{"id":2095,"date":"2016-11-03T17:21:44","date_gmt":"2016-11-03T17:21:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2095"},"modified":"2020-11-19T23:47:26","modified_gmt":"2020-11-19T23:47:26","slug":"introduction-exponential-and-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebrapclc\/chapter\/introduction-exponential-and-logarithmic-equations\/","title":{"raw":"Exponential and Logarithmic Equations","rendered":"Exponential and Logarithmic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve an exponential equation with a common base.<\/li>\r\n \t<li>Rewrite an exponential equation so all terms have a common base then solve.<\/li>\r\n \t<li>Recognize when an exponential equation does not have a solution.<\/li>\r\n \t<li>Use logarithms to solve exponential equations.<\/li>\r\n \t<li>Solve a logarithmic equation algebraically.<\/li>\r\n \t<li>Solve a logarithmic equation graphically.<\/li>\r\n \t<li>Use the one-to-one property of logarithms to solve a logarithmic equation.<\/li>\r\n \t<li>Solve a radioactive decay problem.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn 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.\r\n\r\nUncontrolled 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 and logarithmic equations.\r\n<h2>Exponential Equations<\/h2>\r\nThe first technique we will introduce for solving exponential equations 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>.\r\n\r\nIn 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.\r\n\r\nFor 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>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{3}^{4x - 7}\\hfill &amp; =\\frac{{3}^{2x}}{3}\\hfill &amp; \\hfill \\\\ {3}^{4x - 7}\\hfill &amp; =\\frac{{3}^{2x}}{{3}^{1}}\\hfill &amp; {\\text{Rewrite 3 as 3}}^{1}.\\hfill \\\\ {3}^{4x - 7}\\hfill &amp; ={3}^{2x - 1}\\hfill &amp; \\text{Use the division property of exponents}\\text{.}\\hfill \\\\ 4x - 7\\hfill &amp; =2x - 1\\text{ }\\hfill &amp; \\text{Apply the one-to-one property of exponents}\\text{.}\\hfill \\\\ 2x\\hfill &amp; =6\\hfill &amp; \\text{Subtract 2}x\\text{ and add 7 to both sides}\\text{.}\\hfill \\\\ x\\hfill &amp; =3\\hfill &amp; \\text{Divide by 3}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations<\/h3>\r\nFor any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex],\r\n\r\n[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an exponential equation Of 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>\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 to each other.<\/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 exercises\">\r\n<h3>Example: Solving an Exponential Equation with a Common Base<\/h3>\r\nSolve [latex]{2}^{x - 1}={2}^{2x - 4}[\/latex].\r\n\r\n[reveal-answer q=\"766535\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"766535\"]\r\n\r\n[latex]\\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\\hfill &amp; \\text{The common base is }2.\\hfill \\\\ \\text{ }x - 1=2x - 4\\hfill &amp; \\text{By the one-to-one property the exponents must be equal}.\\hfill \\\\ \\text{ }x=3\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{5}^{2x}={5}^{3x+2}[\/latex].\r\n\r\n[reveal-answer q=\"902679\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"902679\"]\r\n\r\n[latex]x=\u20132[\/latex][\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2637&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Rewriting Equations So All Powers Have the Same Base<\/h3>\r\nSometimes 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.\r\n\r\nFor 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>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}256={4}^{x - 5}\\hfill &amp; \\hfill \\\\ {2}^{8}={\\left({2}^{2}\\right)}^{x - 5}\\hfill &amp; \\text{Rewrite each side as a power with base 2}.\\hfill \\\\ {2}^{8}={2}^{2x - 10}\\hfill &amp; \\text{To take a power of a power, multiply the exponents}.\\hfill \\\\ 8=2x - 10\\hfill &amp; \\text{Apply the one-to-one property of exponents}.\\hfill \\\\ 18=2x\\hfill &amp; \\text{Add 10 to both sides}.\\hfill \\\\ x=9\\hfill &amp; \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it<\/h3>\r\n<ol>\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 to each other.<\/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 exercises\">\r\n<h3>Example: Solving Equations by Rewriting Them to Have a Common Base<\/h3>\r\nSolve [latex]{8}^{x+2}={16}^{x+1}[\/latex].\r\n\r\n[reveal-answer q=\"214040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"214040\"]\r\n\r\n[latex]\\begin{array}{llllll}\\text{ }{8}^{x+2}={16}^{x+1}\\hfill &amp; \\hfill \\\\ {\\left({2}^{3}\\right)}^{x+2}={\\left({2}^{4}\\right)}^{x+1}\\hfill &amp; \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\hfill \\\\ \\text{ }{2}^{3x+6}={2}^{4x+4}\\hfill &amp; \\text{To take a power of a power, multiply the exponents}.\\hfill \\\\ \\text{ }3x+6=4x+4\\hfill &amp; \\text{Use the one-to-one property to set the exponents equal to each other}.\\hfill \\\\ \\text{ }x=2\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{5}^{2x}={25}^{3x+2}[\/latex].\r\n\r\n[reveal-answer q=\"660468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"660468\"]\r\n\r\n[latex]x=\u20131[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2620&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base<\/h3>\r\nSolve [latex]{2}^{5x}=\\sqrt{2}[\/latex].\r\n\r\n[reveal-answer q=\"256816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"256816\"]\r\n\r\n[latex]\\begin{array}{l}{2}^{5x}={2}^{\\frac{1}{2}}\\hfill &amp; \\text{Write the square root of 2 as a power of }2.\\hfill \\\\ 5x=\\frac{1}{2}\\hfill &amp; \\text{Use the one-to-one property}.\\hfill \\\\ x=\\frac{1}{10}\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{5}^{x}=\\sqrt{5}[\/latex].\r\n\r\n[reveal-answer q=\"743764\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"743764\"]\r\n\r\n[latex]x=\\frac{1}{2}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2638&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?<\/strong>\r\n\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining When an\u00a0Equation has No Solution<\/h3>\r\nSolve [latex]{3}^{x+1}=-2[\/latex].\r\n\r\n[reveal-answer q=\"897533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"897533\"]\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<h4>Analysis of the Solution<\/h4>\r\nThe figure below\u00a0shows that the two graphs do not cross so the left side of the equation is never equal to the right side of the equation. Thus the equation has no solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172009\/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\" \/>\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\nSolve [latex]{2}^{x}=-100[\/latex].\r\n\r\n[reveal-answer q=\"161944\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161944\"]\r\n\r\nThe equation has no solution.[\/hidden-answer]\r\n<iframe id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98554&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using Logarithms to Solve Exponential Equations<\/h3>\r\nSometimes 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 that since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equal to <em>a\u00a0<\/em>= <em>b<\/em>,<em>\u00a0<\/em>we may apply logarithms with the same base to both sides of an exponential equation.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an exponential equation Where a common base cannot be found, solve for the unknown<\/h3>\r\n<ol>\r\n \t<li>Apply the logarithm to 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 class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation Containing Powers of Different Bases<\/h3>\r\nSolve [latex]{5}^{x+2}={4}^{x}[\/latex].\r\n\r\n[reveal-answer q=\"837781\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"837781\"]\r\n\r\n[latex]\\begin{array}{l}\\text{ }{5}^{x+2}={4}^{x}\\hfill &amp; \\text{There is no easy way to get the powers to have the same base}.\\hfill \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use the power rule for logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use the distributive property}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill &amp; \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill &amp; \\text{On the left hand side, factor out }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill &amp; \\text{Use the properties of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill &amp; \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{2}^{x}={3}^{x+1}[\/latex].\r\n\r\n[reveal-answer q=\"311643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"311643\"]\r\n\r\n[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom9\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98555&amp;theme=oea&amp;iframe_resize_id=mom9\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/strong>\r\n\r\n<em>Yes. The solution is x = 0.<\/em>\r\n\r\n<\/div>\r\n<h3>Equations Containing [latex]e[\/latex]<\/h3>\r\nOne common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, mathematics, science, engineering, and 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\r\n \t<li>Apply the natural logarithm to 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 class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\nSolve [latex]100=20{e}^{2t}[\/latex].\r\n\r\n[reveal-answer q=\"7965\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7965\"]\r\n\r\n[latex]\\begin{array}{l}100\\hfill &amp; =20{e}^{2t}\\hfill &amp; \\hfill \\\\ 5\\hfill &amp; ={e}^{2t}\\hfill &amp; \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =\\mathrm{ln}{{e}^{2t}}\\hfill &amp; \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =2t\\hfill &amp; \\text{Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill &amp; =\\frac{\\mathrm{ln}5}{2}\\hfill &amp; \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\n&nbsp;\r\n<h4>Analysis of the Solution<\/h4>\r\nUsing 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, then we use a calculator.\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\nSolve [latex]3{e}^{0.5t}=11[\/latex].\r\n\r\n[reveal-answer q=\"585330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"585330\"\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][\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98596&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Does every equation of the form [latex]y=A{e}^{kt}[\/latex] have a solution?<\/strong>\r\n\r\n<em>No. There is a solution when [latex]k\\ne 0[\/latex], and when<\/em>\u00a0[latex]y[\/latex]<em>\u00a0and [latex]<\/em>A<em>[\/latex] 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>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\nSolve [latex]4{e}^{2x}+5=12[\/latex].\r\n\r\n[reveal-answer q=\"480495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"480495\"]\r\n\r\n[latex]\\begin{array}{l}4{e}^{2x}+5=12\\hfill &amp; \\hfill \\\\ 4{e}^{2x}=7\\hfill &amp; \\text{Subtract 5 from both sides}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill &amp; \\text{Divide both sides by 4}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ x=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]3+{e}^{2t}=7{e}^{2t}[\/latex].\r\n\r\n[reveal-answer q=\"326491\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326491\"\r\n\r\n[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129891&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Extraneous Solutions<\/h3>\r\nSometimes 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 taking the logarithm of 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Exponential Functions in Quadratic Form<\/h3>\r\nSolve [latex]{e}^{2x}-{e}^{x}=56[\/latex].\r\n\r\n[reveal-answer q=\"592930\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"592930\"]\r\n\r\n&nbsp;\r\n\r\n[latex]\\begin{array}{l}{e}^{2x}-{e}^{x}=56\\hfill \\\\ {e}^{2x}-{e}^{x}-56=0\\hfill &amp; \\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)=0\\hfill &amp; \\text{Factor by the FOIL method}.\\hfill \\\\ {e}^{x}+7=0\\text{ or }{e}^{x}-8=0 &amp; \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\ {e}^{x}=-7{\\text{ or e}}^{x}=8\\hfill &amp; \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}=8\\hfill &amp; \\text{Reject the equation in which the power equals a negative number}.\\hfill \\\\ x=\\mathrm{ln}8\\hfill &amp; \\text{Solve the equation in which the power equals a positive number}.\\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nWhen 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.\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\nSolve [latex]{e}^{2x}={e}^{x}+2[\/latex].\r\n\r\n[reveal-answer q=\"728419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728419\"]\r\n\r\n[latex]x=\\mathrm{ln}2[\/latex][\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98598&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Does every logarithmic equation have a solution?<\/strong>\r\n\r\n<em>No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.<\/em>\r\n\r\n<\/div>\r\n<h2>Logarithmic Equations<\/h2>\r\nWe have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equal 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.\r\n\r\nFor 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 as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill &amp; \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill &amp; \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill &amp; \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill &amp; \\text{Convert to exponential form}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill &amp; \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill &amp; \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill &amp; \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\r\nFor 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],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Algebra to Solve a Logarithmic Equation<\/h3>\r\nSolve [latex]2\\mathrm{ln}x+3=7[\/latex].\r\n\r\n[reveal-answer q=\"977891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977891\"]\r\n\r\n[latex]\\begin{array}{l}2\\mathrm{ln}x+3=7\\hfill &amp; \\hfill \\\\ \\text{}2\\mathrm{ln}x=4\\hfill &amp; \\text{Subtract 3 from both sides}.\\hfill \\\\ \\text{}\\mathrm{ln}x=2\\hfill &amp; \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}x={e}^{2}\\hfill &amp; \\text{Rewrite in exponential form}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]6+\\mathrm{ln}x=10[\/latex].\r\n\r\n[reveal-answer q=\"183568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"183568\"]\r\n\r\n[latex]x={e}^{4}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\r\nSolve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].\r\n\r\n[reveal-answer q=\"231886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"231886\"]\r\n\r\n[latex]\\begin{array}{l}2\\mathrm{ln}\\left(6x\\right)=7\\hfill &amp; \\hfill \\\\ \\text{}\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill &amp; \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill &amp; \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{}x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill &amp; \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].\r\n\r\n[reveal-answer q=\"62905\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"62905\"]\r\n\r\n[latex]x={e}^{5}-1[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\r\nSolve [latex]\\mathrm{ln}x=3[\/latex].\r\n\r\n[reveal-answer q=\"960461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960461\"]\r\n\r\n[latex]\\begin{array}{l}\\mathrm{ln}x=3\\hfill &amp; \\hfill \\\\ x={e}^{3}\\hfill &amp; \\text{Use the definition of }\\mathrm{ln}\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\nBelow is a\u00a0graph of the equation. On the graph the <em>x<\/em>-coordinate of the point where 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].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172012\/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\" \/> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex] which is approximately (20.0855, 3).[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.\r\n\r\n[reveal-answer q=\"889911\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889911\"]\r\n\r\n[latex]x\\approx 9.97[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\r\nAs 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],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\r\nFor example,\r\n<p 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]<\/p>\r\nSo if [latex]x - 1=8[\/latex], then we can solve for <em>x\u00a0<\/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.\r\n\r\nFor 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>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill &amp; \\hfill \\\\ \\text{}\\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill &amp; \\text{Apply the quotient rule of logarithms}.\\hfill \\\\ \\text{}\\frac{3x - 2}{2}=x+4\\hfill &amp; \\text{Apply the one-to-one property}.\\hfill \\\\ \\text{}3x - 2=2x+8\\hfill &amp; \\text{Multiply both sides of the equation by }2.\\hfill \\\\ \\text{}x=10\\hfill &amp; \\text{Subtract 2}x\\text{ and add 2}.\\hfill \\end{array}[\/latex]<\/p>\r\nTo 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\hfill &amp; \\hfill \\\\ \\text{}\\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(14\\right)\\hfill &amp; \\hfill \\\\ \\text{}\\mathrm{log}\\left(\\frac{28}{2}\\right)=\\mathrm{log}\\left(14\\right)\\hfill &amp; \\text{The solution checks}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\r\nFor 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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\r\nNote, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation containing logarithms, solve it using the one-to-one property<\/h3>\r\n<ol>\r\n \t<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of 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 to each other.<\/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 class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\r\nSolve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)[\/latex].\r\n\r\n[reveal-answer q=\"957758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"957758\"]\r\n\r\n[latex]\\begin{array}{l}\\text{ }\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\hfill &amp; \\hfill \\\\ \\text{ }{x}^{2}=2x+3\\hfill &amp; \\text{Use the one-to-one property of the logarithm}.\\hfill \\\\ \\text{ }{x}^{2}-2x - 3=0\\hfill &amp; \\text{Get zero on one side before factoring}.\\hfill \\\\ \\left(x - 3\\right)\\left(x+1\\right)=0\\hfill &amp; \\text{Factor using FOIL}.\\hfill \\\\ \\text{ }x - 3=0\\text{ or }x+1=0\\hfill &amp; \\text{If a product is zero, one of the factors must be zero}.\\hfill \\\\ \\text{ }x=3\\text{ or }x=-1\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nThere 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 function is still positive.\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\nSolve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}1[\/latex].\r\n\r\n[reveal-answer q=\"807762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"807762\"]\r\n\r\n[latex]x=1[\/latex] or [latex]x=\u20131[\/latex][\/hidden-answer]\r\n<iframe id=\"mom100\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129918&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table 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&gt;0,\\text{ }b\\ne 1, {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[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<h2>Key Concepts<\/h2>\r\n<ul>\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>\u00a0by applying the natural logarithm to 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 of 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 [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the point of intersecting.<\/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<h2>Glossary<\/h2>\r\n&nbsp;\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>extraneous solution<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dd id=\"fs-id1165137838640\"><\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve an exponential equation with a common base.<\/li>\n<li>Rewrite an exponential equation so all terms have a common base then solve.<\/li>\n<li>Recognize when an exponential equation does not have a solution.<\/li>\n<li>Use logarithms to solve exponential equations.<\/li>\n<li>Solve a logarithmic equation algebraically.<\/li>\n<li>Solve a logarithmic equation graphically.<\/li>\n<li>Use the one-to-one property of logarithms to solve a logarithmic equation.<\/li>\n<li>Solve a radioactive decay problem.<\/li>\n<\/ul>\n<\/div>\n<p>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>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 and logarithmic equations.<\/p>\n<h2>Exponential Equations<\/h2>\n<p>The first technique we will introduce for solving exponential equations 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>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>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}{3}^{4x - 7}\\hfill & =\\frac{{3}^{2x}}{3}\\hfill & \\hfill \\\\ {3}^{4x - 7}\\hfill & =\\frac{{3}^{2x}}{{3}^{1}}\\hfill & {\\text{Rewrite 3 as 3}}^{1}.\\hfill \\\\ {3}^{4x - 7}\\hfill & ={3}^{2x - 1}\\hfill & \\text{Use the division property of exponents}\\text{.}\\hfill \\\\ 4x - 7\\hfill & =2x - 1\\text{ }\\hfill & \\text{Apply the one-to-one property of exponents}\\text{.}\\hfill \\\\ 2x\\hfill & =6\\hfill & \\text{Subtract 2}x\\text{ and add 7 to both sides}\\text{.}\\hfill \\\\ x\\hfill & =3\\hfill & \\text{Divide by 3}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations<\/h3>\n<p>For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex],<\/p>\n<p>[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an exponential equation Of 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>\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 to each other.<\/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 exercises\">\n<h3>Example: Solving an Exponential Equation with a Common Base<\/h3>\n<p>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=\"q766535\">Show Solution<\/span><\/p>\n<div id=\"q766535\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\\hfill & \\text{The common base is }2.\\hfill \\\\ \\text{ }x - 1=2x - 4\\hfill & \\text{By the one-to-one property the exponents must be equal}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q902679\">Show Solution<\/span><\/p>\n<div id=\"q902679\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\u20132[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2637&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Rewriting Equations So All Powers Have the Same Base<\/h3>\n<p>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>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}256={4}^{x - 5}\\hfill & \\hfill \\\\ {2}^{8}={\\left({2}^{2}\\right)}^{x - 5}\\hfill & \\text{Rewrite each side as a power with base 2}.\\hfill \\\\ {2}^{8}={2}^{2x - 10}\\hfill & \\text{To take a power of a power, multiply the exponents}.\\hfill \\\\ 8=2x - 10\\hfill & \\text{Apply the one-to-one property of exponents}.\\hfill \\\\ 18=2x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ x=9\\hfill & \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it<\/h3>\n<ol>\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 to each other.<\/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 exercises\">\n<h3>Example: Solving Equations by Rewriting Them to Have a Common Base<\/h3>\n<p>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=\"q214040\">Show Solution<\/span><\/p>\n<div id=\"q214040\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{llllll}\\text{ }{8}^{x+2}={16}^{x+1}\\hfill & \\hfill \\\\ {\\left({2}^{3}\\right)}^{x+2}={\\left({2}^{4}\\right)}^{x+1}\\hfill & \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\hfill \\\\ \\text{ }{2}^{3x+6}={2}^{4x+4}\\hfill & \\text{To take a power of a power, multiply the exponents}.\\hfill \\\\ \\text{ }3x+6=4x+4\\hfill & \\text{Use the one-to-one property to set the exponents equal to each other}.\\hfill \\\\ \\text{ }x=2\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q660468\">Show Solution<\/span><\/p>\n<div id=\"q660468\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\u20131[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2620&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base<\/h3>\n<p>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=\"q256816\">Show Solution<\/span><\/p>\n<div id=\"q256816\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}{2}^{5x}={2}^{\\frac{1}{2}}\\hfill & \\text{Write the square root of 2 as a power of }2.\\hfill \\\\ 5x=\\frac{1}{2}\\hfill & \\text{Use the one-to-one property}.\\hfill \\\\ x=\\frac{1}{10}\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q743764\">Show Solution<\/span><\/p>\n<div id=\"q743764\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{1}{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2638&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><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><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 class=\"textbox exercises\">\n<h3>Example: Determining When an\u00a0Equation has No Solution<\/h3>\n<p>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=\"q897533\">Show Solution<\/span><\/p>\n<div id=\"q897533\" 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<h4>Analysis of the Solution<\/h4>\n<p>The figure below\u00a0shows that the two graphs do not cross so the left side of the equation is never equal to the right side of the equation. Thus the equation has no solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172009\/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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q161944\">Show Solution<\/span><\/p>\n<div id=\"q161944\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation has no solution.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98554&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Using Logarithms to Solve Exponential Equations<\/h3>\n<p>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 that since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equal to <em>a\u00a0<\/em>= <em>b<\/em>,<em>\u00a0<\/em>we may apply logarithms with the same base to both sides of an exponential equation.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an exponential equation Where a common base cannot be found, solve for the unknown<\/h3>\n<ol>\n<li>Apply the logarithm to 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 class=\"textbox exercises\">\n<h3>Example: Solving an Equation Containing Powers of Different Bases<\/h3>\n<p>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=\"q837781\">Show Solution<\/span><\/p>\n<div id=\"q837781\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\text{ }{5}^{x+2}={4}^{x}\\hfill & \\text{There is no easy way to get the powers to have the same base}.\\hfill \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use the power rule for logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use the distributive property}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill & \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill & \\text{On the left hand side, factor out }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill & \\text{Use the properties of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill & \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q311643\">Show Solution<\/span><\/p>\n<div id=\"q311643\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom9\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98555&amp;theme=oea&amp;iframe_resize_id=mom9\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/strong><\/p>\n<p><em>Yes. The solution is x = 0.<\/em><\/p>\n<\/div>\n<h3>Equations Containing [latex]e[\/latex]<\/h3>\n<p>One common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, mathematics, science, engineering, and 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 class=\"textbox\">\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\n<ol>\n<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\n<li>Apply the natural logarithm to 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 class=\"textbox exercises\">\n<h3>Example: Solving an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\n<p>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=\"q7965\">Show Solution<\/span><\/p>\n<div id=\"q7965\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}100\\hfill & =20{e}^{2t}\\hfill & \\hfill \\\\ 5\\hfill & ={e}^{2t}\\hfill & \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =\\mathrm{ln}{{e}^{2t}}\\hfill & \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =2t\\hfill & \\text{Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill & =\\frac{\\mathrm{ln}5}{2}\\hfill & \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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, then we use a calculator.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q585330\">Show Solution<\/span><\/p>\n<div id=\"q&#8221;585330&#8243;\" class=\"hidden-answer\" style=\"display: none\">t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98596&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Does every equation of the form [latex]y=A{e}^{kt}[\/latex] have a solution?<\/strong><\/p>\n<p><em>No. There is a solution when [latex]k\\ne 0[\/latex], and when<\/em>\u00a0[latex]y[\/latex]<em>\u00a0and [latex]<\/em>A<em>[\/latex] 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 class=\"textbox exercises\">\n<h3>Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\n<p>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=\"q480495\">Show Solution<\/span><\/p>\n<div id=\"q480495\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}4{e}^{2x}+5=12\\hfill & \\hfill \\\\ 4{e}^{2x}=7\\hfill & \\text{Subtract 5 from both sides}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill & \\text{Divide both sides by 4}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ x=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q326491\">Show Solution<\/span><\/p>\n<div id=\"q&#8221;326491&#8243;\" class=\"hidden-answer\" style=\"display: none\">t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129891&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Extraneous Solutions<\/h3>\n<p>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 taking the logarithm of 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 class=\"textbox exercises\">\n<h3>Example: Solving Exponential Functions in Quadratic Form<\/h3>\n<p>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=\"q592930\">Show Solution<\/span><\/p>\n<div id=\"q592930\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>[latex]\\begin{array}{l}{e}^{2x}-{e}^{x}=56\\hfill \\\\ {e}^{2x}-{e}^{x}-56=0\\hfill & \\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)=0\\hfill & \\text{Factor by the FOIL method}.\\hfill \\\\ {e}^{x}+7=0\\text{ or }{e}^{x}-8=0 & \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\ {e}^{x}=-7{\\text{ or e}}^{x}=8\\hfill & \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}=8\\hfill & \\text{Reject the equation in which the power equals a negative number}.\\hfill \\\\ x=\\mathrm{ln}8\\hfill & \\text{Solve the equation in which the power equals a positive number}.\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q728419\">Show Solution<\/span><\/p>\n<div id=\"q728419\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\mathrm{ln}2[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98598&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Does every logarithmic equation have a solution?<\/strong><\/p>\n<p><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<h2>Logarithmic Equations<\/h2>\n<p>We have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equal 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>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 as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Convert to exponential form}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Algebra to Solve a Logarithmic Equation<\/h3>\n<p>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=\"q977891\">Show Solution<\/span><\/p>\n<div id=\"q977891\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{}2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3 from both sides}.\\hfill \\\\ \\text{}\\mathrm{ln}x=2\\hfill & \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}x={e}^{2}\\hfill & \\text{Rewrite in exponential form}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q183568\">Show Solution<\/span><\/p>\n<div id=\"q183568\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{4}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\n<p>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=\"q231886\">Show Solution<\/span><\/p>\n<div id=\"q231886\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{}\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{}x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q62905\">Show Solution<\/span><\/p>\n<div id=\"q62905\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{5}-1[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\n<p>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=\"q960461\">Show Solution<\/span><\/p>\n<div id=\"q960461\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\mathrm{ln}x=3\\hfill & \\hfill \\\\ x={e}^{3}\\hfill & \\text{Use the definition of }\\mathrm{ln}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>Below is a\u00a0graph of the equation. On the graph the <em>x<\/em>-coordinate of the point where 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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172012\/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\" \/><\/p>\n<p class=\"wp-caption-text\">The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex] which is approximately (20.0855, 3).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q889911\">Show Solution<\/span><\/p>\n<div id=\"q889911\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx 9.97[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h3>Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<p>For example,<\/p>\n<p 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]<\/p>\n<p>So if [latex]x - 1=8[\/latex], then we can solve for <em>x\u00a0<\/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>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill & \\hfill \\\\ \\text{}\\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill & \\text{Apply the quotient rule of logarithms}.\\hfill \\\\ \\text{}\\frac{3x - 2}{2}=x+4\\hfill & \\text{Apply the one-to-one property}.\\hfill \\\\ \\text{}3x - 2=2x+8\\hfill & \\text{Multiply both sides of the equation by }2.\\hfill \\\\ \\text{}x=10\\hfill & \\text{Subtract 2}x\\text{ and add 2}.\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\hfill & \\hfill \\\\ \\text{}\\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(14\\right)\\hfill & \\hfill \\\\ \\text{}\\mathrm{log}\\left(\\frac{28}{2}\\right)=\\mathrm{log}\\left(14\\right)\\hfill & \\text{The solution checks}.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<p>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 class=\"textbox\">\n<h3>How To: Given an equation containing logarithms, solve it using the one-to-one property<\/h3>\n<ol>\n<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of 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 to each other.<\/li>\n<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\n<p>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=\"q957758\">Show Solution<\/span><\/p>\n<div id=\"q957758\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\text{ }\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\hfill & \\hfill \\\\ \\text{ }{x}^{2}=2x+3\\hfill & \\text{Use the one-to-one property of the logarithm}.\\hfill \\\\ \\text{ }{x}^{2}-2x - 3=0\\hfill & \\text{Get zero on one side before factoring}.\\hfill \\\\ \\left(x - 3\\right)\\left(x+1\\right)=0\\hfill & \\text{Factor using FOIL}.\\hfill \\\\ \\text{ }x - 3=0\\text{ or }x+1=0\\hfill & \\text{If a product is zero, one of the factors must be zero}.\\hfill \\\\ \\text{ }x=3\\text{ or }x=-1\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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 function is still positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q807762\">Show Solution<\/span><\/p>\n<div id=\"q807762\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=1[\/latex] or [latex]x=\u20131[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom100\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129918&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table 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>0,\\text{ }b\\ne 1, {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],<br \/>\n[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<h2>Key Concepts<\/h2>\n<ul>\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>\u00a0by applying the natural logarithm to 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 of 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 [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the point of intersecting.<\/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<h2>Glossary<\/h2>\n<p>&nbsp;<\/p>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>extraneous solution<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt>\n<\/dt>\n<dd id=\"fs-id1165137838640\"><\/dd>\n<\/dl>\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-2095\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 2637, 2620, 2638. <strong>Authored by<\/strong>: Greg Langkamp. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 98554, 98555, 98596. <strong>Authored by<\/strong>: Michael Jenck. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 14406. <strong>Authored by<\/strong>: James Sousa. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 122911. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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