{"id":217,"date":"2023-11-08T16:10:28","date_gmt":"2023-11-08T16:10:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/evaluate-exponential-functions-with-base-e\/"},"modified":"2024-08-14T23:02:41","modified_gmt":"2024-08-14T23:02:41","slug":"7-8-applications-of-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/7-8-applications-of-exponential-functions\/","title":{"raw":"7.8 Applications of Exponential Functions","rendered":"7.8 Applications of Exponential Functions"},"content":{"raw":"
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\r\n

Learning Outcome<\/h3>\r\n
    \r\n \t
  • Solve compound interest applications of exponential functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>
    You may have seen or heard of formulas that are used to calculate compound interest rates, for example the interest in a bank account. These formulas are an example of exponential growth.\u00a0The term compounding<\/em> refers to interest earned not only on the original value but on the accumulated value of the account.\r\n

    The annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the interest rate earned by an investment account per year. The term\u00a0nominal<\/em>\u00a0is used when the compounding occurs a number of times other than once per year. When interest is compounded more than once a year, the effective interest rate ends up being greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\n

    We can calculate the balance in an account with compounding interest using the following compound interest formula.<\/p>\r\n\r\n

    \r\n

    The Compound Interest Formula<\/h3>\r\n

    Compound interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n

    [latex]\\large A\\left(t\\right)=P{\\left(1+\\dfrac{r}{n}\\right)}^{nt}[\/latex]<\/div>\r\n

    where<\/p>\r\n\r\n

      \r\n \t
    • [latex]A(T)[\/latex]\u00a0is the final account value, often called the future value<\/li>\r\n \t
    • [latex]t[\/latex] is time measured in years<\/li>\r\n \t
    • [latex]P[\/latex]\u00a0is the starting value of the account, often called the principal or the present value<\/li>\r\n \t
    • [latex]r[\/latex]\u00a0is the annual percentage rate (APR) expressed as a decimal<\/li>\r\n \t
    • [latex]n[\/latex]\u00a0is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe variable [latex]n[\/latex] appearing in the formula is usually expressed as how often the account generates interest. For example, an account with interest compounding monthly<\/strong> will use [latex]n=12[\/latex] since there are [latex]12[\/latex] months in a year.\r\n\r\nNote that this formula does NOT mean that the full APR is paid out [latex]n[\/latex] times per year. The APR is split between the number of times interest is compounded. For example, if the APR is [latex]6\\%[\/latex] and interest is compounded monthly, then the actual interest received per month is [latex]\\dfrac{6\\%}{12}[\/latex] or [latex]0.5\\%.[\/latex]\r\n\r\nIn our next example, we will calculate the value of an account after\u00a0[latex]10[\/latex] years of interest compounded quarterly.\r\n
      \r\n

      Example<\/h3>\r\nIf we invest\u00a0[latex]$3,000[\/latex] in an investment account paying\u00a0[latex]3\\%[\/latex] interest compounded quarterly, how much will the account be worth in\u00a0[latex]10[\/latex] years?\r\n[reveal-answer q=\"689928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689928\"]\r\n

      Because the principal is [latex]$3,000[\/latex], we set [latex]P=3000.[\/latex] Our interest rate is\u00a0[latex]3\\%[\/latex], so [latex]r=0.03.[\/latex] Because we are compounding quarterly, and there are [latex]4[\/latex] quarters in a year, we set [latex]n=4[\/latex]. We want to know the value of the account in\u00a0[latex]t=10[\/latex] years, so we are looking for [latex]A(10).[\/latex]<\/p>\r\n\r\n

      [latex]\\begin{array}{rl}A\\left(t\\right)&=P\\left(1+\\dfrac{r}{n}\\right)^{nt}\\\\\r\nA\\left(10\\right)&=3000\\left(1+\\dfrac{0.03}{4}\\right)^{4\\cdot 10}\\\\\r\n& \\approx 4045.05\\end{array}[\/latex]<\/div>\r\n \r\n

      The account will be worth about\u00a0[latex]$4,045.05[\/latex] in\u00a0[latex]10[\/latex] years. By default, round answers involving money to the nearest cent.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of using exponential growth to calculate interest compounded quarterly.\r\n\r\nhttps:\/\/youtu.be\/3az4AKvUmmI\r\n\r\nIn our next example, we will use the compound interest formula to solve for the principal.\r\n

      \r\n

      Example<\/h3>\r\nA\u00a0[latex]529[\/latex] Plan is a college-savings plan that allows relatives to invest money to pay for a child\u2019s future college tuition. The account grows tax-free. Lily wants to set up a\u00a0[latex]529[\/latex] account for her new granddaughter and wants the account to grow to\u00a0[latex]$40,000[\/latex] over\u00a0[latex]18[\/latex] years. She believes the account will earn\u00a0[latex]6\\%[\/latex] compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?\r\n[reveal-answer q=\"254680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254680\"]\r\n

      The nominal interest rate is\u00a0[latex]6\\%[\/latex], so [latex]r=0.06[\/latex]. Interest is compounded twice a year, so [latex]n=2[\/latex].<\/p>\r\n

      We want to find the initial investment, [latex]P[\/latex], needed so that in\u00a0[latex]t=18[\/latex] years the account will be worth\u00a0[latex]A(18)=$40,000.[\/latex] Substitute the given values into the compound interest formula and solve for [latex]P[\/latex].<\/p>\r\n\r\n

      [latex]\\begin{array}{rl}A\\left(t\\right)&=P{\\left(1+\\dfrac{r}{n}\\right)}^{nt}\\\\\r\n40,000&=P{\\left(1+\\dfrac{0.06}{2}\\right)}^{2\\left(18\\right)}\\\\\r\n40,000&=P{\\left(1.03\\right)}^{36}\\\\\r\n\\dfrac{40,000}{{\\left(1.03\\right)}^{36}}&=P\\\\\r\nP&\\approx 13,801\\end{array}[\/latex]<\/div>\r\n \r\n

      Lily will need to invest\u00a0[latex]$13,801[\/latex] to have\u00a0[latex]$40,000[\/latex] in\u00a0[latex]18[\/latex] years. Make sure to only round numbers at the very end to avoid rounding errors throughout.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of finding the principal necessary to obtain a future value from compounded interest.\r\n\r\nhttps:\/\/youtu.be\/saq9dF7a4r8\r\n

      Investigating Continuous Compounding<\/h2>\r\nSo far, the examples we have used had interest paid at long and regular intervals. There are some accounts that compound interest much more frequently though. Let's investigate what happens if the frequency of compounding is increased.\r\n\r\nSuppose [latex]\\$1,000[\/latex] is invested in an\u00a0account paying\u00a0[latex]10\\%[\/latex] interest for [latex]3[\/latex] years. What will the final balance be with different values of [latex]n[\/latex]?\r\n\r\nAnnual ([latex]n=1[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{1}\\right)^{(1\\cdot 3)}=\\$1331[\/latex]\r\n\r\nSemi-annual ([latex]n=2[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{2}\\right)^{(2\\cdot 3)}\\approx\\$1340.10[\/latex]\r\n\r\nQuarterly ([latex]n=4[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{4}\\right)^{(4\\cdot 3)}\\approx\\$1344.89[\/latex]\r\n\r\nMonthly ([latex]n=12[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{12}\\right)^{(12\\cdot 3)}\\approx\\$1348.18[\/latex]\r\n\r\nWeekly ([latex]n=52[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{52}\\right)^{(52\\cdot 3)}\\approx\\$1349.47[\/latex]\r\n\r\nDaily ([latex]n=365[\/latex]): [latex]A(3)=1000\\left(1+\\dfrac{0.10}{365}\\right)^{(365\\cdot 3)}\\approx\\$1349.80[\/latex]\r\n\r\nNotice that the final balance is larger in each case, but the gains are getting smaller and it seems there is a limit to how much money we can make just by increasing compounding frequency. This limiting process is called continuous compounding<\/strong>, which you can think of as compounding millions of times per second. There is a formula for this type of compounding.\r\n
      \r\n

      THE CONTINUOUS COMPOUNDING FORMULA<\/h3>\r\n

      Continuously compounding interest,<\/strong>\u00a0which can be though of as interest compounding millions of times per second, can be calculated using the formula<\/p>\r\n\r\n

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

      where<\/p>\r\n\r\n

        \r\n \t
      • [latex]A(T)[\/latex]\u00a0is the final account value<\/li>\r\n \t
      • [latex]t[\/latex] is time measured in years<\/li>\r\n \t
      • [latex]P[\/latex]\u00a0is the starting value of the account, or the principal<\/li>\r\n \t
      • [latex]r[\/latex]\u00a0is the annual percentage rate (APR) expressed as a decimal<\/li>\r\n<\/ul>\r\n<\/div>\r\nNote that this formula is actually very generic and can be applied in any real word application of exponential growth, as long as the growth is happening continuously (for example, population growth).\r\n\r\nIf we revisit our previous example with [latex]\\$1,000[\/latex] now invested with continuous compounding for three years, the final balance is [latex]A(3)=1000e^{0.1\\cdot 3}\\approx \\$1,349.86.[\/latex] The implication is that no matter how often interest is compounded in this example, the final balance after three years will never exceed [latex]\\$1,349.86.[\/latex]. This extreme amount of compounding only earned us [latex]6[\/latex] more cents than the daily compounding.\r\n\r\nIn our next example, we will calculate continuous growth of an account. It is important to note the language that is used in the instructions for interest rate problems. \u00a0You will know to use the continuous<\/em> compounding when it uses the word continuous to describe the compounding instead of a frequency like monthly or quarterly.\r\n
        \r\n

        Example<\/h3>\r\nCredit card interest is often compounded continuously. Suppose a balance of [latex]$2,500[\/latex] is left on a credit card with an APR of [latex]26.9\\%[\/latex] compounded continuously and no additional charges or payments are made to the card. What will the balance be after [latex]3[\/latex] years?\r\n[reveal-answer q=\"33008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"33008\"]\r\n\r\nThe problem mentioned that the compounding is continuous so we use that formula. We set interest rate [latex]r=0.269[\/latex], present value [latex]P=$2,500[\/latex], and time [latex]t=3[\/latex] years.\r\n
        [latex]\\begin{array}{rl}A(t)&=P{e}^{rt}\\\\\r\n&=2500{\\left(e\\right)}^{0.269\\cdot 3}\\\\\r\n&\\approx 5602.94\\end{array}[\/latex]<\/div>\r\n \r\n

        The balance on the credit card is [latex]$5602.94[\/latex] after three years. This is more than double what it started at. If left for three more years, it will more than double again. This is how people get into trouble with credit card debt!<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of interest compounded continuously.\r\n\r\nhttps:\/\/youtu.be\/fEjrYCog_8w\r\n

        Summary<\/h2>\r\nThe Compound Interest Formula can calculate the balance in an account which is gaining interest at regular intervals over time. It grows exponentially because the interest is calculated based on the current account balance rather than the principal, which means it generates interest from the previous interest that has already been earned. If interest is compounded more and more frequently, the limit of this process is called Continuous Compounding and uses a base of [latex]e[\/latex] in the exponential growth.\r\n\r\n<\/section>","rendered":"
        \n
        \n

        Learning Outcome<\/h3>\n
          \n
        • Solve compound interest applications of exponential functions<\/li>\n<\/ul>\n<\/div>\n<\/section>\n
          You may have seen or heard of formulas that are used to calculate compound interest rates, for example the interest in a bank account. These formulas are an example of exponential growth.\u00a0The term compounding<\/em> refers to interest earned not only on the original value but on the accumulated value of the account.<\/p>\n

          The annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the interest rate earned by an investment account per year. The term\u00a0nominal<\/em>\u00a0is used when the compounding occurs a number of times other than once per year. When interest is compounded more than once a year, the effective interest rate ends up being greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\n

          We can calculate the balance in an account with compounding interest using the following compound interest formula.<\/p>\n

          \n

          The Compound Interest Formula<\/h3>\n

          Compound interest<\/strong> can be calculated using the formula<\/p>\n

          [latex]\\large A\\left(t\\right)=P{\\left(1+\\dfrac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n

          where<\/p>\n

            \n
          • [latex]A(T)[\/latex]\u00a0is the final account value, often called the future value<\/li>\n
          • [latex]t[\/latex] is time measured in years<\/li>\n
          • [latex]P[\/latex]\u00a0is the starting value of the account, often called the principal or the present value<\/li>\n
          • [latex]r[\/latex]\u00a0is the annual percentage rate (APR) expressed as a decimal<\/li>\n
          • [latex]n[\/latex]\u00a0is the number of compounding periods in one year.<\/li>\n<\/ul>\n<\/div>\n

            The variable [latex]n[\/latex] appearing in the formula is usually expressed as how often the account generates interest. For example, an account with interest compounding monthly<\/strong> will use [latex]n=12[\/latex] since there are [latex]12[\/latex] months in a year.<\/p>\n

            Note that this formula does NOT mean that the full APR is paid out [latex]n[\/latex] times per year. The APR is split between the number of times interest is compounded. For example, if the APR is [latex]6\\%[\/latex] and interest is compounded monthly, then the actual interest received per month is [latex]\\dfrac{6\\%}{12}[\/latex] or [latex]0.5\\%.[\/latex]<\/p>\n

            In our next example, we will calculate the value of an account after\u00a0[latex]10[\/latex] years of interest compounded quarterly.<\/p>\n

            \n

            Example<\/h3>\n

            If we invest\u00a0[latex]$3,000[\/latex] in an investment account paying\u00a0[latex]3\\%[\/latex] interest compounded quarterly, how much will the account be worth in\u00a0[latex]10[\/latex] years?<\/p>\n

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

            Because the principal is [latex]$3,000[\/latex], we set [latex]P=3000.[\/latex] Our interest rate is\u00a0[latex]3\\%[\/latex], so [latex]r=0.03.[\/latex] Because we are compounding quarterly, and there are [latex]4[\/latex] quarters in a year, we set [latex]n=4[\/latex]. We want to know the value of the account in\u00a0[latex]t=10[\/latex] years, so we are looking for [latex]A(10).[\/latex]<\/p>\n

            [latex]\\begin{array}{rl}A\\left(t\\right)&=P\\left(1+\\dfrac{r}{n}\\right)^{nt}\\\\ A\\left(10\\right)&=3000\\left(1+\\dfrac{0.03}{4}\\right)^{4\\cdot 10}\\\\ & \\approx 4045.05\\end{array}[\/latex]<\/div>\n

             <\/p>\n

            The account will be worth about\u00a0[latex]$4,045.05[\/latex] in\u00a0[latex]10[\/latex] years. By default, round answers involving money to the nearest cent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

            The following video shows an example of using exponential growth to calculate interest compounded quarterly.<\/p>\n