Section 3: Applications of Exponential Functions

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 8.3 Applications of Exponential Functions 789 Version: Fall 2007 8.3 Applications of Exponential Functions In the preceding section, we examined a population growth problem in which the popu- lation grew at a fixed percentage each year. In that case, we found that the population can be described by an exponential function. A similar analysis will show that any process in which a quantity grows by a fixed percentage each year (or each day, hour, etc.) can be modeled by an exponential function. Compound interest is a good example of such a process. Discrete Compound Interest If you put money in a savings account, then the bank will pay you interest (a percentage of your account balance) at the end of each time period, typically one month or one day. For example, if the time period is one month, this process is called monthly compounding . The term compounding refers to the fact that interest is added to your account each month and then in subsequent months you earn interest on the interest. If the time period is one day, it’s called daily compounding . Let’s look at monthly compounding in more detail. Suppose that you deposit $100 in your account, and the bank pays interest at an annual rate of 5%. Let the function P ( t ) represent the amount of money that you have in your account at time t , where we measure t in years. We will start time at t = 0 when the initial amount, called the principal , is $100. In other words, P (0) = 100 . In the discussion that follows, we will compute the account balance at the end of each month. Since one month is 1/12 of a year, P ( 1 / 12 ) represents the balance at the end of the first month, P ( 2 / 12 ) represents the balance at the end of the second month, etc. At the end of the first month, interest is added to the account balance. Since the annual interest rate 5%, the monthly interest rate is 5%/12, or .05/12 in decimal form. Although we could approximate .05/12 by a decimal, it will be more useful, as well as more accurate, to leave it in this form. Therefore, at the end of the first month, the interest earned will be 100( . 05 / 12) , so the total amount will be P ( 1 / 12 ) = 100 + 100 . 05 12 = 100 1 + . 05 12 . (1) Now at the end of the second month, you will have the amount that you started that month with, namely P ( 1 / 12 ) , plus another month’s worth of interest on that amount. Therefore, the total amount will be P ( 2 / 12 ) = P ( 1 / 12 ) + P ( 1 / 12 ) . 05 12 = P ( 1 / 12 ) 1 + . 05 12 . (2) If we replace P ( 1 / 12 ) in equation (2) with the result found in equation (1) , then Copyrighted material. See: 1
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790 Chapter 8 Exponential and Logarithmic Functions Version: Fall 2007 P ( 2 / 12 ) = 100 1 + . 05 12 1 + . 05 12 = 100 1 + . 05 12 2 . (3) Let’s iterate one more month. At the end of the third month, you will have the amount that you started that month with, namely P ( 2 / 12 ) , plus another month’s worth of interest on that amount. Therefore, the total amount will be P ( 3 / 12 ) = P ( 2 / 12 ) + P ( 2 / 12 ) . 05 12 = P ( 2 / 12 ) 1 + . 05 12 . (4) However, if we replace P ( 2 / 12 ) in equation (4) with the result found in equation (3) , then P ( 3 / 12 ) = 100 1 + . 05 12 2 1 + . 05 12 = 100 1 + . 05 12 3 . (5) The pattern should now be clear. The amount of money you will have in the account at the end of m
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