\/latex]\u00a0is the concentration of hydrogen ion in the solution.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\text{pH} &amp; = -\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)\\hfill \\\\ \\text{} &amp; =\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)\\hfill \\end{array}[\/latex]<\/p>\r\n[latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex]\u00a0is equal to\u00a0[latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] due to\u00a0one of\u00a0the logarithm properties we will examine in this section.\r\n<h2>Properties of Logarithms<\/h2>\r\nRecall 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\{\\mathrm{log}}_{b}b=1\\end{array}[\/latex]<\/p>\r\nFor 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].\r\n\r\nNext, we have the inverse property.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x&gt;0\\hfill \\end{array}[\/latex]<\/p>\r\nFor 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].\r\n\r\nTo 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].\r\n\r\nFinally, we have the <strong>one-to-one<\/strong> property.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/p>\r\nWe 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>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x=2x+5\\hfill &amp; \\text{Set the arguments equal}\\text{.}\\hfill \\\\ x=5\\hfill &amp; \\text{Subtract 2}x\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\nBut 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\u00a0logarithms on the left side of the equation.\r\n<h3>Using the Product Rule for Logarithms<\/h3>\r\nRecall 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.\r\n\r\nGiven 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\r\n<p 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].<\/p>\r\nLet [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\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right)\\hfill &amp; \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right)\\hfill &amp; \\text{Apply the product rule for exponents}.\\hfill \\\\ \\hfill &amp; =m+n\\hfill &amp; \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\hfill &amp; \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Product Rule for Logarithms<\/h3>\r\nThe <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.\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 class=\"textbox exercises\">\r\n<h3>Example: Using the Product Rule for Logarithms<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"603492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"603492\"]\r\n\r\nWe begin by writing\u00a0an equal\u00a0equation by summing the logarithms of each factor.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(30x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)={\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The final expansion looks like this. Note how the factor [latex]30x[\/latex] can be expanded into the sum of two logarithms:<\/p>\r\n<p style=\"text-align: left;\">[latex]{\\mathrm{log}}_{3}\\left(30\\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 class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].\r\n\r\n[reveal-answer q=\"829261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"829261\"]\r\n\r\n[latex]{\\mathrm{log}}_{b}8+{\\mathrm{log}}_{b}k[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=63342&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using the Quotient Rule for Logarithms<\/h3>\r\nFor 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.\r\n\r\nGiven 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\r\n<p 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].<\/p>\r\nLet [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\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\hfill &amp; \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill &amp; \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill &amp; =m-n\\hfill &amp; \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill &amp; \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/p>\r\nFor 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\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\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)\\hfill &amp; \\text{Factor the numerator and denominator}.\\hfill \\\\ &amp; \\text{}=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill &amp; \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/p>\r\nNext we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\mathrm{log}\\left(\\frac{2x}{3}\\right) &amp; =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{} &amp; =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Quotient Rule for Logarithms<\/h3>\r\nThe <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.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Using the Quotient Rule for Logarithms<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"582435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"582435\"]\r\n\r\nFirst we note that the quotient is factored and in lowest terms, so we apply the quotient rule.\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\nNotice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\\\text{}= \\left[{\\mathrm{log}}_{2}\\left(15\\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]\\hfill \\\\ \\text{}={\\mathrm{log}}_{2}\\left(15\\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)\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThere 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.\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\nExpand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"442008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"442008\"]\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][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129646&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using the Power Rule for Logarithms<\/h3>\r\nWe have 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:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left({x}^{2}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x\\hfill \\\\ \\hfill &amp; =2{\\mathrm{log}}_{b}x\\hfill \\end{array}[\/latex]<\/p>\r\nNotice 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,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}100={10}^{2}, \\hfill &amp; \\sqrt{3}={3}^{\\frac{1}{2}}, \\hfill &amp; \\frac{1}{e}={e}^{-1}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power Rule for Logarithms<\/h3>\r\nThe <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.\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 class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Expanding a Logarithm with Powers<\/h3>\r\nRewrite [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].\r\n\r\n[reveal-answer q=\"979765\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979765\"]\r\n\r\nThe 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.\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 class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]\\mathrm{ln}{x}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"383972\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"383972\"]\r\n\r\n[latex]2\\mathrm{ln}x[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16926&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\nRewrite [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.\r\n\r\n[reveal-answer q=\"984289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984289\"]\r\n\r\nExpressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].\r\n\r\nNext we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\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 class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"947582\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947582\"]\r\n\r\n[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=63350&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Expanding Logarithms<\/h2>\r\nTaken together, the product rule, quotient rule, and power rule are often called \"properties of logs.\" Sometimes we apply more than one rule in order to expand an expression. For example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\r\nWe 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:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/p>\r\nWe can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.\r\n\r\nWith practice, we can look at a logarithmic expression and expand it mentally and then just 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a combination of the rules for logarithms to expand a logarithm<\/h3>\r\nRewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.\r\n\r\n[reveal-answer q=\"526416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"526416\"]\r\n\r\nFirst, because we have a quotient of two expressions, we can use the quotient rule:\r\n\r\n[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]\r\n\r\nThen seeing the product in the first term, we use the product rule:\r\n\r\n[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]\r\n\r\nFinally, we use the power rule on the first term:\r\n\r\n[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]\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\nExpand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"722800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"722800\"]\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<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35034&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nIn the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\r\nExpand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"914877\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"914877\"]\r\n\r\n[latex]\\begin{array}{l}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill &amp; =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill &amp; =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"2296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"2296\"]\r\n\r\n[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129752&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/strong>\r\n\r\n<em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em>\r\n\r\n<\/div>\r\nNow we will provide\u00a0some examples that will require careful attention.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding Complex Logarithmic Expressions<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"324399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324399\"]\r\n\r\nWe can expand by applying the product and quotient rules.\r\n\r\n[latex]\\begin{array}{lllllll}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)\\hfill &amp; ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill &amp; \\text{Apply the product and quotient rule}.\\hfill \\\\ \\hfill &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)\\hfill &amp; {\\text{Simplify by writing 64 as 2}}^{6}.\\hfill \\\\ \\hfill &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)\\hfill &amp; \\text{Apply the power rule}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"767425\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"767425\"]\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][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129764&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Condensing Logarithms<\/h2>\r\nWe 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.\r\n<div class=\"textbox\">\r\n<h3>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>\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>From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power Rule in Reverse<\/h3>\r\nUse the power rule for logs to rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] as a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"959478\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"959478\"]\r\n\r\nBecause 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:\r\n\r\n[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\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\nUse the power rule for logs to rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] as a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"720709\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720709\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}16[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next few examples we will use a combination of logarithm rules to condense logarithms.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\r\nWrite [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.\r\n\r\n[reveal-answer q=\"484876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484876\"]\r\n\r\nFrom left to right, since we have the addition of two logs, we first use the product rule:\r\n\r\n[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]\r\n\r\nThis simplifies our original expression to:\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]\r\n\r\nUsing the quotient rule:\r\n\r\n[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]\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\nCondense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].\r\n\r\n[reveal-answer q=\"52020\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"52020\"]\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 simplifying the fraction to lowest terms.[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129766&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Condensing Complex Logarithmic Expressions<\/h3>\r\nCondense [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].\r\n\r\n[reveal-answer q=\"841660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841660\"]\r\n\r\nWe apply the power rule first:\r\n\r\n[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]\r\n\r\nFrom left to right, since we have the addition of two logs, we apply the product rule to the sum:\r\n\r\n[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]\r\n\r\nFinally we apply the quotient rule to the difference:\r\n\r\n[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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rewriting as a Single Logarithm<\/h3>\r\nRewrite [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.\r\n\r\n[reveal-answer q=\"598036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598036\"]\r\n\r\nWe apply the power rule first:\r\n\r\n[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]\r\n\r\nFrom left to right, since we have the difference of two logs, we apply the quotient rule to the difference:\r\n\r\n[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(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]\r\n\r\nFinally we apply the product rule to the sum:\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{{x}^{2}}{{\\left(3x+5\\right)}^{{x}^{-1}}}}{{\\left(x+5\\right)}^{4}}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [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.\r\n\r\n[reveal-answer q=\"812624\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812624\"]\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nCondense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"622494\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"622494\"]\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 as [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].[\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129768&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Applications of Properties of Logarithms<\/h3>\r\nIn chemistry, pH is a measure of how acidic or basic a liquid is. It is essentially a measure of the concentration of hydrogen ions in a solution. The scale for measuring pH is standardized across the world, the scientific community having agreed upon its values and methods for acquiring them.\r\n\r\nMeasurements of pH can help scientists, farmers, doctors, and engineers solve problems and identify sources of problems.\r\n<p style=\"text-align: center;\">pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, [latex]a_{H}+[\/latex], in a solution.\r\n[latex]\\text{pH} =-\\log _{10}(a_{{\\text{H}}^{+}})=\\log _{10}\\left({\\frac {1}{a_{{\\text{H}}^{+}}}}\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">For example, a solution with a hydrogen ion activity of [latex]2.5\u00d7{10}^{-6}[\/latex] (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of [latex]\\log_{10}\\left(\\frac{1}{2.5\u00d7{10}^{-6}}\\right)=5.6[\/latex]<\/p>\r\n<p style=\"text-align: left;\">In the next examples, we will solve some problems involving pH.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying Properties of Logs<\/h3>\r\nRecall 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?\r\n\r\n[reveal-answer q=\"92345\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"92345\"]\r\n\r\nSuppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]\\text{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 [latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]\r\n\r\nUsing the product rule of logs\r\n\r\n[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]\r\n\r\nSince [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is\r\n\r\n[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]\r\n\r\nWhen the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.\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\nHow does the pH change when the concentration of positive hydrogen ions is decreased by half?\r\n\r\n[reveal-answer q=\"569571\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569571\"]The pH increases by about 0.301.[\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129783&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Change-of-Base Formula for Logarithms<\/h2>\r\nMost calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.\r\n\r\nTo derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.\r\n\r\nGiven any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex], we show\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\r\nLet [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill &amp; ={\\mathrm{log}}_{n}M\\hfill &amp; \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill &amp; ={\\mathrm{log}}_{n}M \\hfill &amp; \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\r\nFor example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill &amp; =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill &amp; \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill &amp; \\approx 2.2266\\text{ }\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Change-of-Base Formula<\/h3>\r\nThe <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.\r\n\r\nFor any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\r\nIt follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a logarithm Of the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10 and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\r\n \t<li>Rewrite the log as a quotient using the change-of-base formula:\r\n<ul>\r\n \t<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\r\n \t<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\r\nChange [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.\r\n\r\n[reveal-answer q=\"303162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303162\"]\r\n\r\nBecause we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base\u00a0<em>n\u00a0<\/em>= <em>e<\/em>.\r\n\r\nWe rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}M\\hfill &amp; =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}\\hfill \\\\ {\\mathrm{log}}_{5}3\\hfill &amp; =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nChange [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.\r\n\r\n[reveal-answer q=\"7928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7928\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86013&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we change common logarithms to natural logarithms?<\/strong>\r\n\r\n<em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.\r\n\r\n[reveal-answer q=\"448676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"448676\"]\r\n\r\nAccording to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.\r\n\r\n[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill &amp; \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.\r\n\r\n[reveal-answer q=\"732930\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"732930\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35015&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n<h2>Key Equations<\/h2>\r\n<table 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<h2>Key Concepts<\/h2>\r\n<ul>\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, quotient rule, and 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 or [latex]e[\/latex]\u00a0as the quotient of natural or common logs. A calculator can then be used to evaluate it.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135397912\" class=\"definition\">\r\n \t<dt><strong>change-of-base formula<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135397918\">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-id1165135397926\" class=\"definition\">\r\n \t<dt><strong>power rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135397932\">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-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>product rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">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-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>quotient rule for logarithms<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">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<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><\/dt>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Rewrite a logarithmic expression using the power rule, product rule, or quotient rule.<\/li>\n<li>Expand logarithmic expressions using a combination of logarithm rules.<\/li>\n<li>Condense logarithmic expressions using logarithm rules.<\/li>\n<li>Expand a logarithm\u00a0using a combination of logarithm rules.<\/li>\n<li>Condense a logarithmic expression into one logarithm.<\/li>\n<li>Rewrite logarithms with a different base using the change of base formula.<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 254px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03170558\/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\">The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)<\/p>\n<\/div>\n<p>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>\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>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 [latex]a[\/latex]\u00a0is the concentration of hydrogen ion in the solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\text{pH} & = -\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)\\hfill \\\\ \\text{} & =\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)\\hfill \\end{array}[\/latex]<\/p>\n<p>[latex]-\\mathrm{log}\\left(\\left[{H}^{+}\\right]\\right)[\/latex]\u00a0is equal to\u00a0[latex]\\mathrm{log}\\left(\\frac{1}{\\left[{H}^{+}\\right]}\\right)[\/latex] due to\u00a0one of\u00a0the logarithm properties we will examine in this section.<\/p>\n<h2>Properties of Logarithms<\/h2>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\{\\mathrm{log}}_{b}b=1\\end{array}[\/latex]<\/p>\n<p>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>Next, we have the inverse property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x>0\\hfill \\end{array}[\/latex]<\/p>\n<p>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>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>Finally, we have the <strong>one-to-one<\/strong> property.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x=2x+5\\hfill & \\text{Set the arguments equal}\\text{.}\\hfill \\\\ x=5\\hfill & \\text{Subtract 2}x\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>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\u00a0logarithms on the left side of the equation.<\/p>\n<h3>Using the Product Rule for Logarithms<\/h3>\n<p>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>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<p 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].<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}{\\mathrm{log}}_{b}\\left(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m}{b}^{n}\\right)\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m+n}\\right)\\hfill & \\text{Apply the product rule for exponents}.\\hfill \\\\ \\hfill & =m+n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Product Rule for Logarithms<\/h3>\n<p>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 class=\"textbox exercises\">\n<h3>Example: Using the Product Rule for Logarithms<\/h3>\n<p>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=\"q603492\">Show Solution<\/span><\/p>\n<div id=\"q603492\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by writing\u00a0an equal\u00a0equation 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(30x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)={\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">The final expansion looks like this. Note how the factor [latex]30x[\/latex] can be expanded into the sum of two logarithms:<\/p>\n<p style=\"text-align: left;\">[latex]{\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q829261\">Show Solution<\/span><\/p>\n<div id=\"q829261\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{b}8+{\\mathrm{log}}_{b}k[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=63342&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Quotient Rule for Logarithms<\/h3>\n<p>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>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<p 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].<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill & \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill & =m-n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\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)\\hfill & \\text{Factor the numerator and denominator}.\\hfill \\\\ & \\text{}=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\mathrm{log}\\left(\\frac{2x}{3}\\right) & =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{} & =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Quotient Rule for Logarithms<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms<\/h3>\n<ol>\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 class=\"textbox exercises\">\n<h3>Example: Using the Quotient Rule for Logarithms<\/h3>\n<p>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=\"q582435\">Show Solution<\/span><\/p>\n<div id=\"q582435\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) \\\\\\text{}= \\left[{\\mathrm{log}}_{2}\\left(15\\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]\\hfill \\\\ \\text{}={\\mathrm{log}}_{2}\\left(15\\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)\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q442008\">Show Solution<\/span><\/p>\n<div id=\"q442008\" 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><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129646&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Power Rule for Logarithms<\/h3>\n<p>We have 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<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left({x}^{2}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x\\hfill \\\\ \\hfill & =2{\\mathrm{log}}_{b}x\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{lll}100={10}^{2}, \\hfill & \\sqrt{3}={3}^{\\frac{1}{2}}, \\hfill & \\frac{1}{e}={e}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Power Rule for Logarithms<\/h3>\n<p>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 class=\"textbox\">\n<h3>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>\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 class=\"textbox exercises\">\n<h3>Example: Expanding a Logarithm with Powers<\/h3>\n<p>Rewrite [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=\"q979765\">Show Solution<\/span><\/p>\n<div id=\"q979765\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Rewrite [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=\"q383972\">Show Solution<\/span><\/p>\n<div id=\"q383972\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{ln}x[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16926&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n<p>Rewrite [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=\"q984289\">Show Solution<\/span><\/p>\n<div id=\"q984289\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Rewrite [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=\"q947582\">Show Solution<\/span><\/p>\n<div id=\"q947582\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=63350&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Expanding Logarithms<\/h2>\n<p>Taken together, the product rule, quotient rule, and power rule are often called &#8220;properties of logs.&#8221; Sometimes we apply more than one rule in order to expand an expression. For example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/p>\n<p>We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n<p>With practice, we can look at a logarithmic expression and expand it mentally and then just 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 class=\"textbox exercises\">\n<h3>Example: Using a combination of the rules for logarithms to expand a logarithm<\/h3>\n<p>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=\"q526416\">Show Solution<\/span><\/p>\n<div id=\"q526416\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\n<p>[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>Then seeing the product in the first term, we use the product rule:<\/p>\n<p>[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>Finally, we use the power rule on the first term:<\/p>\n<p>[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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q722800\">Show Solution<\/span><\/p>\n<div id=\"q722800\" 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<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35034&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>In the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\n<p>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=\"q914877\">Show Solution<\/span><\/p>\n<div id=\"q914877\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill & =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q2296\">Show Solution<\/span><\/p>\n<div id=\"q2296\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129752&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/strong><\/p>\n<p><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<p>Now we will provide\u00a0some examples that will require careful attention.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding Complex Logarithmic Expressions<\/h3>\n<p>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=\"q324399\">Show Solution<\/span><\/p>\n<div id=\"q324399\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can expand by applying the product and quotient rules.<\/p>\n<p>[latex]\\begin{array}{lllllll}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)\\hfill & ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the product and quotient rule}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & {\\text{Simplify by writing 64 as 2}}^{6}.\\hfill \\\\ \\hfill & =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the power rule}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q767425\">Show Solution<\/span><\/p>\n<div id=\"q767425\" 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><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129764&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Condensing Logarithms<\/h2>\n<p>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 class=\"textbox\">\n<h3>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>\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>From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power Rule in Reverse<\/h3>\n<p>Use the power rule for logs to rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] as 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=\"q959478\">Show Solution<\/span><\/p>\n<div id=\"q959478\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the power rule for logs to rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] as 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=\"q720709\">Show Solution<\/span><\/p>\n<div id=\"q720709\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}16[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In our next few examples we will use a combination of logarithm rules to condense logarithms.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\n<p>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=\"q484876\">Show Solution<\/span><\/p>\n<div id=\"q484876\" class=\"hidden-answer\" style=\"display: none\">\n<p>From left to right, since we have the addition of two logs, we first use the product rule:<\/p>\n<p>[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>This simplifies our original expression to:<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\n<p>Using the quotient rule:<\/p>\n<p>[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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q52020\">Show Solution<\/span><\/p>\n<div id=\"q52020\" 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 simplifying the fraction to lowest terms.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129766&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Condensing Complex Logarithmic Expressions<\/h3>\n<p>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=\"q841660\">Show Solution<\/span><\/p>\n<div id=\"q841660\" class=\"hidden-answer\" style=\"display: none\">\n<p>We apply the power rule first:<\/p>\n<p>[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>From left to right, since we have the addition of two logs, we apply the product rule to the sum:<\/p>\n<p>[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>Finally we apply the quotient rule to the difference:<\/p>\n<p>[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 class=\"textbox exercises\">\n<h3>Example: Rewriting as a Single Logarithm<\/h3>\n<p>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=\"q598036\">Show Solution<\/span><\/p>\n<div id=\"q598036\" class=\"hidden-answer\" style=\"display: none\">\n<p>We apply the power rule first:<\/p>\n<p>[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>From left to right, since we have the difference of two logs, we apply the quotient rule to the difference:<\/p>\n<p>[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(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p>Finally we apply the product rule to the sum:<\/p>\n<p>[latex]\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{{x}^{2}}{{\\left(3x+5\\right)}^{{x}^{-1}}}}{{\\left(x+5\\right)}^{4}}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q812624\">Show Solution<\/span><\/p>\n<div id=\"q812624\" 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><\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>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=\"q622494\">Show Solution<\/span><\/p>\n<div id=\"q622494\" 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 as [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129768&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Applications of Properties of Logarithms<\/h3>\n<p>In chemistry, pH is a measure of how acidic or basic a liquid is. It is essentially a measure of the concentration of hydrogen ions in a solution. The scale for measuring pH is standardized across the world, the scientific community having agreed upon its values and methods for acquiring them.<\/p>\n<p>Measurements of pH can help scientists, farmers, doctors, and engineers solve problems and identify sources of problems.<\/p>\n<p style=\"text-align: center;\">pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, [latex]a_{H}+[\/latex], in a solution.<br \/>\n[latex]\\text{pH} =-\\log _{10}(a_{{\\text{H}}^{+}})=\\log _{10}\\left({\\frac {1}{a_{{\\text{H}}^{+}}}}\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">For example, a solution with a hydrogen ion activity of [latex]2.5\u00d7{10}^{-6}[\/latex] (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of [latex]\\log_{10}\\left(\\frac{1}{2.5\u00d7{10}^{-6}}\\right)=5.6[\/latex]<\/p>\n<p style=\"text-align: left;\">In the next examples, we will solve some problems involving pH.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Applying Properties of Logs<\/h3>\n<p>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=\"q92345\">Show Solution<\/span><\/p>\n<div id=\"q92345\" class=\"hidden-answer\" style=\"display: none\">\n<p>Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]\\text{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 [latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\n<p>Using the product rule of logs<\/p>\n<p>[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>Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\n<p>[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\n<p>When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q569571\">Show Solution<\/span><\/p>\n<div id=\"q569571\" class=\"hidden-answer\" style=\"display: none\">The pH increases by about 0.301.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129783&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Change-of-Base Formula for Logarithms<\/h2>\n<p>Most calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.<\/p>\n<p>To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\n<p>Given any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex], we show<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\n<p>Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill & ={\\mathrm{log}}_{n}M\\hfill & \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill & ={\\mathrm{log}}_{n}M \\hfill & \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\n<p>For example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill & =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill & \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill & \\approx 2.2266\\text{ }\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Change-of-Base Formula<\/h3>\n<p>The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\n<p>For any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\n<p>It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a logarithm Of the form [latex]{\\mathrm{log}}_{b}M[\/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[\/latex], where [latex]n\\ne 1[\/latex]<\/h3>\n<ol>\n<li>Determine the new base <em>n<\/em>, remembering that the common log, [latex]\\mathrm{log}\\left(x\\right)[\/latex], has base 10 and the natural log, [latex]\\mathrm{ln}\\left(x\\right)[\/latex], has base <em>e<\/em>.<\/li>\n<li>Rewrite the log as a quotient using the change-of-base formula:\n<ul>\n<li>The numerator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>M<\/em>.<\/li>\n<li>The denominator of the quotient will be a logarithm with base <em>n<\/em>\u00a0and argument <em>b<\/em>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs<\/h3>\n<p>Change [latex]{\\mathrm{log}}_{5}3[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303162\">Show Solution<\/span><\/p>\n<div id=\"q303162\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we will be expressing [latex]{\\mathrm{log}}_{5}3[\/latex] as a quotient of natural logarithms, the new base\u00a0<em>n\u00a0<\/em>= <em>e<\/em>.<\/p>\n<p>We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}M\\hfill & =\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}\\hfill \\\\ {\\mathrm{log}}_{5}3\\hfill & =\\frac{\\mathrm{ln}3}{\\mathrm{ln}5}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Change [latex]{\\mathrm{log}}_{0.5}8[\/latex] to a quotient of natural logarithms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7928\">Show Solution<\/span><\/p>\n<div id=\"q7928\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86013&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\n<p><em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q448676\">Show Solution<\/span><\/p>\n<div id=\"q448676\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.<\/p>\n<p>[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill & \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q732930\">Show Solution<\/span><\/p>\n<div id=\"q732930\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35015&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<h2>Key Equations<\/h2>\n<table 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<h2>Key Concepts<\/h2>\n<ul>\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, quotient rule, and 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 or [latex]e[\/latex]\u00a0as the quotient of natural or common logs. A calculator can then be used to evaluate it.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135397912\" class=\"definition\">\n<dt><strong>change-of-base formula<\/strong><\/dt>\n<dd id=\"fs-id1165135397918\">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-id1165135397926\" class=\"definition\">\n<dt><strong>power rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165135397932\">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-id1165137838635\" class=\"definition\">\n<dt><strong>product rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">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-id1165137838635\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>quotient rule for logarithms<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><\/dt>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2076\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Change of Base Graphically. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/umnz24xgl1\">https:\/\/www.desmos.com\/calculator\/umnz24xgl1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Hydrochloric Acid pH paper. <strong>Authored by<\/strong>: David Berardan. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Hydrochloric_acid_01.jpg\">https:\/\/commons.wikimedia.org\/wiki\/File:Hydrochloric_acid_01.jpg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Question ID 63350. <strong>Authored by<\/strong>: Brin, Leon. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>pH. <strong>Authored by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/PH\">https:\/\/en.wikipedia.org\/wiki\/PH<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Question ID 35034, 35015. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 129752, 129768, 129766, 129783. <strong>Authored by<\/strong>: Day, Alyson. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/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":21,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Hydrochloric Acid pH paper\",\"author\":\"David Berardan\",\"organization\":\"\",\"url\":\"https:\/\/commons.wikimedia.org\/wiki\/File:Hydrochloric_acid_01.jpg\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 63350\",\"author\":\"Brin, Leon\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"pH\",\"author\":\"Wikipedia\",\"organization\":\"\",\"url\":\"https:\/\/en.wikipedia.org\/wiki\/PH\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 35034, 35015\",\"author\":\"Jim Smart\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 129752, 129768, 129766, 129783\",\"author\":\"Day, Alyson\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Change of Base Graphically\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.desmos.com\/calculator\/umnz24xgl1\",\"project\":\"\",\"license\":\"pd\",\"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-2076","chapter","type-chapter","status-publish","hentry"],"part":1964,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/2076","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/2076\/revisions"}],"predecessor-version":[{"id":5048,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/2076\/revisions\/5048"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/parts\/1964"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapters\/2076\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/media?parent=2076"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/pressbooks\/v2\/chapter-type?post=2076"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/contributor?post=2076"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/wp-json\/wp\/v2\/license?post=2076"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}