{"id":15945,"date":"2019-09-25T21:23:00","date_gmt":"2019-09-25T21:23:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/evaluate-exponential-functions\/"},"modified":"2024-05-02T16:55:57","modified_gmt":"2024-05-02T16:55:57","slug":"evaluate-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/evaluate-exponential-functions\/","title":{"raw":"Exponential Equations with Unlike Bases","rendered":"Exponential Equations with Unlike Bases"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\r\n \t<li>Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for [latex]t[\/latex]<\/li>\r\n \t<li>Recognize when there may be extraneous solutions or no solutions for exponential equations<\/li>\r\n<\/ul>\r\n<\/div>\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, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] can be rewritten as\u00a0<em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.\r\n\r\nIn our first example we will use the laws of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Notice how we rewrite the exponential terms as logarithms first.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{5}^{x+2}={4}^{x}[\/latex].\r\n[reveal-answer q=\"281663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"281663\"]\r\n\r\nThere is no easy way to get the powers to have the same base for this equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }\\mathrm{ln}\\left({5}^{x+2}\\right)=\\mathrm{ln}\\left({4}^{x}\\right)\\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 laws of 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 and 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 an }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill &amp; \\text{Use the laws 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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn general we can solve exponential equations whose terms do not have like bases in the following way:\r\n<ol id=\"fs-id1165137784632\">\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\u00a0[latex]10[\/latex], use the common logarithm.<\/li>\r\n \t<li>If none of the terms in the equation has base\u00a0[latex]10[\/latex], 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\nThe following video provides more examples of solving exponential equations.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=5R5mKpLsfYg[\/embed]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?\r\n\r\nUse the text box below to formulate an answer or example before you look at the solution.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"608971\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"608971\"]\r\n\r\nYes. The solution is\u00a0[latex]x = 0[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Equations Containing [latex]e[\/latex]<\/h2>\r\n<section id=\"fs-id1165137469838\">Base <em>e\u00a0<\/em>is a very common base found in science, finance, and engineering applications.\u00a0When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it. Earlier, we introduced a formula that models continuous growth,\u00a0[latex]y=A{e}^{kt}[\/latex]. This formula is found in business, finance, and many biological and physical science applications. In our next example, we will show how to solve this equation for [latex]t[\/latex], the elapsed time for the behavior in question.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]100=20{e}^{2t}[\/latex].\r\n[reveal-answer q=\"714083\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"714083\"]\r\n\r\n[latex]\\begin{array}{c}100\\hfill &amp; =20{e}^{2t}\\hfill &amp; \\hfill \\\\ 5\\hfill &amp; ={e}^{2t}\\hfill &amp; \\text{Divide by the coefficient of the base raised to a power}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =2t\\hfill &amp; \\text{Take ln of both sides}\\text{. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions to simplify the equation}\\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\nUsing the laws of logarithms, 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.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example using the continuous growth formula, we have to do a couple steps of algebra to get it in a form that can be solved.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]4{e}^{2x}+5=12[\/latex].\r\n[reveal-answer q=\"593052\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"593052\"]\r\n\r\n[latex]\\begin{array}{c}4{e}^{2x}+5=12\\hfill &amp; \\hfill \\\\ 4{e}^{2x}=7\\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill &amp; \\text{Divide by the coefficient of the base raised to a power}.\\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<h2>Exponential Equations with No Solutions or Extraneous Solutions<\/h2>\r\n<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-solving-radical-equations\/\">We have seen in earlier lessons on solving equations that there are some equations where a solution does not exist and or that have extraneous solutions<\/a>. We will explore such examples with exponential equations, but first, take a minute to think about when a solution to an exponential equation might not exist.\r\n<div id=\"fs-id1165137705083\" class=\"commentary\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nWhen might an equation of the form [latex]y=A{e}^{kt}[\/latex] have no solution? Write your thoughts or an example in the textbox below before you check the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"483016\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483016\"]\r\n\r\nAn equation of this form does not have a solution if y and A have different signs. An example of this is [latex]2=-3{e}^{t}[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur next example helps to illustrate that not every equation of the form [latex]y=A{e}^{kt}[\/latex] has a solution.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3{e}^{2x}=-6[\/latex].\r\n[reveal-answer q=\"593055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"593055\"]\r\n\r\n[latex]\\begin{array}{c}3{e}^{2x}=-6\\hfill &amp; \\hfill \\\\ {e}^{2x}=-2\\hfill &amp; \\text{Divide by the coefficient of the base raised to a power}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(-2\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\end{array}[\/latex]\r\nThis equation has no solution because the argument of a logarithm cannot be negative.\u00a0 We also could have recognized in the previous step that this equation has no solution because [latex]e[\/latex] raised to any power will always produce a positive result.\r\n\r\n<strong>Answer<\/strong>\r\nNo solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137665482\">\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 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\nIn the next example, we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions but reject one that does not satisfy the original equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{e}^{2x}-{e}^{x}=56[\/latex].\r\n[reveal-answer q=\"152760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"152760\"]\r\n\r\n[latex]\\begin{array}{c}{e}^{2x}-{e}^{x}=56\\hfill &amp; \\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 by taking the natural log of both sides}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>\u00a0Analysis of the Solution<\/h3>\r\n<div id=\"Example_04_06_08\" class=\"example\">\r\n<div id=\"fs-id1165137828517\" class=\"exercise\">\r\n<div id=\"fs-id1165137806430\" class=\"commentary\">\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\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nDoes every logarithmic equation have a solution? Write your ideas, or a counter example, in the box below.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"810736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"810736\"]\r\n\r\nNo. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nThe inverse operation of exponentiation is the logarithm, so we can use logarithms to solve exponential equations whose terms do not have the same bases. This is similar to using multiplication to \"undo\" division or addition to \"undo\" subtraction. It is important to check exponential equations for extraneous solutions or no solutions.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\n<li>Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for [latex]t[\/latex]<\/li>\n<li>Recognize when there may be extraneous solutions or no solutions for exponential equations<\/li>\n<\/ul>\n<\/div>\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, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] can be rewritten as\u00a0<em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.<\/p>\n<p>In our first example we will use the laws of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Notice how we rewrite the exponential terms as logarithms first.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q281663\">Show Solution<\/span><\/p>\n<div id=\"q281663\" class=\"hidden-answer\" style=\"display: none\">\n<p>There is no easy way to get the powers to have the same base for this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }\\mathrm{ln}\\left({5}^{x+2}\\right)=\\mathrm{ln}\\left({4}^{x}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use laws of 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 and 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 an }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill & \\text{Use the laws 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<p>In general we can solve exponential equations whose terms do not have like bases in the following way:<\/p>\n<ol id=\"fs-id1165137784632\">\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\u00a0[latex]10[\/latex], use the common logarithm.<\/li>\n<li>If none of the terms in the equation has base\u00a0[latex]10[\/latex], 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<p>The following video provides more examples of solving exponential equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solving Exponential Equations - Part 2 of 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5R5mKpLsfYg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/p>\n<p>Use the text box below to formulate an answer or example before you look at the solution.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q608971\">Show Solution<\/span><\/p>\n<div id=\"q608971\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes. The solution is\u00a0[latex]x = 0[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Equations Containing [latex]e[\/latex]<\/h2>\n<section id=\"fs-id1165137469838\">Base <em>e\u00a0<\/em>is a very common base found in science, finance, and engineering applications.\u00a0When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it. Earlier, we introduced a formula that models continuous growth,\u00a0[latex]y=A{e}^{kt}[\/latex]. This formula is found in business, finance, and many biological and physical science applications. In our next example, we will show how to solve this equation for [latex]t[\/latex], the elapsed time for the behavior in question.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q714083\">Show Solution<\/span><\/p>\n<div id=\"q714083\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{c}100\\hfill & =20{e}^{2t}\\hfill & \\hfill \\\\ 5\\hfill & ={e}^{2t}\\hfill & \\text{Divide by the coefficient of the base raised to a power}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =2t\\hfill & \\text{Take ln of both sides}\\text{. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions to simplify the equation}\\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>Using the laws of logarithms, 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<p>In our next example using the continuous growth formula, we have to do a couple steps of algebra to get it in a form that can be solved.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q593052\">Show Solution<\/span><\/p>\n<div id=\"q593052\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{c}4{e}^{2x}+5=12\\hfill & \\hfill \\\\ 4{e}^{2x}=7\\hfill & \\text{Combine like terms}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill & \\text{Divide by the coefficient of the base raised to a power}.\\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<h2>Exponential Equations with No Solutions or Extraneous Solutions<\/h2>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-solving-radical-equations\/\">We have seen in earlier lessons on solving equations that there are some equations where a solution does not exist and or that have extraneous solutions<\/a>. We will explore such examples with exponential equations, but first, take a minute to think about when a solution to an exponential equation might not exist.<\/p>\n<div id=\"fs-id1165137705083\" class=\"commentary\">\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>When might an equation of the form [latex]y=A{e}^{kt}[\/latex] have no solution? Write your thoughts or an example in the textbox below before you check the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483016\">Show Solution<\/span><\/p>\n<div id=\"q483016\" class=\"hidden-answer\" style=\"display: none\">\n<p>An equation of this form does not have a solution if y and A have different signs. An example of this is [latex]2=-3{e}^{t}[\/latex].\n<\/p><\/div>\n<\/div>\n<\/div>\n<p>Our next example helps to illustrate that not every equation of the form [latex]y=A{e}^{kt}[\/latex] has a solution.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]3{e}^{2x}=-6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q593055\">Show Solution<\/span><\/p>\n<div id=\"q593055\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{c}3{e}^{2x}=-6\\hfill & \\hfill \\\\ {e}^{2x}=-2\\hfill & \\text{Divide by the coefficient of the base raised to a power}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(-2\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\end{array}[\/latex]<br \/>\nThis equation has no solution because the argument of a logarithm cannot be negative.\u00a0 We also could have recognized in the previous step that this equation has no solution because [latex]e[\/latex] raised to any power will always produce a positive result.<\/p>\n<p><strong>Answer<\/strong><br \/>\nNo solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137665482\">\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 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<p>In the next example, we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions but reject one that does not satisfy the original equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q152760\">Show Solution<\/span><\/p>\n<div id=\"q152760\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{c}{e}^{2x}-{e}^{x}=56\\hfill & \\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 by taking the natural log of both sides}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>\u00a0Analysis of the Solution<\/h3>\n<div id=\"Example_04_06_08\" class=\"example\">\n<div id=\"fs-id1165137828517\" class=\"exercise\">\n<div id=\"fs-id1165137806430\" class=\"commentary\">\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 class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Does every logarithmic equation have a solution? Write your ideas, or a counter example, in the box below.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810736\">Show Solution<\/span><\/p>\n<div id=\"q810736\" class=\"hidden-answer\" style=\"display: none\">\n<p>No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>The inverse operation of exponentiation is the logarithm, so we can use logarithms to solve exponential equations whose terms do not have the same bases. This is similar to using multiplication to &#8220;undo&#8221; division or addition to &#8220;undo&#8221; subtraction. It is important to check exponential equations for extraneous solutions or no solutions.<\/p>\n<\/div>\n<\/div>\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-15945\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"fb0e0e8d6ce24ba6acac79cd1cb1c5a7, 90c9f00aeea244ea931c2c1462806318, db205c3e4d40454cab47b197a12c4635, 463cb272cf6c4cbd944ce3f3970610f5 ","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-15945","chapter","type-chapter","status-publish","hentry"],"part":15967,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15945","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15945\/revisions"}],"predecessor-version":[{"id":20537,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15945\/revisions\/20537"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/15967"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15945\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=15945"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=15945"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=15945"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=15945"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}