{"id":3509,"date":"2016-08-05T03:57:12","date_gmt":"2016-08-05T03:57:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=3509"},"modified":"2019-07-24T21:33:41","modified_gmt":"2019-07-24T21:33:41","slug":"introduction-natural-and-common-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/introduction-natural-and-common-logarithms\/","title":{"raw":"Product Rule for Logarithms","rendered":"Product Rule for Logarithms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define properties of logarithms, and use them to solve equations<\/li>\r\n \t<li>Define and use the product rule for logarithms<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137597501\">Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\r\n\r\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]<\/div>\r\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive\u00a0and\u00a0that the logarithmic and exponential functions \"undo\" each other. This means that logarithms have properties similar to exponents. Some important properties of logarithms are given in this section.<\/p>\r\nFirst, we will introduce some basic properties of logarithms followed by examples with integer arguments to help you get familiar with the relationship between exponents and logarithms.\r\n<div class=\"textbox\">\r\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">Zero and Identity Exponent Rule for Logarithms and Exponentials<\/h3>\r\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}1=0[\/latex], b&gt;[latex]0[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}b=1[\/latex], b&gt;[latex]0[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p style=\"text-align: left;\">Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for\u00a0the following:<\/p>\r\n<p style=\"text-align: left;\">1. [latex]{\\mathrm{log}}_{5}1=0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">2. [latex]{\\mathrm{log}}_{5}5=1[\/latex]\r\n[reveal-answer q=\"622755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"622755\"]<\/p>\r\n<p style=\"text-align: left;\">1.[latex]{\\mathrm{log}}_{5}1=0[\/latex] \u00a0since [latex]{5}^{0}=1[\/latex]<\/p>\r\n<p style=\"text-align: left;\">2.[latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: left;\">Exponential and logarithmic functions are inverses of each other, and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">Inverse Property of Logarithms and Exponentials<\/h3>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x, x&gt;0, b&gt;0, b\\ne1\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate:\r\n\r\n1.[latex]\\mathrm{log}\\left(100\\right)[\/latex]\r\n\r\n2.[latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex]\r\n[reveal-answer q=\"804776\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"804776\"]\r\n\r\n1. 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\n2. 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\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAnother property that can help us simplify logarithms is the one-to-one property. Essentially, this property states that if two logarithms have the same base, then their arguments - the stuff inside - are also equal to each other.\r\n<div class=\"textbox\">\r\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">The O<strong>ne-To-One<\/strong>\u00a0Property of Logarithms<\/h3>\r\n<p id=\"fs-id1165137592421\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for\u00a0[latex]x[\/latex].\r\n[reveal-answer q=\"964386\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"964386\"]\r\n\r\nIn order for this equation to be true, we must find a value for x such that [latex]3x=2x+5[\/latex]\r\n\r\n[latex]\\begin{array}{c}3x=2x+5\\hfill &amp; \\text{Set the arguments equal to each other}\\text{.}\\hfill \\\\ x=5\\hfill &amp; \\text{Subtract 2}x\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\nCheck your answer by substituting\u00a0[latex]5[\/latex] for\u00a0[latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{\\mathrm{log}}_{3}\\left(3\\cdot5\\right)={\\mathrm{log}}_{3}\\left(2\\cdot5+5\\right)\\\\{\\mathrm{log}}_{3}\\left(15\\right)={\\mathrm{log}}_{3}\\left(15\\right)\\end{array}[\/latex]<\/p>\r\nThis is a true statement, so we must have found the correct value for\u00a0[latex]x[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWhat if we had started with\u00a0the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.\r\n\r\nTo recap, the properties of logarithms and exponentials that can help us understand, simplify, and solve these types of functions more easily include:\r\n<ul>\r\n \t<li>Zero and Identity Exponent Rule:\u00a0[latex]{\\mathrm{log}}_{b}1=0[\/latex], [latex]b\\gt0[\/latex] and [latex]{\\mathrm{log}}_{b}b=1[\/latex], [latex]b\\gt0[\/latex]<\/li>\r\n \t<li>Inverse Property: [latex]\\begin{array}{c}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x&gt;0\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>One-To-One\u00a0Property:\u00a0[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/li>\r\n<\/ul>\r\n<h2>The Product Rule for Logarithms<\/h2>\r\n<\/div>\r\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of like bases raised to exponents by adding the exponents: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the <strong>product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3 class=\"title\" style=\"text-align: left;\">The Product Rule for Logarithms<\/h3>\r\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b&gt;0[\/latex]<\/p>\r\n\r\n<\/div>\r\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<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}{\\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]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137749030\">Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUsing the product rule for logarithms, rewrite the logarithm of a product as the sum of logarithms of its factors.\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]\r\n[reveal-answer q=\"132225\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"132225\"]\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will first factor the argument of a logarithm before expanding it with the product rule.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].\r\n[reveal-answer q=\"583943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583943\"]\r\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing\u00a0[latex]30[\/latex] as a product of primes.<\/p>\r\n\r\n<div id=\"eip-id1165137725082\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithm of each factor.<\/p>\r\n\r\n<div id=\"eip-id1165137836500\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Analysis of the Solution<\/h3>\r\nIt is tempting to use the distributive property when you see\u00a0an expression like\u00a0[latex]\\left(30x\\left(3x+4\\right)\\right)[\/latex], but in this case, it is better to leave the argument of this logarithm as a product since you can then use the product rule for logarithms to simplify the expression.\r\n\r\nThe following video provides more examples of using the product rule to expand logarithms.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=_mqlFChBT9k&amp;feature=youtu.be[\/embed]\r\n<h2 id=\"fs-id1165134223340\">Summary<\/h2>\r\nLogarithms have properties that can help us simplify and solve expressions and equations that contain logarithms. Exponentials and logarithms are inverses of each other; therefore, we can define the product rule for logarithms. We can use this to simplify or solve expressions with logarithms. Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms as follows:\r\n<ol id=\"fs-id1165137748303\">\r\n \t<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\r\n \t<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\r\n<\/ol>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define properties of logarithms, and use them to solve equations<\/li>\n<li>Define and use the product rule for logarithms<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137597501\">Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/div>\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive\u00a0and\u00a0that the logarithmic and exponential functions &#8220;undo&#8221; each other. This means that logarithms have properties similar to exponents. Some important properties of logarithms are given in this section.<\/p>\n<p>First, we will introduce some basic properties of logarithms followed by examples with integer arguments to help you get familiar with the relationship between exponents and logarithms.<\/p>\n<div class=\"textbox\">\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">Zero and Identity Exponent Rule for Logarithms and Exponentials<\/h3>\n<div id=\"eip-id1165135349439\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}1=0[\/latex], b&gt;[latex]0[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}b=1[\/latex], b&gt;[latex]0[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p style=\"text-align: left;\">Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for\u00a0the following:<\/p>\n<p style=\"text-align: left;\">1. [latex]{\\mathrm{log}}_{5}1=0[\/latex]<\/p>\n<p style=\"text-align: left;\">2. [latex]{\\mathrm{log}}_{5}5=1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q622755\">Show Solution<\/span><\/p>\n<div id=\"q622755\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">1.[latex]{\\mathrm{log}}_{5}1=0[\/latex] \u00a0since [latex]{5}^{0}=1[\/latex]<\/p>\n<p style=\"text-align: left;\">2.[latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p style=\"text-align: left;\">Exponential and logarithmic functions are inverses of each other, and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.<\/p>\n<div class=\"textbox\">\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">Inverse Property of Logarithms and Exponentials<\/h3>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x, x>0, b>0, b\\ne1\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Evaluate:<\/p>\n<p>1.[latex]\\mathrm{log}\\left(100\\right)[\/latex]<\/p>\n<p>2.[latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q804776\">Show Solution<\/span><\/p>\n<div id=\"q804776\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. 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>2. 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<\/div>\n<\/div>\n<\/div>\n<p>Another property that can help us simplify logarithms is the one-to-one property. Essentially, this property states that if two logarithms have the same base, then their arguments &#8211; the stuff inside &#8211; are also equal to each other.<\/p>\n<div class=\"textbox\">\n<h3 class=\"equation unnumbered\" style=\"text-align: left;\">The O<strong>ne-To-One<\/strong>\u00a0Property of Logarithms<\/h3>\n<p id=\"fs-id1165137592421\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/p>\n<\/div>\n<div id=\"eip-186\" class=\"equation unnumbered\" style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for\u00a0[latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q964386\">Show Solution<\/span><\/p>\n<div id=\"q964386\" class=\"hidden-answer\" style=\"display: none\">\n<p>In order for this equation to be true, we must find a value for x such that [latex]3x=2x+5[\/latex]<\/p>\n<p>[latex]\\begin{array}{c}3x=2x+5\\hfill & \\text{Set the arguments equal to each other}\\text{.}\\hfill \\\\ x=5\\hfill & \\text{Subtract 2}x\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>Check your answer by substituting\u00a0[latex]5[\/latex] for\u00a0[latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{\\mathrm{log}}_{3}\\left(3\\cdot5\\right)={\\mathrm{log}}_{3}\\left(2\\cdot5+5\\right)\\\\{\\mathrm{log}}_{3}\\left(15\\right)={\\mathrm{log}}_{3}\\left(15\\right)\\end{array}[\/latex]<\/p>\n<p>This is a true statement, so we must have found the correct value for\u00a0[latex]x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>What if we had started with\u00a0the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\n<p>To recap, the properties of logarithms and exponentials that can help us understand, simplify, and solve these types of functions more easily include:<\/p>\n<ul>\n<li>Zero and Identity Exponent Rule:\u00a0[latex]{\\mathrm{log}}_{b}1=0[\/latex], [latex]b\\gt0[\/latex] and [latex]{\\mathrm{log}}_{b}b=1[\/latex], [latex]b\\gt0[\/latex]<\/li>\n<li>Inverse Property: [latex]\\begin{array}{c}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x>0\\hfill \\end{array}[\/latex]<\/li>\n<li>One-To-One\u00a0Property:\u00a0[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/li>\n<\/ul>\n<h2>The Product Rule for Logarithms<\/h2>\n<\/div>\n<p id=\"fs-id1165137455738\">Recall that we use the <em>product rule of exponents<\/em> to combine the product of like bases raised to exponents by adding the exponents: [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<div class=\"textbox\">\n<h3 class=\"title\" style=\"text-align: left;\">The Product Rule for Logarithms<\/h3>\n<p id=\"fs-id1165135344994\">The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/p>\n<\/div>\n<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<div id=\"eip-54\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}{\\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]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137749030\">Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Using the product rule for logarithms, rewrite the logarithm of a product as the sum of logarithms of its factors.<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q132225\">Show Solution<\/span><\/p>\n<div id=\"q132225\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will first factor the argument of a logarithm before expanding it with the product rule.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q583943\">Show Solution<\/span><\/p>\n<div id=\"q583943\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137676251\">We begin by factoring the argument completely, expressing\u00a0[latex]30[\/latex] as a product of primes.<\/p>\n<div id=\"eip-id1165137725082\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/div>\n<p id=\"fs-id1165137438435\">Next we write the equivalent equation by summing the logarithm of each factor.<\/p>\n<div id=\"eip-id1165137836500\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Analysis of the Solution<\/h3>\n<p>It is tempting to use the distributive property when you see\u00a0an expression like\u00a0[latex]\\left(30x\\left(3x+4\\right)\\right)[\/latex], but in this case, it is better to leave the argument of this logarithm as a product since you can then use the product rule for logarithms to simplify the expression.<\/p>\n<p>The following video provides more examples of using the product rule to expand logarithms.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Expand Logarithms Using the Product Rule for Logs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_mqlFChBT9k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 id=\"fs-id1165134223340\">Summary<\/h2>\n<p>Logarithms have properties that can help us simplify and solve expressions and equations that contain logarithms. Exponentials and logarithms are inverses of each other; therefore, we can define the product rule for logarithms. We can use this to simplify or solve expressions with logarithms. Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms as follows:<\/p>\n<ol id=\"fs-id1165137748303\">\n<li>Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\n<li>Write the equivalent expression by summing the logarithms of each factor.<\/li>\n<\/ol>\n","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"16237b65-abb2-4502-8863-440e3c299649","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3509","chapter","type-chapter","status-publish","hentry"],"part":3515,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/3509","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":25,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/3509\/revisions"}],"predecessor-version":[{"id":5650,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/3509\/revisions\/5650"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/parts\/3515"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/3509\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/media?parent=3509"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=3509"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/contributor?post=3509"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/license?post=3509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}