{"id":3520,"date":"2016-08-05T05:30:12","date_gmt":"2016-08-05T05:30:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&p=3520"},"modified":"2019-07-24T21:34:45","modified_gmt":"2019-07-24T21:34:45","slug":"evaluate-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/evaluate-exponential-functions\/","title":{"raw":"Exponential Equations with Unlike Bases","rendered":"Exponential Equations with Unlike Bases"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\r\n \t
  • Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for t<\/li>\r\n \t
  • 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\u00a0a\u00a0<\/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. Note how we rewrite the exponential terms as logarithms first.\r\n
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    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

    [latex]\\begin{array}{c}\\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 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>\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

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    1. Apply the logarithm to both sides of the equation.\r\n
        \r\n \t
      • If one of the terms in the equation has base\u00a0[latex]10[\/latex], use the common logarithm.<\/li>\r\n \t
      • 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
      • 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
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        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

        \u00a0Equations Containing [latex]e[\/latex]<\/h2>\r\n
        Base e\u00a0<\/em>is a very common base found in science, finance, and engineering applications.\u00a0When we have an equation with a base e<\/em>\u00a0on either side, we can use the 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
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        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 & =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}\\text{.}\\hfill \\\\ t\\hfill & =\\frac{\\mathrm{ln}5}{2}\\hfill & \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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        Analysis of the Solution<\/h3>\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, we use a calculator.\r\n\r\n<\/div>\r\n
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        Think About It<\/h3>\r\nDoes every equation of the form [latex]y=A{e}^{kt}[\/latex] have a 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\nNo. There is a solution when [latex]k\\ne 0[\/latex] and when y and A are either both\u00a0[latex]0[\/latex] or neither\u00a0[latex]0[\/latex] and they have the same sign. An example of an equation of this form that has no solution is [latex]2=-3{e}^{t}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will provide one more example using the continuous growth formula, but this time we have to do a couple steps of algebra to get it in a form that can be solved.\r\n
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        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 & \\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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>
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        Extraneous Solutions<\/h2>\r\n

        Sometimes the methods used to solve an equation introduce an 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

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        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 & \\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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        \u00a0Analysis of the Solution<\/h3>\r\n
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        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

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        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

        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><\/section>","rendered":"
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        Learning Outcomes<\/h3>\n
          \n
        • Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\n
        • Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for t<\/li>\n
        • Recognize when there may be extraneous solutions or no solutions for exponential equations<\/li>\n<\/ul>\n<\/div>\n

          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\u00a0a\u00a0<\/em>= b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.<\/p>\n

          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. Note how we rewrite the exponential terms as logarithms first.<\/p>\n

          \n

          Example<\/h3>\n

          Solve [latex]{5}^{x+2}={4}^{x}[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          \n

          There is no easy way to get the powers to have the same base for this equation.<\/p>\n

          [latex]\\begin{array}{c}\\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 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

          In general we can solve exponential equations whose terms do not have like bases in the following way:<\/p>\n

            \n
          1. Apply the logarithm to both sides of the equation.\n
              \n
            • If one of the terms in the equation has base\u00a0[latex]10[\/latex], use the common logarithm.<\/li>\n
            • If none of the terms in the equation has base\u00a0[latex]10[\/latex], use the natural logarithm.<\/li>\n<\/ul>\n<\/li>\n
            • Use the rules of logarithms to solve for the unknown.<\/li>\n<\/ol>\n

              The following video provides more examples of solving exponential equations.<\/p>\n