{"id":113,"date":"2025-02-13T22:43:55","date_gmt":"2025-02-13T22:43:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models\/"},"modified":"2026-03-18T21:29:40","modified_gmt":"2026-03-18T21:29:40","slug":"exponential-and-logarithmic-models","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models\/","title":{"raw":"Exponential and Logarithmic Models: Learn It 1","rendered":"Exponential and Logarithmic Models: Learn It 1"},"content":{"raw":"
\r\n
\r\n
    \r\n \t
  • Model exponential growth and decay.<\/li>\r\n \t
  • Use Newton\u2019s Law of Cooling.<\/li>\r\n \t
  • Use logistic-growth models.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n

    Exponential Growth and Decay<\/h2>\r\nIn mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze.\r\n\r\nIn the case of rapid growth, we may choose the exponential growth function, [latex]A={A}_{0}{e}^{rt}[\/latex], where [latex]{A}_{0}[\/latex] (used to be labeled as [latex]P[\/latex]) is equal to the value at time zero, [latex]e[\/latex] is Euler\u2019s constant, and [latex]r[\/latex] is a positive constant that determines the rate (percentage) of growth.\r\n\r\nWe may use the exponential growth<\/strong> function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.\r\n\r\nOn the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and [latex]e[\/latex]\u00a0is Euler\u2019s constant. Now [latex]k[\/latex]\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.\r\n\r\n
    An exponential function of the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:\r\n
      \r\n \t
    • one-to-one function<\/li>\r\n \t
    • horizontal asymptote: [latex]y\u00a0= 0[\/latex]<\/li>\r\n \t
    • domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t
    • range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t
    • [latex]x[\/latex] intercept: none<\/li>\r\n \t
    • [latex]y[\/latex]-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\r\n \t
    • increasing if [latex]k\u00a0> 0[\/latex]<\/li>\r\n \t
    • decreasing if [latex]k\u00a0< 0[\/latex]<\/li>\r\n<\/ul>\r\nAn exponential function models exponential growth when [latex]k\u00a0> 0[\/latex] and exponential decay when [latex]k\u00a0< 0[\/latex].\r\n\r\n<\/section>
      Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude<\/strong> is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal.\r\n[latex]\\\\[\/latex]\r\nFor example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is [latex]40,113,497,200,000[\/latex] kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/section>
      A population of bacteria doubles every hour. If the culture started with [latex]10[\/latex] bacteria, graph the population as a function of time.[reveal-answer q=\"151444\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151444\"]When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find [latex]r[\/latex], use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from [latex]10[\/latex] to [latex]20[\/latex]. The formula is derived as follows:\r\n

      [latex]\\begin{array}{l}\\text{ }20=10{e}^{r\\cdot 1}\\hfill & \\hfill \\\\ \\text{ }2={e}^{r}\\hfill & \\text{Divide both sides by 10}\\hfill \\\\ \\mathrm{ln}2=r\\hfill & \\text{Take the natural logarithm of both sides}\\hfill \\end{array}[\/latex]<\/p>\r\nso [latex]r=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex]. The graph is shown below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex].<\/b>[\/caption]Analysis of the Solution\r\n<\/strong>[latex]\\\\[\/latex]\r\n<\/strong>The population of bacteria after ten hours is [latex]10,240[\/latex]. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[\/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[\/latex], so we could say that the population has increased by three orders of magnitude in ten hours.[\/hidden-answer]<\/section>\r\n

      Calculating Doubling Time<\/h3>\r\nFor growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time<\/strong>.\r\n\r\nGiven the basic exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].\r\n\r\nThe formula is derived as follows:\r\n

      [latex]\\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\\hfill & \\hfill \\\\ 2={e}^{kt}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}2=kt\\hfill & \\text{Take the natural logarithm of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{Divide by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\r\nThus the doubling time is\r\n

      [latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/p>\r\n\r\n

      According to Moore\u2019s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.[reveal-answer q=\"567900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567900\"]The formula is derived as follows:\r\n

      [latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}2}{r}\\hfill & \\text{The doubling time formula}.\\hfill \\\\ 2=\\frac{\\mathrm{ln}2}{r}\\hfill & \\text{Use a doubling time of two years}.\\hfill \\\\ r=\\frac{\\mathrm{ln}2}{2}\\hfill & \\text{Multiply by }r\\text{ and divide by 2}.\\hfill \\\\ A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}\\hfill & \\text{Substitute }r\\text{ into the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}r}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm_question hide_question_numbers=1]321544[\/ohm_question]<\/section><\/div>\r\n
      \r\n
      \r\n \t
      <\/dd>\r\n<\/dl>\r\n<\/section>","rendered":"
      \n
      \n
      \n
        \n
      • Model exponential growth and decay.<\/li>\n
      • Use Newton\u2019s Law of Cooling.<\/li>\n
      • Use logistic-growth models.<\/li>\n<\/ul>\n<\/div>\n<\/section>\n

        Exponential Growth and Decay<\/h2>\n

        In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze.<\/p>\n

        In the case of rapid growth, we may choose the exponential growth function, [latex]A={A}_{0}{e}^{rt}[\/latex], where [latex]{A}_{0}[\/latex] (used to be labeled as [latex]P[\/latex]) is equal to the value at time zero, [latex]e[\/latex] is Euler\u2019s constant, and [latex]r[\/latex] is a positive constant that determines the rate (percentage) of growth.<\/p>\n

        We may use the exponential growth<\/strong> function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.<\/p>\n

        On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and [latex]e[\/latex]\u00a0is Euler\u2019s constant. Now [latex]k[\/latex]\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.<\/p>\n

        An exponential function of the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:<\/p>\n
          \n
        • one-to-one function<\/li>\n
        • horizontal asymptote: [latex]y\u00a0= 0[\/latex]<\/li>\n
        • domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n
        • range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n
        • [latex]x[\/latex] intercept: none<\/li>\n
        • [latex]y[\/latex]-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\n
        • increasing if [latex]k\u00a0> 0[\/latex]<\/li>\n
        • decreasing if [latex]k\u00a0< 0[\/latex]<\/li>\n<\/ul>\n

          An exponential function models exponential growth when [latex]k\u00a0> 0[\/latex] and exponential decay when [latex]k\u00a0< 0[\/latex].\n\n<\/section>\n

          Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude<\/strong> is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal.
          \n[latex]\\\\[\/latex]
          \nFor example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is [latex]40,113,497,200,000[\/latex] kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/section>\n
          A population of bacteria doubles every hour. If the culture started with [latex]10[\/latex] bacteria, graph the population as a function of time.<\/p>\n