{"id":15973,"date":"2019-12-05T05:17:09","date_gmt":"2019-12-05T05:17:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/chapter\/evaluate-exponential-functions-with-base-e\/"},"modified":"2019-12-05T05:17:31","modified_gmt":"2019-12-05T05:17:31","slug":"evaluate-exponential-functions-with-base-e","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/chapter\/evaluate-exponential-functions-with-base-e\/","title":{"raw":"Exponential Functions with Base e","rendered":"Exponential Functions with Base e"},"content":{"raw":"\n
\n
\n

Learning Outcome<\/h3>\n
    \n \t
  • Evaluate exponential functions with base e<\/em><\/li>\n<\/ul>\n<\/div>\nAs we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.\n

    Examine the value of [latex]$1[\/latex] invested at [latex]100\\%[\/latex] interest for [latex]1[\/latex] year, compounded at various frequencies.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Frequency<\/th>\n[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\nValue<\/th>\n<\/tr>\n<\/thead>\n
    Annually<\/td>\n[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n[latex]$2[\/latex]<\/td>\n<\/tr>\n
    Semiannually<\/td>\n[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n[latex]$2.25[\/latex]<\/td>\n<\/tr>\n
    Quarterly<\/td>\n[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n[latex]$2.441406[\/latex]<\/td>\n<\/tr>\n
    Monthly<\/td>\n[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n[latex]$2.613035[\/latex]<\/td>\n<\/tr>\n
    Daily<\/td>\n[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n[latex]$2.714567[\/latex]<\/td>\n<\/tr>\n
    Hourly<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n[latex]$2.718127[\/latex]<\/td>\n<\/tr>\n
    Once per minute<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n[latex]$2.718279[\/latex]<\/td>\n<\/tr>\n
    Once per second<\/td>\n[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n[latex]$2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    These values appear to be approaching a limit as n<\/em> increases. In fact, as n<\/em> gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.<\/p>\n\n

    \n

    A General Note: The Number [latex]e[\/latex]<\/h3>\n

    The letter e<\/em> represents the irrational number<\/p>\n\n

    [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\n
    as n increases without bound<\/div>\n

    The letter e <\/em>is used as a base for many real-world exponential models. To work with base e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n\n<\/div>\nIn our first example, we will use a calculator to find powers of e.<\/em>\n

    \n

    Example<\/h3>\nCalculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.\n[reveal-answer q=\"465847\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"465847\"]\n\nOn a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [e<\/em>^(]. Type [latex]3.14[\/latex] and then close parenthesis, (]). Press [ENTER]. Rounding to [latex]5[\/latex] decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an \"Exp\" button, which is used to enter numbers in scientific notation. It is not used to find powers of e<\/em>.[\/hidden-answer]\n\n<\/div>\n<\/section>
    \n

    Investigating Continuous Growth<\/h2>\n

    So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e <\/em>is used as the base for exponential functions. Exponential models that use e<\/em> as the base are called continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n\n

    \n

    The Continuous Growth\/Decay Formula<\/h3>\n

    For all real numbers r, t<\/em>, and all positive numbers a<\/em>, continuous growth or decay is represented by the formula<\/p>\n\n

    [latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n

    where<\/p>\n\n

      \n \t
    • a<\/em> is the initial value,<\/li>\n \t
    • r<\/em> is the continuous growth or decay rate per unit time,<\/li>\n \t
    • and t<\/em> is the elapsed time.<\/li>\n<\/ul>\n

      If r <\/em>>[latex]0[\/latex], then the formula represents continuous growth. If r <\/em>< [latex]0[\/latex], then the formula represents continuous decay.<\/p>\n

      For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n\n

      [latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n

      where<\/p>\n\n

        \n \t
      • P<\/em> is the principal or the initial amount invested,<\/li>\n \t
      • r<\/em> is the growth or interest rate per unit time,<\/li>\n \t
      • and t<\/em> is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\nIn our next example, we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems.  You will know to use the continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding.  In previous examples we asked that you find an amount based on quarterly or monthly compounding where, in that case, you used the compound<\/em> interest formula.\n
        \n

        Example<\/h3>\nA person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?\n[reveal-answer q=\"33008\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"33008\"]\n\nSince the account is growing in value, this is a continuous compounding problem with growth rate r <\/em>=[latex]0.10[\/latex]. The initial investment was [latex]$1,000[\/latex], so P <\/em>=[latex]1000[\/latex]. We use the continuous compounding formula to find the value after t <\/em>=[latex]1[\/latex] year:\n
        [latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n

        The account is worth [latex]$1,105.17[\/latex] after one year.<\/p>\n[\/hidden-answer]\n\n<\/div>\nIn the following video, we show another example of interest compounded continuously.\n\nhttps:\/\/youtu.be\/fEjrYCog_8w\n

        \n

        How To: Given the initial value, rate of growth or decay, and time [latex]t[\/latex], solve a continuous growth or decay function<\/h3>\n
          \n \t
        1. Use the information in the problem to determine a<\/em>, the initial value of the function.<\/li>\n \t
        2. Use the information in the problem to determine the growth rate r<\/em>.\n
            \n \t
          1. If the problem refers to continuous growth, then r <\/em>> [latex]0[\/latex].<\/li>\n \t
          2. If the problem refers to continuous decay, then r <\/em>< [latex]0[\/latex].<\/li>\n<\/ol>\n<\/li>\n \t
          3. Use the information in the problem to determine the time t<\/em>.<\/li>\n \t
          4. Substitute the given information into the continuous growth formula and solve for A<\/em>(t<\/em>).<\/li>\n<\/ol>\n<\/div>\nIn our next example, we will calculate continuous decay. Pay attention to the rate - it is negative which means we are considering a situation where an amount decreases or decays.\n
            \n

            Example<\/h3>\nRadon-222 decays at a continuous rate of [latex]17.3\\%[\/latex] per day. How much will [latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in [latex]3[\/latex] days?\n[reveal-answer q=\"995802\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"995802\"]\n\nSince the substance is decaying, the rate, [latex]17.3\\%[\/latex], is negative. So, r <\/em>= [latex]\u20130.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so a <\/em>= [latex]100[\/latex]. We use the continuous decay formula to find the value after t <\/em>=[latex]3[\/latex] days:\n
            [latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)} & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n

            So [latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.<\/p>\n[\/hidden-answer]\n\n<\/div>\nIn the following video, we show an example of calculating the remaining amount of a radioactive substance after it decays for a length of time.\n\nhttps:\/\/youtu.be\/Vyl3NcTGRAo\n

            Summary<\/h2>\nContinuous growth or decay functions are of the form [latex]A\\left(t\\right)=a{e}^{rt}[\/latex]. If r <\/em>> [latex]0[\/latex], then the formula represents continuous growth. If r <\/em>< [latex]0[\/latex], then the formula represents continuous decay. For business applications, the continuous growth formula is called the continuous compounding formula and takes the form [latex]A\\left(t\\right)=P{e}^{rt}[\/latex].\n\n<\/section>\n","rendered":"
            \n
            \n

            Learning Outcome<\/h3>\n
              \n
            • Evaluate exponential functions with base e<\/em><\/li>\n<\/ul>\n<\/div>\n

              As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.<\/p>\n

              Examine the value of [latex]$1[\/latex] invested at [latex]100\\%[\/latex] interest for [latex]1[\/latex] year, compounded at various frequencies.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
              Frequency<\/th>\n[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\nValue<\/th>\n<\/tr>\n<\/thead>\n
              Annually<\/td>\n[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n[latex]$2[\/latex]<\/td>\n<\/tr>\n
              Semiannually<\/td>\n[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n[latex]$2.25[\/latex]<\/td>\n<\/tr>\n
              Quarterly<\/td>\n[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n[latex]$2.441406[\/latex]<\/td>\n<\/tr>\n
              Monthly<\/td>\n[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n[latex]$2.613035[\/latex]<\/td>\n<\/tr>\n
              Daily<\/td>\n[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n[latex]$2.714567[\/latex]<\/td>\n<\/tr>\n
              Hourly<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n[latex]$2.718127[\/latex]<\/td>\n<\/tr>\n
              Once per minute<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n[latex]$2.718279[\/latex]<\/td>\n<\/tr>\n
              Once per second<\/td>\n[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n[latex]$2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              These values appear to be approaching a limit as n<\/em> increases. In fact, as n<\/em> gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.<\/p>\n

              \n

              A General Note: The Number [latex]e[\/latex]<\/h3>\n

              The letter e<\/em> represents the irrational number<\/p>\n

              [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\n
              as n increases without bound<\/div>\n

              The letter e <\/em>is used as a base for many real-world exponential models. To work with base e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n

              In our first example, we will use a calculator to find powers of e.<\/em><\/p>\n

              \n

              Example<\/h3>\n

              Calculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.<\/p>\n

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

              On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [e<\/em>^(]. Type [latex]3.14[\/latex] and then close parenthesis, (]). Press [ENTER]. Rounding to [latex]5[\/latex] decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of e<\/em>.<\/div>\n<\/div>\n<\/div>\n<\/section>\n

              \n

              Investigating Continuous Growth<\/h2>\n

              So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e <\/em>is used as the base for exponential functions. Exponential models that use e<\/em> as the base are called continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n

              \n

              The Continuous Growth\/Decay Formula<\/h3>\n

              For all real numbers r, t<\/em>, and all positive numbers a<\/em>, continuous growth or decay is represented by the formula<\/p>\n

              [latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n

              where<\/p>\n

                \n
              • a<\/em> is the initial value,<\/li>\n
              • r<\/em> is the continuous growth or decay rate per unit time,<\/li>\n
              • and t<\/em> is the elapsed time.<\/li>\n<\/ul>\n

                If r <\/em>>[latex]0[\/latex], then the formula represents continuous growth. If r <\/em>< [latex]0[\/latex], then the formula represents continuous decay.<\/p>\n

                For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n

                [latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n

                where<\/p>\n

                  \n
                • P<\/em> is the principal or the initial amount invested,<\/li>\n
                • r<\/em> is the growth or interest rate per unit time,<\/li>\n
                • and t<\/em> is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\n

                  In our next example, we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems.  You will know to use the continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding.  In previous examples we asked that you find an amount based on quarterly or monthly compounding where, in that case, you used the compound<\/em> interest formula.<\/p>\n

                  \n

                  Example<\/h3>\n

                  A person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?<\/p>\n

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

                  Since the account is growing in value, this is a continuous compounding problem with growth rate r <\/em>=[latex]0.10[\/latex]. The initial investment was [latex]$1,000[\/latex], so P <\/em>=[latex]1000[\/latex]. We use the continuous compounding formula to find the value after t <\/em>=[latex]1[\/latex] year:<\/p>\n

                  [latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n

                  The account is worth [latex]$1,105.17[\/latex] after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                  In the following video, we show another example of interest compounded continuously.<\/p>\n