{"id":279,"date":"2023-06-21T13:22:51","date_gmt":"2023-06-21T13:22:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/define-and-identify-exponential-and-logarithmic-properties\/"},"modified":"2024-01-09T16:18:34","modified_gmt":"2024-01-09T16:18:34","slug":"define-and-identify-exponential-and-logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/define-and-identify-exponential-and-logarithmic-properties\/","title":{"raw":"R5.5   Define and Identify Exponential and Logarithmic Properties","rendered":"R5.5   Define and Identify Exponential and Logarithmic Properties"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li>Define the identity and zero properties of exponents and logarithms.<\/li>\r\n \t<li>Define the inverse property of exponents and logarithms.<\/li>\r\n \t<li>Define the one-to-one properties of exponents and logarithms.<\/li>\r\n \t<li>Define the product and quotient properties of exponents and logarithms.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen you learned how to solve linear equations, you were likely introduced to the properties of real numbers. These properties help us know what the rules are for isolating and combining numbers and variables. For example, it is advantageous to know that multiplication and division \"undo\" each other when you want to solve an equation for a variable that is multiplied by a number. You may have also been introduced to properties and rules for writing and using exponents. In this section, we will show you how the properties and rules for logarithms compare to those for exponents.\r\n\r\nRecall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:\r\n<p style=\"text-align: center;\">[latex]\\log_b\\left(x\\right)=y\\ \\Leftrightarrow \\ {b}^{y}=x,\\ \\text{}b&gt;0,\\ b\\ne 1[\/latex]<\/p>\r\nThat is, to say that the\u00a0<em>logarithm to base b of <\/em>\u00a0[latex]x[\/latex]<em> is\u00a0<\/em>[latex]y[\/latex]\u00a0is equivalent to saying that\u00a0[latex]y[\/latex]\u00a0<em>is the exponent on the base b that produces\u00a0<\/em>[latex]x[\/latex].\r\n\r\nNote that the base <em>b<\/em>\u00a0is always a positive number other than [latex]1[\/latex]\u00a0and\u00a0that the logarithmic and exponential forms \"undo\" each other. This means that logarithms have properties similar to exponents. Let's compare several such properties side by side.\r\n<h2>Identity and Zero Properties of Exponents and Logarithms<\/h2>\r\n<div class=\"textbox\">\r\n<h3>Zero Exponent Rule for Logarithms and Exponentials<\/h3>\r\nRecall that for exponents,\r\n<p style=\"text-align: center;\">[latex]b^0=1[\/latex].<\/p>\r\nUsing the definition of a logarithm, we can rewrite that statement in logarithm form:\r\n<p style=\"text-align: center;\">[latex]\\log_b1=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Identity Exponent Rule for Logarithms and Exponentials<\/h3>\r\nRecall that for exponents,\r\n<p style=\"text-align: center;\">[latex]b^1 = b[\/latex]<\/p>\r\nUsing the definition of a logarithm, we can rewrite that statement in logarithm form:\r\n<p style=\"text-align: center;\">[latex]\\log_bb=1[\/latex]<\/p>\r\n\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<h2>Inverse Property of Logarithms and Exponents<\/h2>\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>Inverse Property of Logarithms and Exponentials<\/h3>\r\nRecall that we can use the definition of a logarithm to rewrite it as an exponent. That is\r\n<p style=\"text-align: center;\">[latex]\\log_bx=y \\ \\Leftrightarrow b^y=x[\/latex].<\/p>\r\nBy extension,\r\n<p style=\"text-align: center;\">[latex]\\log_bb^x=x \\qquad[\/latex] and [latex]\\qquad b^{\\log_bx}=x[\/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\n<h2>One-to-One Property of Logarithms and Exponents<\/h2>\r\nConsider the mathematical statement [latex]2^x=2^y[\/latex]. The only possible way this could be a true statement is if the variables [latex]x[\/latex] and [latex]y[\/latex] are equivalent. This statement is an example of the <strong>one-to-one property of exponents<\/strong>. More formally, it states\r\n<p style=\"text-align: center;\">[latex]a^m = a^n \\Rightarrow m=n[\/latex].<\/p>\r\nIn more general terms, if like bases raised to perhaps different exponents are given to be equivalent, then the exponents themselves must also be equivalent.\r\n\r\nThe same idea applies to logarithms. If two logarithms of like base but with perhaps different arguments are given to be equivalent, then the arguments themselves must also be equivalent. More formally, the\u00a0<strong>one-to-one property of logarithms\u00a0<\/strong>states\r\n<p style=\"text-align: center;\">[latex]\\log_b(M) = \\log_b(N) \\Rightarrow M = N[\/latex].<\/p>\r\n\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\n<h2>The Product and Quotient Rules for Logarithms<\/h2>\r\nRecall that we use the <em>product rule of exponents<\/em> to combine the product of like bases raised to exponents by adding the exponents:\r\n<p style=\"text-align: center;\">[latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex].<\/p>\r\nWe 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<div class=\"textbox\">\r\n<h3>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)[\/latex]<\/p>\r\n\r\n<\/div>\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\nFor quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to simplify division of like bases raised to powers by subtracting the exponents:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x^a}{x^b}={x}^{a-b}[\/latex].<\/p>\r\nThe <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<div id=\"fs-id1165137733855\" class=\"note textbox\">\r\n<h3 class=\"title\">\u00a0The 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<div style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\dfrac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nExpand the following expression using the quotient rule for logarithms.\r\n\r\n[latex]\\mathrm{log}\\left(\\dfrac{2{x}^{2}+6x}{3x+9}\\right)[\/latex]\r\n[reveal-answer q=\"942932\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942932\"]\r\n<p id=\"fs-id1165137474733\">Factoring and canceling, we get:<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(\\dfrac{2{x}^{2}+6x}{3x+9}\\right) &amp; =\\mathrm{log}\\left(\\dfrac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill &amp; \\text{Factor the numerator and denominator}.\\hfill \\\\ &amp; =\\mathrm{log}\\left(\\dfrac{2x}{3}\\right)\\hfill &amp; \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<div class=\"equation unnumbered\">\r\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\r\n\r\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(\\dfrac{2x}{3}\\right) &amp; =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ &amp; =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLater in the module, you'll extend these properties and rules to more complicated examples and use them to write compact logarithms in expanded form and contract those already expanded. You'll also obtain the power rule for logarithms, which will be particularly helpful when writing and solving exponential models of real-world situations.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li>Define the identity and zero properties of exponents and logarithms.<\/li>\n<li>Define the inverse property of exponents and logarithms.<\/li>\n<li>Define the one-to-one properties of exponents and logarithms.<\/li>\n<li>Define the product and quotient properties of exponents and logarithms.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>When you learned how to solve linear equations, you were likely introduced to the properties of real numbers. These properties help us know what the rules are for isolating and combining numbers and variables. For example, it is advantageous to know that multiplication and division &#8220;undo&#8221; each other when you want to solve an equation for a variable that is multiplied by a number. You may have also been introduced to properties and rules for writing and using exponents. In this section, we will show you how the properties and rules for logarithms compare to those for exponents.<\/p>\n<p>Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b\\left(x\\right)=y\\ \\Leftrightarrow \\ {b}^{y}=x,\\ \\text{}b>0,\\ b\\ne 1[\/latex]<\/p>\n<p>That is, to say that the\u00a0<em>logarithm to base b of <\/em>\u00a0[latex]x[\/latex]<em> is\u00a0<\/em>[latex]y[\/latex]\u00a0is equivalent to saying that\u00a0[latex]y[\/latex]\u00a0<em>is the exponent on the base b that produces\u00a0<\/em>[latex]x[\/latex].<\/p>\n<p>Note that the base <em>b<\/em>\u00a0is always a positive number other than [latex]1[\/latex]\u00a0and\u00a0that the logarithmic and exponential forms &#8220;undo&#8221; each other. This means that logarithms have properties similar to exponents. Let&#8217;s compare several such properties side by side.<\/p>\n<h2>Identity and Zero Properties of Exponents and Logarithms<\/h2>\n<div class=\"textbox\">\n<h3>Zero Exponent Rule for Logarithms and Exponentials<\/h3>\n<p>Recall that for exponents,<\/p>\n<p style=\"text-align: center;\">[latex]b^0=1[\/latex].<\/p>\n<p>Using the definition of a logarithm, we can rewrite that statement in logarithm form:<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b1=0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Identity Exponent Rule for Logarithms and Exponentials<\/h3>\n<p>Recall that for exponents,<\/p>\n<p style=\"text-align: center;\">[latex]b^1 = b[\/latex]<\/p>\n<p>Using the definition of a logarithm, we can rewrite that statement in logarithm form:<\/p>\n<p style=\"text-align: center;\">[latex]\\log_bb=1[\/latex]<\/p>\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<h2>Inverse Property of Logarithms and Exponents<\/h2>\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>Inverse Property of Logarithms and Exponentials<\/h3>\n<p>Recall that we can use the definition of a logarithm to rewrite it as an exponent. That is<\/p>\n<p style=\"text-align: center;\">[latex]\\log_bx=y \\ \\Leftrightarrow b^y=x[\/latex].<\/p>\n<p>By extension,<\/p>\n<p style=\"text-align: center;\">[latex]\\log_bb^x=x \\qquad[\/latex] and [latex]\\qquad b^{\\log_bx}=x[\/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<h2>One-to-One Property of Logarithms and Exponents<\/h2>\n<p>Consider the mathematical statement [latex]2^x=2^y[\/latex]. The only possible way this could be a true statement is if the variables [latex]x[\/latex] and [latex]y[\/latex] are equivalent. This statement is an example of the <strong>one-to-one property of exponents<\/strong>. More formally, it states<\/p>\n<p style=\"text-align: center;\">[latex]a^m = a^n \\Rightarrow m=n[\/latex].<\/p>\n<p>In more general terms, if like bases raised to perhaps different exponents are given to be equivalent, then the exponents themselves must also be equivalent.<\/p>\n<p>The same idea applies to logarithms. If two logarithms of like base but with perhaps different arguments are given to be equivalent, then the arguments themselves must also be equivalent. More formally, the\u00a0<strong>one-to-one property of logarithms\u00a0<\/strong>states<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b(M) = \\log_b(N) \\Rightarrow M = N[\/latex].<\/p>\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<h2>The Product and Quotient Rules for Logarithms<\/h2>\n<p>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:<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex].<\/p>\n<p>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>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)[\/latex]<\/p>\n<\/div>\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>For quotients, we have a similar rule for logarithms. Recall that we use the <em>quotient rule of exponents<\/em> to simplify division of like bases raised to powers by subtracting the exponents:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x^a}{x^b}={x}^{a-b}[\/latex].<\/p>\n<p>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<div id=\"fs-id1165137733855\" class=\"note textbox\">\n<h3 class=\"title\">\u00a0The 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<div style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\dfrac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Expand the following expression using the quotient rule for logarithms.<\/p>\n<p>[latex]\\mathrm{log}\\left(\\dfrac{2{x}^{2}+6x}{3x+9}\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q942932\">Show Solution<\/span><\/p>\n<div id=\"q942932\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137474733\">Factoring and canceling, we get:<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(\\dfrac{2{x}^{2}+6x}{3x+9}\\right) & =\\mathrm{log}\\left(\\dfrac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill & \\text{Factor the numerator and denominator}.\\hfill \\\\ & =\\mathrm{log}\\left(\\dfrac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<div class=\"equation unnumbered\">\n<p id=\"fs-id1165137805392\">Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\n<div id=\"eip-46\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(\\dfrac{2x}{3}\\right) & =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ & =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Later in the module, you&#8217;ll extend these properties and rules to more complicated examples and use them to write compact logarithms in expanded form and contract those already expanded. You&#8217;ll also obtain the power rule for logarithms, which will be particularly helpful when writing and solving exponential models of real-world situations.<\/p>\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-279\">\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><\/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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":40,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-279","chapter","type-chapter","status-publish","hentry"],"part":251,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/279","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions"}],"predecessor-version":[{"id":1534,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions\/1534"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/251"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/279\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=279"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=279"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=279"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}