Section 2: Exponential Functions

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

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Section 8.2 Exponential Functions 771 Version: Fall 2007 8.2 Exponential Functions Let’s suppose that the current population of the city of Pleasantville is 10 000 and that the population is growing at a rate of 2% per year. In order to analyze the population growth over a period of years, we’ll try to develop a formula for the population as a function of time, and then graph the result. First, note that at the end of one year, the population increase is 2% of 10 000, or 200 people. We would now have 10 200 people in Pleasantville. At the end of the second year, take another 2% of 10 200, which is an increase of 204 people, for a total of 10 404. Because the increase each year is not constant, the graph of population versus time cannot be a line. Hence, our eventual population function will not be linear. To develop our population formula, we start by letting the function P ( t ) represent the population of Pleasantville at time t , where we measure t in years. We will start time at t = 0 when the initial population of Pleasantville is 10 000. In other words, P (0) = 10 000 . The key to understanding this example is the fact that the population increases by 2% each year. We are making an assumption here that this overall growth accounts for births, deaths, and people coming into and leaving Pleasantville. That is, at the end of the first year, the population of Pleasantville will be 102% of the initial population. Thus, P (1) = 1 . 02 P (0) = 1 . 02(10 000) . (1) We could multiply out the right side of this equation, but it will actually be more useful to leave it in its current form. Now each year the population increases by 2%. Therefore, at the end of the second year, the population will be 102% of the population at the end of the first year. In other words, P (2) = 1 . 02 P (1) . (2) If we replace P (1) in equation (2) with the result found in equation (1) , then P (2) = (1 . 02)(1 . 02)(10 000) = (1 . 02) 2 (10 000) . (3) Let’s iterate one more year. At the end of the third year, the population will be 102% of the population at the end of the second year, so P (3) = 1 . 02 P (2) . (4) However, if we replace P (2) in equation (4) with the result found in equation (3) , we obtain P (3) = (1 . 02)(1 . 02) 2 (10 000) = (1 . 02) 3 (10 000) . (5) The pattern should now be clear. The population at the end of t years is given by the function Copyrighted material. See: 1

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772 Chapter 8 Exponential and Logarithmic Functions Version: Fall 2007 P ( t ) = (1 . 02) t (10 000) . It is traditional in mathematics and science to place the initial population in front in this formula, writing instead P ( t ) = 10 000(1 . 02) t . (6) Our function P ( t ) is defined by equation (6) for all positive integers { 1 , 2 , 3 , . . . } , and P (0) = 10 000 , the initial population. Figure 1 shows a plot of our function. Although points are plotted only at integer values of t from 0 to 40, that’s enough to show the trend of the population over time. The population starts at 10 000, increases over time, and the yearly increase (the difference in population from one year to the next) also gets larger as time passes.
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