{"id":102,"date":"2021-06-04T18:08:58","date_gmt":"2021-06-04T18:08:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/exponential-and-logarithmic-equations\/"},"modified":"2021-06-14T23:53:14","modified_gmt":"2021-06-14T23:53:14","slug":"exponential-and-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/exponential-and-logarithmic-equations\/","title":{"raw":"5.6 Exponential and Logarithmic Equations","rendered":"5.6 Exponential and Logarithmic Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use like bases to solve exponential equations.<\/li>\r\n \t<li>Use logarithms to solve exponential equations.<\/li>\r\n \t<li>Use the definition of a logarithm to solve logarithmic equations.<\/li>\r\n \t<li>Use the one-to-one property of logarithms to solve logarithmic equations.<\/li>\r\n \t<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<figure id=\"CNX_Precalc_Figure_04_06_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_06_0012.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"488\" height=\"324\" \/> <b>Figure 1.<\/b> Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the \"rabbit plague.\" (credit: Richard Taylor, Flickr)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137871518\">In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.<\/p>\r\n<p id=\"fs-id1165135695212\">Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.<\/p>\r\n\r\n<h2>Use like bases to solve exponential equations<\/h2>\r\n<p id=\"fs-id1165134354674\">The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers <em>b<\/em>, <em>S<\/em>, and <em>T<\/em>, where [latex]b&gt;0,\\text{ }b\\ne 1[\/latex], [latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S\u00a0<\/em>= <em>T<\/em>.<\/p>\r\n<p id=\"fs-id1165137551370\">In other words, when an <strong>exponential equation<\/strong> has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.<\/p>\r\n<p id=\"fs-id1165135192889\">For example, consider the equation [latex]{3}^{4x - 7}=\\frac{{3}^{2x}}{3}[\/latex]. To solve for <i>x<\/i>, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-435\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{3}^{4x - 7} &amp; =\\frac{{3}^{2x}}{3} \\\\ {3}^{4x - 7}&amp; =\\frac{{3}^{2x}}{{3}^{1}} &amp;&amp; {\\text{Rewrite 3 as 3}}^{1}.\\\\ {3}^{4x - 7}&amp; ={3}^{2x - 1}&amp;&amp; \\text{Use the division property of exponents.} \\\\ 4x - 7&amp; =2x - 1 &amp;&amp; \\text{Apply the one-to-one property of exponents.} \\\\ 2x&amp; =6 &amp;&amp; \\text{Subtract 2}x\\text{ and add 7 to both sides.} \\\\ x &amp; =3&amp;&amp; \\text{Divide by 3.} \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137634286\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations<\/h3>\r\n<p id=\"fs-id1165135188636\">For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137702126\" class=\"equation\" style=\"text-align: center;\">[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137540162\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137426079\">How To: Given an exponential equation with the form [latex]{b}^{S}={b}^{T}[\/latex], where <i>S<\/i>\u00a0and\u00a0<em>T<\/em>\u00a0are algebraic expressions with an unknown, solve for the unknown.<\/h3>\r\n<ol id=\"fs-id1165137749890\">\r\n \t<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the exponents equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_01\" class=\"example\">\r\n<div id=\"fs-id1165137561722\" class=\"exercise\">\r\n<div id=\"fs-id1165135502074\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Solving an Exponential Equation with a Common Base<\/h3>\r\n<p id=\"fs-id1165137696680\">Solve [latex]{2}^{x - 1}={2}^{2x - 4}[\/latex].<\/p>\r\n[reveal-answer q=\"13280\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13280\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {2}^{x - 1}&amp;={2}^{2x - 4}&amp;&amp; \\text{The common base is }2. \\\\ x - 1&amp;=2x - 4&amp;&amp; \\text{By the one-to-one property the exponents must be equal}. \\\\ x&amp;=3 &amp;&amp; \\text{Solve for }x. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137533181\">Solve [latex]{5}^{2x}={5}^{3x+2}[\/latex].<\/p>\r\n[reveal-answer q=\"477277\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477277\"]\r\n\r\n<em>x\u00a0<\/em>= \u20132\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]14358[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Rewriting Equations So All Powers Have the Same Base<\/span>\r\n<div id=\"fs-id1165137730366\" class=\"solution\"><section id=\"fs-id1165137748966\"><section id=\"fs-id1165137667260\">\r\n<p id=\"fs-id1165137725147\">Sometimes the <strong>common base<\/strong> for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.<\/p>\r\n<p id=\"fs-id1165137784867\">For example, consider the equation [latex]256={4}^{x - 5}[\/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-687\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}256&amp;={4}^{x - 5}\\\\ {2}^{8}&amp;={\\left({2}^{2}\\right)}^{x - 5}&amp;&amp; \\text{Rewrite each side as a power with base 2}. \\\\ {2}^{8}&amp;={2}^{2x - 10}&amp;&amp; \\text{Use the one-to-one property of exponents}.\\\\ 8&amp;=2x - 10&amp;&amp; \\text{Apply the one-to-one property of exponents}. \\\\ 18&amp;=2x&amp;&amp; \\text{Add 10 to both sides}. \\\\ x&amp;=9&amp;&amp; \\text{Divide by 2}. \\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165135321969\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135414434\">How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it.<\/h3>\r\n<ol id=\"fs-id1165137663646\">\r\n \t<li>Rewrite each side in the equation as a power with a common base.<\/li>\r\n \t<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the exponents equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Solving Equations by Rewriting Them to Have a Common Base<\/h3>\r\n<p id=\"fs-id1165137454536\">Solve [latex]{8}^{x+2}={16}^{x+1}[\/latex].<\/p>\r\n[reveal-answer q=\"330675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"330675\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{8}^{x+2}&amp;={16}^{x+1}\\\\ {\\left({2}^{3}\\right)}^{x+2}&amp;={\\left({2}^{4}\\right)}^{x+1}&amp;&amp; \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\\\ {2}^{3x+6}&amp;={2}^{4x+4}&amp;&amp; \\text{To take a power of a power, multiply exponents}.\\\\ 3x+6&amp;=4x+4&amp;&amp; \\text{Use the one-to-one property to set the exponents equal}. \\\\ x&amp;=2&amp;&amp; \\text{Solve for }x.\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/section><\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137423021\">Solve [latex]{5}^{2x}={25}^{3x+2}[\/latex].<\/p>\r\n[reveal-answer q=\"285633\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"285633\"]\r\n\r\n<em>x\u00a0<\/em>= \u20131\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base<\/h3>\r\n<p id=\"fs-id1165137637470\">Solve [latex]{2}^{5x}=\\sqrt{2}[\/latex].<\/p>\r\n[reveal-answer q=\"567205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567205\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{2}^{5x}&amp;={2}^{\\frac{1}{2}}&amp;&amp; \\text{Write the square root of 2 as a power of }2.\\\\ 5x&amp;=\\frac{1}{2}&amp;&amp; \\text{Use the one-to-one property}. \\\\ x&amp;=\\frac{1}{10}&amp;&amp; \\text{Solve for }x. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137834466\">Solve [latex]{5}^{x}=\\sqrt{5}[\/latex].<\/p>\r\n[reveal-answer q=\"589572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589572\"]\r\n\r\n[latex]x=\\frac{1}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]38453[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135369638\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165137547216\"><strong>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?<\/strong><\/p>\r\n<p id=\"fs-id1165137435646\"><em>No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_04\" class=\"example\">\r\n<div id=\"fs-id1165137405247\" class=\"exercise\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Solving an Equation with Positive and Negative Powers<\/h3>\r\n<p id=\"fs-id1165137805073\">Solve [latex]{3}^{x+1}=-2[\/latex].<\/p>\r\n[reveal-answer q=\"662476\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"662476\"]\r\n\r\nThis equation has no solution. There is no real value of <em>x<\/em>\u00a0that will make the equation a true statement because any power of a positive number is positive.\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165137578263\">The figure below\u00a0shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010830\/CNX_Precalc_Figure_04_06_0022.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"487\" height=\"438\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137849213\" class=\"commentary\">\r\n<div class=\"mceTemp\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137847264\">Solve [latex]{2}^{x}=-100[\/latex].<\/p>\r\n[reveal-answer q=\"932786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932786\"]\r\n\r\nThe equation has no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Use logarithms to solve exponential equations<\/h2>\r\n<section id=\"fs-id1165137641700\">\r\n<p id=\"fs-id1165137939689\">Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equivalent to <em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.<\/p>\r\n\r\n<div id=\"fs-id1165137939925\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137673524\">How To: Given an exponential equation in which a common base cannot be found, solve for the unknown.<\/h3>\r\n<ol id=\"fs-id1165137784632\">\r\n \t<li>Apply the logarithm of both sides of the equation.\r\n<ul id=\"fs-id1165137824134\">\r\n \t<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\r\n \t<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the rules of logarithms to solve for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_05\" class=\"example\">\r\n<div id=\"fs-id1165137653913\" class=\"exercise\">\r\n<div id=\"fs-id1165137715230\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Solving an Equation Containing Powers of Different Bases<\/h3>\r\n<p id=\"fs-id1165137549801\">Solve [latex]{5}^{x+2}={4}^{x}[\/latex].<\/p>\r\n[reveal-answer q=\"982586\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982586\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{5}^{x+2}&amp;={4}^{x}&amp;&amp; \\text{There is no easy way to get the powers to have the same base}.\\\\ \\mathrm{ln}{5}^{x+2}&amp;=\\mathrm{ln}{4}^{x}&amp;&amp; \\text{Take ln of both sides}. \\\\ \\left(x+2\\right)\\mathrm{ln}5&amp;=x\\mathrm{ln}4&amp;&amp; \\text{Use laws of logs}.\\\\ x\\mathrm{ln}5+2\\mathrm{ln}5&amp;=x\\mathrm{ln}4&amp;&amp; \\text{Use the distributive law}. \\\\ x\\mathrm{ln}5-x\\mathrm{ln}4&amp;=-2\\mathrm{ln}5&amp;&amp; \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}. \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)&amp;=-2\\mathrm{ln}5&amp;&amp; \\text{On the left hand side, factor out an }x.\\\\ x\\mathrm{ln}\\left(\\frac{5}{4}\\right)&amp;=\\mathrm{ln}\\left(\\frac{1}{25}\\right)&amp;&amp; \\text{Use the laws of logs}.\\\\ x&amp;=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}&amp;&amp; \\text{Divide by the coefficient of }x. \\end{align}[\/latex]<\/p>\r\nYou can also finish the equation from [latex]x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5[\/latex] by dividing by the parentheses, giving an equivalent solution of:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x=\\frac{-2\\mathrm{ln}5}{\\mathrm{ln}5-\\mathrm{ln}4}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137402175\">Solve [latex]{2}^{x}={3}^{x+1}[\/latex].<\/p>\r\n[reveal-answer q=\"839428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"839428\"]\r\n\r\n[latex]x=\\frac{\\mathrm{ln}(3)}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]14365[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137470141\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1165137693320\"><strong>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/strong><\/p>\r\n<p id=\"fs-id1165135696730\"><em>Yes. The solution is x = 0.<\/em><\/p>\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137469838\">\r\n<h2>Equations Containing [latex]e[\/latex]<\/h2>\r\n<p id=\"fs-id1165137606150\">One common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it.<\/p>\r\n\r\n<div id=\"fs-id1165137727175\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137656412\">How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for\u00a0<em>t<\/em>.<\/h3>\r\n<ol id=\"fs-id1165135189859\">\r\n \t<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\r\n \t<li>Apply the natural logarithm of both sides of the equation.<\/li>\r\n \t<li>Divide both sides of the equation by <em>k<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_06\" class=\"example\">\r\n<div id=\"fs-id1165137846466\" class=\"exercise\">\r\n<div id=\"fs-id1165137846468\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Solve an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\n<p id=\"fs-id1165137581969\">Solve [latex]100=20{e}^{2t}[\/latex].<\/p>\r\n[reveal-answer q=\"783045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"783045\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}100 &amp; =20{e}^{2t}\\\\ 5 &amp; ={e}^{2t}&amp;&amp; \\text{Divide by the coefficient of the power.}\\\\ \\mathrm{ln}5&amp; =2t&amp;&amp; \\text{Take ln of both sides. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions.}\\\\ t&amp; =\\frac{\\mathrm{ln}5}{2}&amp;&amp; \\text{Divide by the coefficient of }t\\text{.}\\end{align}[\/latex]<\/p>\r\n\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165137666936\">Using laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, we use a calculator.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137431561\">Solve [latex]3{e}^{0.5t}=11[\/latex].<\/p>\r\n[reveal-answer q=\"845001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"845001\"]\r\n\r\n[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137532335\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1165135593037\"><strong>Does every equation of the form<\/strong> [latex]y=A{e}^{kt}[\/latex] <strong>have a solution?<\/strong><\/p>\r\n<p id=\"fs-id1165137423829\"><em>No. There is a solution when [latex]k\\ne 0[\/latex], and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}[\/latex].<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_07\" class=\"example\">\r\n<div id=\"fs-id1165137828228\" class=\"exercise\">\r\n<div id=\"fs-id1165137704438\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\n<p id=\"fs-id1165134223341\">Solve [latex]4{e}^{2x}+5=12[\/latex].<\/p>\r\n[reveal-answer q=\"758221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"758221\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}4{e}^{2x}+5&amp;=12\\\\ 4{e}^{2x}&amp;=7&amp;&amp; \\text{Combine like terms}.\\\\ {e}^{2x}&amp;=\\frac{7}{4}&amp;&amp; \\text{Divide by the coefficient of the power}. \\\\ 2x&amp;=\\mathrm{ln}\\left(\\frac{7}{4}\\right)&amp;&amp; \\text{Take ln of both sides}. \\\\ x&amp;=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)&amp;&amp; \\text{Solve for }x.\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137936640\">Solve [latex]3+{e}^{2t}=7{e}^{2t}[\/latex].<\/p>\r\n[reveal-answer q=\"346324\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346324\"]\r\n\r\n[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137665482\">\r\n<h2>Extraneous Solutions<\/h2>\r\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an <strong>extraneous solution<\/strong>, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.<\/p>\r\n\r\n<div id=\"Example_04_06_08\" class=\"example\">\r\n<div id=\"fs-id1165137828517\" class=\"exercise\">\r\n<div id=\"fs-id1165137828519\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Solving Exponential Functions in Quadratic Form<\/h3>\r\n<p id=\"fs-id1165137443096\">Solve [latex]{e}^{2x}-{e}^{x}=56[\/latex].<\/p>\r\n[reveal-answer q=\"351850\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"351850\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{e}^{2x}-{e}^{x} &amp; =56 \\\\ {e}^{2x}-{e}^{x}-56&amp; =0&amp;&amp; \\text{Get one side of the equation equal to zero}.\\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right) &amp; =0 &amp;&amp; \\text{Factor by the FOIL method}. \\\\ {e}^{x}+7&amp; =0{ or }{e}^{x}-8=0 &amp;&amp; \\text{If a product is zero, then one factor must be zero}.\\\\ {e}^{x}&amp; =-7{\\text{ or e}}^{x}=8&amp;&amp; \\text{Isolate the exponentials}. \\\\ {e}^{x}&amp; =8&amp;&amp; \\text{Reject the equation in which the power equals a negative number}. \\\\ x&amp; =\\mathrm{ln}8&amp;&amp; \\text{Solve the equation in which the power equals a positive number}. \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137806436\">When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[\/latex] because a positive number never equals a negative number. The solution [latex]x=\\mathrm{ln}\\left(-7\\right)[\/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134378708\">Solve [latex]{e}^{2x}={e}^{x}+2[\/latex].<\/p>\r\n[reveal-answer q=\"101352\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"101352\"]\r\n\r\n[latex]x=\\mathrm{ln}2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137470353\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"eip-id1165135393411\"><strong>Does every logarithmic equation have a solution?<\/strong><\/p>\r\n<p id=\"fs-id1165137922606\"><em>No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.<\/em><\/p>\r\n\r\n<\/div>\r\n<\/section><\/section>\r\n<h2>\u00a0Use the definition of a logarithm to solve logarithmic equations<\/h2>\r\n<p id=\"fs-id1165137862553\">We have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\r\n<p id=\"fs-id1165134148350\">For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for <em>x<\/em>:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3 \\\\ {\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3 \\\\ {\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\\\ {2}^{3}=6x - 10 \\\\ 8=6x - 10\\\\ 18=6x\\\\ x=3 \\end{gathered} \\begin{align}\\text{ } \\\\ &amp;\\text{Apply the product rule of logarithms}. \\\\&amp;\\text{Distribute}. \\\\&amp;\\text{Apply the definition of a logarithm}. \\\\&amp;\\text{Calculate }{2}^{3}. \\\\&amp;\\text{Add 10 to both sides}. \\\\&amp;\\text{Divide by 6}.\\end{align} [\/latex]<\/div>\r\n<div id=\"fs-id1165135516879\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\r\n<p id=\"fs-id1165137824007\">For any algebraic expression <em>S<\/em> and real numbers <em>b<\/em> and <em>c<\/em>, where [latex]b&gt;0,\\text{ }b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137732219\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{if and only if}{b}^{c}=S[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"Example_04_06_09\" class=\"example\">\r\n<div id=\"fs-id1165137841585\" class=\"exercise\">\r\n<div id=\"fs-id1165137725474\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Using Algebra to Solve a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165135152135\">Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\r\n[reveal-answer q=\"475091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"475091\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\mathrm{ln}x+3&amp;=7\\\\ 2\\mathrm{ln}x&amp;=4&amp;&amp; \\text{Subtract 3}. \\\\ \\mathrm{ln}x&amp;=2&amp;&amp; \\text{Divide by 2}.\\\\ x&amp;={e}^{2}&amp;&amp; \\text{Rewrite in exponential form}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137437380\">Solve [latex]6+\\mathrm{ln}x=10[\/latex].<\/p>\r\n[reveal-answer q=\"565722\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565722\"]\r\n\r\n[latex]x={e}^{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]74833[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_10\" class=\"example\">\r\n<div id=\"fs-id1165137482839\" class=\"exercise\">\r\n<div id=\"fs-id1165137557104\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\r\n<p id=\"fs-id1165137557109\">Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].<\/p>\r\n[reveal-answer q=\"321060\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"321060\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\mathrm{ln}\\left(6x\\right)&amp;=7 \\\\ \\mathrm{ln}\\left(6x\\right)&amp;=\\frac{7}{2}&amp;&amp; \\text{Divide by 2}. \\\\ 6x&amp;={e}^{\\left(\\frac{7}{2}\\right)} &amp;&amp; \\text{Use the definition of }\\mathrm{ln}. \\\\ x&amp;=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}&amp;&amp; \\text{Divide by 6}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137862630\">Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].<\/p>\r\n[reveal-answer q=\"832682\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"832682\"]\r\n\r\n[latex]x={e}^{5}-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]74835[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_11\" class=\"example\">\r\n<div id=\"fs-id1165137531831\" class=\"exercise\">\r\n<div id=\"fs-id1165137805003\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165137805008\">Solve [latex]\\mathrm{ln}x=3[\/latex].<\/p>\r\n[reveal-answer q=\"998818\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"998818\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{ln}x&amp;=3 \\\\ x&amp;={e}^{3}&amp;&amp; \\text{Use the definition of the natural logarithm.} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137443165\">Figure 2\u00a0represents the graph of the equation. On the graph, the <em>x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\r\n\r\n<figure class=\"small\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010830\/CNX_Precalc_Figure_04_06_0032.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/><\/figure>\r\n<figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\r\n<div style=\"text-align: center;\"><strong>Figure 2.<\/strong> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/div>\r\n<\/figcaption><\/figure>\r\n[\/hidden-answer]<span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134375706\">Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\r\n[reveal-answer q=\"174340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174340\"]\r\n\r\n[latex]x\\approx 9.97[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Use the one-to-one property of logarithms to solve logarithmic equations<\/h2>\r\n<section id=\"fs-id1165137755280\">\r\n<p id=\"fs-id1165135237092\">As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers <em>x\u00a0<\/em>&gt; 0, <em>S\u00a0<\/em>&gt; 0, <em>T\u00a0<\/em>&gt; 0 and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"eip-674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex].<\/div>\r\n<p id=\"eip-625\">For example,<\/p>\r\n\r\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex].<\/div>\r\n<p id=\"fs-id1165137874962\">So, if [latex]x - 1=8[\/latex], then we can solve for <em>x<\/em>, and we get <em>x\u00a0<\/em>= 9. To check, we can substitute <em>x\u00a0<\/em>= 9 into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.<\/p>\r\n<p id=\"fs-id1165135161040\">For example, consider the equation [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-227\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)&amp;=\\mathrm{log}\\left(x+4\\right) \\\\ \\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)&amp;=\\mathrm{log}\\left(x+4\\right) &amp;&amp;\\text{Apply the quotient rule of logarithms.}\\\\ \\frac{3x - 2}{2}&amp;=x+4 &amp;&amp;\\text{Apply the one to one property of a logarithm}. \\\\ 3x - 2&amp;=2x+8 &amp;&amp; \\text{Multiply by 2.}\\\\ x&amp;=10 &amp;&amp;\\text{Subtract 2}x\\text{ and add 2}. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135172191\">To check the result, substitute <em>x\u00a0<\/em>= 10 into [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex].<\/p>\r\n\r\n<div id=\"eip-316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)&amp;=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\\\ \\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)&amp;=\\mathrm{log}\\left(14\\right) \\\\ \\mathrm{log}\\left(\\frac{28}{2}\\right)&amp;=\\mathrm{log}\\left(14\\right) &amp;&amp; \\text{The solution checks}.\\end{align}[\/latex]<\/div>\r\n<div id=\"fs-id1165137538838\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\r\n<p id=\"fs-id1165137538845\">For any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165135154034\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/div>\r\n<p id=\"fs-id1165137592108\">Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137592114\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137726533\">How To: Given an equation containing logarithms, solve it using the one-to-one property.<\/h3>\r\n<ol id=\"fs-id1165137698132\">\r\n \t<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the arguments equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_12\" class=\"example\">\r\n<div id=\"fs-id1165137451547\" class=\"exercise\">\r\n<div id=\"fs-id1165137451549\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\r\n<p id=\"fs-id1165137849361\">Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"5314\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"5314\"]\r\n\r\n[latex]\\begin{align}&amp;\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\\\ &amp;{x}^{2}=2x+3&amp;&amp; \\text{Use the one-to-one property of the logarithm}. \\\\ &amp;{x}^{2}-2x - 3=0&amp;&amp; \\text{Get zero on one side before factoring}.\\\\ &amp;\\left(x - 3\\right)\\left(x+1\\right)=0&amp;&amp; \\text{Factor using FOIL}. \\\\ &amp;x - 3=0\\text{ or }x+1=0&amp;&amp; \\text{If a product is zero, one of the factors must be zero}. \\\\ &amp;x=3\\text{ or }x=-1 &amp;&amp; \\text{Solve for }x. \\end{align}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165134505611\">There are two solutions: <em>x\u00a0<\/em>= 3 or <em>x\u00a0<\/em>= \u20131. The solution <em>x\u00a0<\/em>= \u20131 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137650837\">Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}1[\/latex].<\/p>\r\n[reveal-answer q=\"530996\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530996\"]\r\n\r\n<em>x\u00a0<\/em>= 1 or <em>x\u00a0<\/em>= \u20131\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]35089[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">\u00a0Solve applied problems involving exponential and logarithmic equations<\/span>\r\n\r\n<\/section><section id=\"fs-id1165137828382\">\r\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\r\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\r\n\r\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\">Substance<\/th>\r\n<th style=\"text-align: center;\">Use<\/th>\r\n<th style=\"text-align: center;\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\r\n\r\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t} \\\\ &amp;A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}} \\\\ &amp;A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}} \\\\ &amp;A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137408740\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137408743\">\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\r\n \t<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\r\n \t<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\r\n<\/ul>\r\n<div id=\"Example_04_06_13\" class=\"example\">\r\n<div id=\"fs-id1165137628651\" class=\"exercise\">\r\n<div id=\"fs-id1165137628653\" class=\"problem textbox shaded\">\r\n<h3>Example 13: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\r\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\r\n[reveal-answer q=\"860588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"860588\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t \\\\ &amp;900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&amp;&amp; \\text{After 10% decays, 900 grams are left}. \\\\ &amp;0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&amp;&amp; \\text{Divide by 1000}. \\\\ &amp;\\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)&amp;&amp; \\text{Take ln of both sides}. \\\\ &amp;\\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t&amp;&amp; \\text{ln}\\left({e}^{M}\\right)=M \\\\ &amp;t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}&amp;&amp; \\text{Solve for }t. \\\\ &amp;t\\approx \\text{106,979,777 years}\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137426966\">How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\r\n[reveal-answer q=\"196779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196779\"]\r\n\r\n[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><span style=\"color: #1d1d1d; font-size: 1.5em; font-weight: bold;\">Key Equations<\/span>\r\n\r\n<section id=\"fs-id1165135255941\" class=\"key-equations\">\r\n<table id=\"fs-id2868258\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>One-to-one property for exponential functions<\/td>\r\n<td>For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>\u00a0and any positive real number <em>b<\/em>, where[latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S<\/em>\u00a0=\u00a0<em>T<\/em>.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of a logarithm<\/td>\r\n<td>For any algebraic expression <em>S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex],[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One-to-one property for logarithmic functions<\/td>\r\n<td>For any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],\r\n<div><\/div>\r\n[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex] if and only if <em>S<\/em> =\u00a0<em>T<\/em>.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165137431646\" class=\"key-concepts\">\r\n<h1>Key Concepts<\/h1>\r\n<ul id=\"fs-id1165137450888\">\r\n \t<li>We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.<\/li>\r\n \t<li>When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.<\/li>\r\n \t<li>When we are given an exponential equation where the bases are <em>not<\/em> explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.<\/li>\r\n \t<li>When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.<\/li>\r\n \t<li>We can solve exponential equations with base <em>e<\/em>, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\r\n \t<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\r\n \t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex], and solve for the unknown.<\/li>\r\n \t<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and <em>y\u00a0<\/em>= <em>c<\/em> on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the intersecting point.<\/li>\r\n \t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex], where <em>S<\/em>\u00a0and <em>T<\/em>\u00a0are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation <em>S\u00a0<\/em>= <em>T<\/em>\u00a0for the unknown.<\/li>\r\n \t<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135149311\" class=\"definition\">\r\n \t<dt><strong>extraneous solution<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use like bases to solve exponential equations.<\/li>\n<li>Use logarithms to solve exponential equations.<\/li>\n<li>Use the definition of a logarithm to solve logarithmic equations.<\/li>\n<li>Use the one-to-one property of logarithms to solve logarithmic equations.<\/li>\n<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\n<\/ul>\n<\/div>\n<figure id=\"CNX_Precalc_Figure_04_06_001\" class=\"small\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010829\/CNX_Precalc_Figure_04_06_0012.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"488\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the &#8220;rabbit plague.&#8221; (credit: Richard Taylor, Flickr)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137871518\">In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.<\/p>\n<p id=\"fs-id1165135695212\">Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.<\/p>\n<h2>Use like bases to solve exponential equations<\/h2>\n<p id=\"fs-id1165134354674\">The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers <em>b<\/em>, <em>S<\/em>, and <em>T<\/em>, where [latex]b>0,\\text{ }b\\ne 1[\/latex], [latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S\u00a0<\/em>= <em>T<\/em>.<\/p>\n<p id=\"fs-id1165137551370\">In other words, when an <strong>exponential equation<\/strong> has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.<\/p>\n<p id=\"fs-id1165135192889\">For example, consider the equation [latex]{3}^{4x - 7}=\\frac{{3}^{2x}}{3}[\/latex]. To solve for <i>x<\/i>, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for <em>x<\/em>:<\/p>\n<div id=\"eip-435\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{3}^{4x - 7} & =\\frac{{3}^{2x}}{3} \\\\ {3}^{4x - 7}& =\\frac{{3}^{2x}}{{3}^{1}} && {\\text{Rewrite 3 as 3}}^{1}.\\\\ {3}^{4x - 7}& ={3}^{2x - 1}&& \\text{Use the division property of exponents.} \\\\ 4x - 7& =2x - 1 && \\text{Apply the one-to-one property of exponents.} \\\\ 2x& =6 && \\text{Subtract 2}x\\text{ and add 7 to both sides.} \\\\ x & =3&& \\text{Divide by 3.} \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137634286\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations<\/h3>\n<p id=\"fs-id1165135188636\">For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165137702126\" class=\"equation\" style=\"text-align: center;\">[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137540162\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137426079\">How To: Given an exponential equation with the form [latex]{b}^{S}={b}^{T}[\/latex], where <i>S<\/i>\u00a0and\u00a0<em>T<\/em>\u00a0are algebraic expressions with an unknown, solve for the unknown.<\/h3>\n<ol id=\"fs-id1165137749890\">\n<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\n<li>Use the one-to-one property to set the exponents equal.<\/li>\n<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_01\" class=\"example\">\n<div id=\"fs-id1165137561722\" class=\"exercise\">\n<div id=\"fs-id1165135502074\" class=\"problem textbox shaded\">\n<h3>Example 1: Solving an Exponential Equation with a Common Base<\/h3>\n<p id=\"fs-id1165137696680\">Solve [latex]{2}^{x - 1}={2}^{2x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13280\">Show Solution<\/span><\/p>\n<div id=\"q13280\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align} {2}^{x - 1}&={2}^{2x - 4}&& \\text{The common base is }2. \\\\ x - 1&=2x - 4&& \\text{By the one-to-one property the exponents must be equal}. \\\\ x&=3 && \\text{Solve for }x. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137533181\">Solve [latex]{5}^{2x}={5}^{3x+2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477277\">Show Solution<\/span><\/p>\n<div id=\"q477277\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>x\u00a0<\/em>= \u20132<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14358\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14358&theme=oea&iframe_resize_id=ohm14358\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Rewriting Equations So All Powers Have the Same Base<\/span><\/p>\n<div id=\"fs-id1165137730366\" class=\"solution\">\n<section id=\"fs-id1165137748966\">\n<section id=\"fs-id1165137667260\">\n<p id=\"fs-id1165137725147\">Sometimes the <strong>common base<\/strong> for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.<\/p>\n<p id=\"fs-id1165137784867\">For example, consider the equation [latex]256={4}^{x - 5}[\/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for <em>x<\/em>:<\/p>\n<div id=\"eip-687\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}256&={4}^{x - 5}\\\\ {2}^{8}&={\\left({2}^{2}\\right)}^{x - 5}&& \\text{Rewrite each side as a power with base 2}. \\\\ {2}^{8}&={2}^{2x - 10}&& \\text{Use the one-to-one property of exponents}.\\\\ 8&=2x - 10&& \\text{Apply the one-to-one property of exponents}. \\\\ 18&=2x&& \\text{Add 10 to both sides}. \\\\ x&=9&& \\text{Divide by 2}. \\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165135321969\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135414434\">How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it.<\/h3>\n<ol id=\"fs-id1165137663646\">\n<li>Rewrite each side in the equation as a power with a common base.<\/li>\n<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\n<li>Use the one-to-one property to set the exponents equal.<\/li>\n<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Solving Equations by Rewriting Them to Have a Common Base<\/h3>\n<p id=\"fs-id1165137454536\">Solve [latex]{8}^{x+2}={16}^{x+1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q330675\">Show Solution<\/span><\/p>\n<div id=\"q330675\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}{8}^{x+2}&={16}^{x+1}\\\\ {\\left({2}^{3}\\right)}^{x+2}&={\\left({2}^{4}\\right)}^{x+1}&& \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\\\ {2}^{3x+6}&={2}^{4x+4}&& \\text{To take a power of a power, multiply exponents}.\\\\ 3x+6&=4x+4&& \\text{Use the one-to-one property to set the exponents equal}. \\\\ x&=2&& \\text{Solve for }x.\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137423021\">Solve [latex]{5}^{2x}={25}^{3x+2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q285633\">Show Solution<\/span><\/p>\n<div id=\"q285633\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>x\u00a0<\/em>= \u20131<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base<\/h3>\n<p id=\"fs-id1165137637470\">Solve [latex]{2}^{5x}=\\sqrt{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567205\">Show Solution<\/span><\/p>\n<div id=\"q567205\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}{2}^{5x}&={2}^{\\frac{1}{2}}&& \\text{Write the square root of 2 as a power of }2.\\\\ 5x&=\\frac{1}{2}&& \\text{Use the one-to-one property}. \\\\ x&=\\frac{1}{10}&& \\text{Solve for }x. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137834466\">Solve [latex]{5}^{x}=\\sqrt{5}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q589572\">Show Solution<\/span><\/p>\n<div id=\"q589572\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm38453\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38453&theme=oea&iframe_resize_id=ohm38453\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1165135369638\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165137547216\"><strong>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?<\/strong><\/p>\n<p id=\"fs-id1165137435646\"><em>No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.<\/em><\/p>\n<\/div>\n<div id=\"Example_04_06_04\" class=\"example\">\n<div id=\"fs-id1165137405247\" class=\"exercise\">\n<div class=\"textbox shaded\">\n<h3>Example 4: Solving an Equation with Positive and Negative Powers<\/h3>\n<p id=\"fs-id1165137805073\">Solve [latex]{3}^{x+1}=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q662476\">Show Solution<\/span><\/p>\n<div id=\"q662476\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation has no solution. There is no real value of <em>x<\/em>\u00a0that will make the equation a true statement because any power of a positive number is positive.<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165137578263\">The figure below\u00a0shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010830\/CNX_Precalc_Figure_04_06_0022.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"487\" height=\"438\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137849213\" class=\"commentary\">\n<div class=\"mceTemp\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137847264\">Solve [latex]{2}^{x}=-100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932786\">Show Solution<\/span><\/p>\n<div id=\"q932786\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation has no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Use logarithms to solve exponential equations<\/h2>\n<section id=\"fs-id1165137641700\">\n<p id=\"fs-id1165137939689\">Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equivalent to <em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.<\/p>\n<div id=\"fs-id1165137939925\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137673524\">How To: Given an exponential equation in which a common base cannot be found, solve for the unknown.<\/h3>\n<ol id=\"fs-id1165137784632\">\n<li>Apply the logarithm of both sides of the equation.\n<ul id=\"fs-id1165137824134\">\n<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\n<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\n<\/ul>\n<\/li>\n<li>Use the rules of logarithms to solve for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_05\" class=\"example\">\n<div id=\"fs-id1165137653913\" class=\"exercise\">\n<div id=\"fs-id1165137715230\" class=\"problem textbox shaded\">\n<h3>Example 5: Solving an Equation Containing Powers of Different Bases<\/h3>\n<p id=\"fs-id1165137549801\">Solve [latex]{5}^{x+2}={4}^{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q982586\">Show Solution<\/span><\/p>\n<div id=\"q982586\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}{5}^{x+2}&={4}^{x}&& \\text{There is no easy way to get the powers to have the same base}.\\\\ \\mathrm{ln}{5}^{x+2}&=\\mathrm{ln}{4}^{x}&& \\text{Take ln of both sides}. \\\\ \\left(x+2\\right)\\mathrm{ln}5&=x\\mathrm{ln}4&& \\text{Use laws of logs}.\\\\ x\\mathrm{ln}5+2\\mathrm{ln}5&=x\\mathrm{ln}4&& \\text{Use the distributive law}. \\\\ x\\mathrm{ln}5-x\\mathrm{ln}4&=-2\\mathrm{ln}5&& \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}. \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)&=-2\\mathrm{ln}5&& \\text{On the left hand side, factor out an }x.\\\\ x\\mathrm{ln}\\left(\\frac{5}{4}\\right)&=\\mathrm{ln}\\left(\\frac{1}{25}\\right)&& \\text{Use the laws of logs}.\\\\ x&=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}&& \\text{Divide by the coefficient of }x. \\end{align}[\/latex]<\/p>\n<p>You can also finish the equation from [latex]x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5[\/latex] by dividing by the parentheses, giving an equivalent solution of:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x=\\frac{-2\\mathrm{ln}5}{\\mathrm{ln}5-\\mathrm{ln}4}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137402175\">Solve [latex]{2}^{x}={3}^{x+1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q839428\">Show Solution<\/span><\/p>\n<div id=\"q839428\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{\\mathrm{ln}(3)}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14365\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14365&theme=oea&iframe_resize_id=ohm14365\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1165137470141\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1165137693320\"><strong>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/strong><\/p>\n<p id=\"fs-id1165135696730\"><em>Yes. The solution is x = 0.<\/em><\/p>\n<\/div>\n<section id=\"fs-id1165137469838\">\n<h2>Equations Containing [latex]e[\/latex]<\/h2>\n<p id=\"fs-id1165137606150\">One common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it.<\/p>\n<div id=\"fs-id1165137727175\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137656412\">How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for\u00a0<em>t<\/em>.<\/h3>\n<ol id=\"fs-id1165135189859\">\n<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\n<li>Apply the natural logarithm of both sides of the equation.<\/li>\n<li>Divide both sides of the equation by <em>k<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_06\" class=\"example\">\n<div id=\"fs-id1165137846466\" class=\"exercise\">\n<div id=\"fs-id1165137846468\" class=\"problem textbox shaded\">\n<h3>Example 6: Solve an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\n<p id=\"fs-id1165137581969\">Solve [latex]100=20{e}^{2t}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q783045\">Show Solution<\/span><\/p>\n<div id=\"q783045\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}100 & =20{e}^{2t}\\\\ 5 & ={e}^{2t}&& \\text{Divide by the coefficient of the power.}\\\\ \\mathrm{ln}5& =2t&& \\text{Take ln of both sides. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions.}\\\\ t& =\\frac{\\mathrm{ln}5}{2}&& \\text{Divide by the coefficient of }t\\text{.}\\end{align}[\/latex]<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165137666936\">Using laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, we use a calculator.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137431561\">Solve [latex]3{e}^{0.5t}=11[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q845001\">Show Solution<\/span><\/p>\n<div id=\"q845001\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137532335\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1165135593037\"><strong>Does every equation of the form<\/strong> [latex]y=A{e}^{kt}[\/latex] <strong>have a solution?<\/strong><\/p>\n<p id=\"fs-id1165137423829\"><em>No. There is a solution when [latex]k\\ne 0[\/latex], and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}[\/latex].<\/em><\/p>\n<\/div>\n<div id=\"Example_04_06_07\" class=\"example\">\n<div id=\"fs-id1165137828228\" class=\"exercise\">\n<div id=\"fs-id1165137704438\" class=\"problem textbox shaded\">\n<h3>Example 7: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\n<p id=\"fs-id1165134223341\">Solve [latex]4{e}^{2x}+5=12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q758221\">Show Solution<\/span><\/p>\n<div id=\"q758221\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}4{e}^{2x}+5&=12\\\\ 4{e}^{2x}&=7&& \\text{Combine like terms}.\\\\ {e}^{2x}&=\\frac{7}{4}&& \\text{Divide by the coefficient of the power}. \\\\ 2x&=\\mathrm{ln}\\left(\\frac{7}{4}\\right)&& \\text{Take ln of both sides}. \\\\ x&=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)&& \\text{Solve for }x.\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137936640\">Solve [latex]3+{e}^{2t}=7{e}^{2t}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346324\">Show Solution<\/span><\/p>\n<div id=\"q346324\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137665482\">\n<h2>Extraneous Solutions<\/h2>\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an <strong>extraneous solution<\/strong>, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.<\/p>\n<div id=\"Example_04_06_08\" class=\"example\">\n<div id=\"fs-id1165137828517\" class=\"exercise\">\n<div id=\"fs-id1165137828519\" class=\"problem textbox shaded\">\n<h3>Example 8: Solving Exponential Functions in Quadratic Form<\/h3>\n<p id=\"fs-id1165137443096\">Solve [latex]{e}^{2x}-{e}^{x}=56[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q351850\">Show Solution<\/span><\/p>\n<div id=\"q351850\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}{e}^{2x}-{e}^{x} & =56 \\\\ {e}^{2x}-{e}^{x}-56& =0&& \\text{Get one side of the equation equal to zero}.\\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right) & =0 && \\text{Factor by the FOIL method}. \\\\ {e}^{x}+7& =0{ or }{e}^{x}-8=0 && \\text{If a product is zero, then one factor must be zero}.\\\\ {e}^{x}& =-7{\\text{ or e}}^{x}=8&& \\text{Isolate the exponentials}. \\\\ {e}^{x}& =8&& \\text{Reject the equation in which the power equals a negative number}. \\\\ x& =\\mathrm{ln}8&& \\text{Solve the equation in which the power equals a positive number}. \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137806436\">When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[\/latex] because a positive number never equals a negative number. The solution [latex]x=\\mathrm{ln}\\left(-7\\right)[\/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134378708\">Solve [latex]{e}^{2x}={e}^{x}+2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q101352\">Show Solution<\/span><\/p>\n<div id=\"q101352\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\mathrm{ln}2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137470353\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"eip-id1165135393411\"><strong>Does every logarithmic equation have a solution?<\/strong><\/p>\n<p id=\"fs-id1165137922606\"><em>No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.<\/em><\/p>\n<\/div>\n<\/section>\n<\/section>\n<h2>\u00a0Use the definition of a logarithm to solve logarithmic equations<\/h2>\n<p id=\"fs-id1165137862553\">We have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\n<p id=\"fs-id1165134148350\">For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for <em>x<\/em>:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3 \\\\ {\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3 \\\\ {\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\\\ {2}^{3}=6x - 10 \\\\ 8=6x - 10\\\\ 18=6x\\\\ x=3 \\end{gathered} \\begin{align}\\text{ } \\\\ &\\text{Apply the product rule of logarithms}. \\\\&\\text{Distribute}. \\\\&\\text{Apply the definition of a logarithm}. \\\\&\\text{Calculate }{2}^{3}. \\\\&\\text{Add 10 to both sides}. \\\\&\\text{Divide by 6}.\\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165135516879\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\n<p id=\"fs-id1165137824007\">For any algebraic expression <em>S<\/em> and real numbers <em>b<\/em> and <em>c<\/em>, where [latex]b>0,\\text{ }b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165137732219\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{if and only if}{b}^{c}=S[\/latex]<\/div>\n<\/div>\n<div id=\"Example_04_06_09\" class=\"example\">\n<div id=\"fs-id1165137841585\" class=\"exercise\">\n<div id=\"fs-id1165137725474\" class=\"problem textbox shaded\">\n<h3>Example 9: Using Algebra to Solve a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165135152135\">Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q475091\">Show Solution<\/span><\/p>\n<div id=\"q475091\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\mathrm{ln}x+3&=7\\\\ 2\\mathrm{ln}x&=4&& \\text{Subtract 3}. \\\\ \\mathrm{ln}x&=2&& \\text{Divide by 2}.\\\\ x&={e}^{2}&& \\text{Rewrite in exponential form}. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137437380\">Solve [latex]6+\\mathrm{ln}x=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565722\">Show Solution<\/span><\/p>\n<div id=\"q565722\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm74833\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74833&theme=oea&iframe_resize_id=ohm74833\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_04_06_10\" class=\"example\">\n<div id=\"fs-id1165137482839\" class=\"exercise\">\n<div id=\"fs-id1165137557104\" class=\"problem textbox shaded\">\n<h3>Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\n<p id=\"fs-id1165137557109\">Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q321060\">Show Solution<\/span><\/p>\n<div id=\"q321060\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\mathrm{ln}\\left(6x\\right)&=7 \\\\ \\mathrm{ln}\\left(6x\\right)&=\\frac{7}{2}&& \\text{Divide by 2}. \\\\ 6x&={e}^{\\left(\\frac{7}{2}\\right)} && \\text{Use the definition of }\\mathrm{ln}. \\\\ x&=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}&& \\text{Divide by 6}. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137862630\">Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q832682\">Show Solution<\/span><\/p>\n<div id=\"q832682\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{5}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm74835\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74835&theme=oea&iframe_resize_id=ohm74835\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_04_06_11\" class=\"example\">\n<div id=\"fs-id1165137531831\" class=\"exercise\">\n<div id=\"fs-id1165137805003\" class=\"problem textbox shaded\">\n<h3>Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165137805008\">Solve [latex]\\mathrm{ln}x=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q998818\">Show Solution<\/span><\/p>\n<div id=\"q998818\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{ln}x&=3 \\\\ x&={e}^{3}&& \\text{Use the definition of the natural logarithm.} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137443165\">Figure 2\u00a0represents the graph of the equation. On the graph, the <em>x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\n<figure class=\"small\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010830\/CNX_Precalc_Figure_04_06_0032.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\n<strong>Figure 2.<\/strong> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134375706\">Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q174340\">Show Solution<\/span><\/p>\n<div id=\"q174340\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx 9.97[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Use the one-to-one property of logarithms to solve logarithmic equations<\/h2>\n<section id=\"fs-id1165137755280\">\n<p id=\"fs-id1165135237092\">As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers <em>x\u00a0<\/em>&gt; 0, <em>S\u00a0<\/em>&gt; 0, <em>T\u00a0<\/em>&gt; 0 and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<div id=\"eip-674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex].<\/div>\n<p id=\"eip-625\">For example,<\/p>\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex].<\/div>\n<p id=\"fs-id1165137874962\">So, if [latex]x - 1=8[\/latex], then we can solve for <em>x<\/em>, and we get <em>x\u00a0<\/em>= 9. To check, we can substitute <em>x\u00a0<\/em>= 9 into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.<\/p>\n<p id=\"fs-id1165135161040\">For example, consider the equation [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for <em>x<\/em>:<\/p>\n<div id=\"eip-227\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)&=\\mathrm{log}\\left(x+4\\right) \\\\ \\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)&=\\mathrm{log}\\left(x+4\\right) &&\\text{Apply the quotient rule of logarithms.}\\\\ \\frac{3x - 2}{2}&=x+4 &&\\text{Apply the one to one property of a logarithm}. \\\\ 3x - 2&=2x+8 && \\text{Multiply by 2.}\\\\ x&=10 &&\\text{Subtract 2}x\\text{ and add 2}. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135172191\">To check the result, substitute <em>x\u00a0<\/em>= 10 into [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex].<\/p>\n<div id=\"eip-316\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)&=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\\\ \\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)&=\\mathrm{log}\\left(14\\right) \\\\ \\mathrm{log}\\left(\\frac{28}{2}\\right)&=\\mathrm{log}\\left(14\\right) && \\text{The solution checks}.\\end{align}[\/latex]<\/div>\n<div id=\"fs-id1165137538838\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\n<p id=\"fs-id1165137538845\">For any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165135154034\" class=\"equation\" style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/div>\n<p id=\"fs-id1165137592108\">Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.<\/p>\n<\/div>\n<div id=\"fs-id1165137592114\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137726533\">How To: Given an equation containing logarithms, solve it using the one-to-one property.<\/h3>\n<ol id=\"fs-id1165137698132\">\n<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex].<\/li>\n<li>Use the one-to-one property to set the arguments equal.<\/li>\n<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_12\" class=\"example\">\n<div id=\"fs-id1165137451547\" class=\"exercise\">\n<div id=\"fs-id1165137451549\" class=\"problem textbox shaded\">\n<h3>Example 12: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\n<p id=\"fs-id1165137849361\">Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q5314\">Show Solution<\/span><\/p>\n<div id=\"q5314\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\\\ &{x}^{2}=2x+3&& \\text{Use the one-to-one property of the logarithm}. \\\\ &{x}^{2}-2x - 3=0&& \\text{Get zero on one side before factoring}.\\\\ &\\left(x - 3\\right)\\left(x+1\\right)=0&& \\text{Factor using FOIL}. \\\\ &x - 3=0\\text{ or }x+1=0&& \\text{If a product is zero, one of the factors must be zero}. \\\\ &x=3\\text{ or }x=-1 && \\text{Solve for }x. \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165134505611\">There are two solutions: <em>x\u00a0<\/em>= 3 or <em>x\u00a0<\/em>= \u20131. The solution <em>x\u00a0<\/em>= \u20131 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137650837\">Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530996\">Show Solution<\/span><\/p>\n<div id=\"q530996\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>x\u00a0<\/em>= 1 or <em>x\u00a0<\/em>= \u20131<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm35089\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35089&theme=oea&iframe_resize_id=ohm35089\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">\u00a0Solve applied problems involving exponential and logarithmic equations<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165137828382\">\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center;\">Substance<\/th>\n<th style=\"text-align: center;\">Use<\/th>\n<th style=\"text-align: center;\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t} \\\\ &A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}} \\\\ &A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}} \\\\ &A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137408740\">where<\/p>\n<ul id=\"fs-id1165137408743\">\n<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\n<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\n<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\n<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\n<\/ul>\n<div id=\"Example_04_06_13\" class=\"example\">\n<div id=\"fs-id1165137628651\" class=\"exercise\">\n<div id=\"fs-id1165137628653\" class=\"problem textbox shaded\">\n<h3>Example 13: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q860588\">Show Solution<\/span><\/p>\n<div id=\"q860588\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t \\\\ &900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&& \\text{After 10% decays, 900 grams are left}. \\\\ &0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&& \\text{Divide by 1000}. \\\\ &\\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)&& \\text{Take ln of both sides}. \\\\ &\\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t&& \\text{ln}\\left({e}^{M}\\right)=M \\\\ &t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}&& \\text{Solve for }t. \\\\ &t\\approx \\text{106,979,777 years}\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137426966\">How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196779\">Show Solution<\/span><\/p>\n<div id=\"q196779\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p><span style=\"color: #1d1d1d; font-size: 1.5em; font-weight: bold;\">Key Equations<\/span><\/p>\n<section id=\"fs-id1165135255941\" class=\"key-equations\">\n<table id=\"fs-id2868258\" summary=\"...\">\n<tbody>\n<tr>\n<td>One-to-one property for exponential functions<\/td>\n<td>For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>\u00a0and any positive real number <em>b<\/em>, where[latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S<\/em>\u00a0=\u00a0<em>T<\/em>.<\/td>\n<\/tr>\n<tr>\n<td>Definition of a logarithm<\/td>\n<td>For any algebraic expression <em>S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex],[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>One-to-one property for logarithmic functions<\/td>\n<td>For any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<div><\/div>\n<p>[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex] if and only if <em>S<\/em> =\u00a0<em>T<\/em>.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165137431646\" class=\"key-concepts\">\n<h1>Key Concepts<\/h1>\n<ul id=\"fs-id1165137450888\">\n<li>We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When we are given an exponential equation where the bases are <em>not<\/em> explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.<\/li>\n<li>We can solve exponential equations with base <em>e<\/em>, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\n<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex], and solve for the unknown.<\/li>\n<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and <em>y\u00a0<\/em>= <em>c<\/em> on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the intersecting point.<\/li>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex], where <em>S<\/em>\u00a0and <em>T<\/em>\u00a0are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation <em>S\u00a0<\/em>= <em>T<\/em>\u00a0for the unknown.<\/li>\n<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135149311\" class=\"definition\">\n<dt><strong>extraneous solution<\/strong><\/dt>\n<dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\n<\/dl>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-102\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":23485,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-102","chapter","type-chapter","status-publish","hentry"],"part":96,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/102\/revisions"}],"predecessor-version":[{"id":679,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/102\/revisions\/679"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/parts\/96"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/102\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/media?parent=102"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=102"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/contributor?post=102"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/license?post=102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}