Section 8.2 Exponential Functions 771 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) = 1.02P (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) = 1.02P (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) = 1.02P (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 1 Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ Version: Fall 2007 772 Chapter 8 Exponential and Logarithmic Functions 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. P (population) 23000 P (t)=10 000(1.02)t 10000 t (years) 0 40 Figure 1. Graph of population P (t) of Pleasantville for t = 0, 1, 2, 3,... I Example 7. We can now use the function P (t) to predict the population in later years. Assuming that the growth rate of 2% continues, what will the population of Pleasantville be after 40 years? What will it be after 100 years? Substitute t = 40 and t = 100 into equation (6). The population in 40 years will be P (40) = 10 000(1.02)40 ≈ 22 080, and the population in 100 years will be P (100) = 10 000(1.02)100 ≈ 72 446. What would be different if we had started with a population of 12 000? By tracing over our previous steps, it should be easy to see that the new formula would be P (t) = 12 000(1.02)t. Similarly, if the growth rate had been 3% per year instead of 2%, then we would have ended up with the formula P (t) = 10 000(1.03)t. Version: Fall 2007 Section 8.2 Exponential Functions 773 Thus, by letting P0 represent the initial population, and r represent the growth rate (in decimal form), we can generalize the formula to t P (t) = P0(1 + r) . (8) Note that our formula for the function P (t) is different from the previous functions that we’ve studied so far, in that the input variable t is part of the exponent in the formula. Thus, this is a new type of function. Now let’s contrast the situation in Pleasantville with the population dynamics of Ghosttown. Ghosttown also starts with a population of 10 000, but several factories have closed, so some people are leaving for better opportunities. In this case, the population of Ghosttown is decreasing at a rate of 2% per year. We’ll again 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 decrease is 2% of 10 000, or 200 people. We would now have 9 800 people left in Ghosttown. At the end of the second year, take another 2% of 9 800, which is a decrease of 196 people, for a total of 9 604. As before, because the decrease each year is not constant, the graph of population versus time cannot be a line, so our eventual population function will not be linear. Now let the function P (t) represent the population of Ghosttown at time t, where we measure t in years. The initial population of Ghosttown at t = 0 is 10 000, so P (0) = 10 000. Since the population decreases by 2% each year, at the end of the first year the population of Ghosttown will be 98% of the initial population. Thus, P (1) = 0.98P (0) = 0.98(10 000). (9) Each year the population deccreases by 2%. Therefore, at the end of the second year, the population will be 98% of the population at the end of the first year. In other words, P (2) = 0.98P (1). (10) If we replace P (1) in equation (10) with the result found in equation (9), then P (2) = (0.98)(0.98)(10 000) = (0.98)2(10 000). (11) Let’s iterate one more year. At the end of the third year, the population will be 98% of the population at the end of the second year, so P (3) = 0.98P (2). (12) However, if we replace P (2) in equation (12) with the result found in equation (11), we obtain P (3) = (0.98)(0.98)2(10 000) = (0.98)3(10 000). (13) The pattern should now be clear. The population at the end of t years is given by the function Version: Fall 2007 774 Chapter 8 Exponential and Logarithmic Functions P (t) = (0.98)t(10 000), or equivalently, P (t) = 10 000(0.98)t. (14) Our function P (t) is defined by equation (14) for all positive integers {1, 2, 3,...}, and P (0) = 10 000, the initial population. Figure 2 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, decreases over time, and the yearly decrease (the difference in population from one year to the next) also gets smaller as time passes. P (population) 10000 5000 P (t)=10 000(0.98)t t (years) 0 0 40 Figure 2. Graph of population P (t) of Ghosttown for t = 0, 1, 2, 3,... I Example 15. Assuming that the rate of decrease continues at 2%, predict the population of Ghosttown after 40 years and after 100 years. Substitute t = 40 and t = 100 into equation (14). The population in 40 years will be P (40) = 10 000(0.98)40 ≈ 4457, and the population in 100 years will be P (100) = 10 000(0.98)100 ≈ 1326. Note that if we had instead started with a population of 9 000, for example, then the new formula would be P (t) = 9 000(0.98)t. Similarly, if the rate of decrease had been 5% per year instead of 2%, then we would have ended up with the formula P (t) = 10 000(0.95)t. Thus, by letting P0 represent the initial population, and r represent the growth rate (in decimal form), we can generalize the formula to t P (t) = P0(1 − r) . (16) Version: Fall 2007 Section 8.2 Exponential Functions 775 Definition As noted before, our functions P (t) in our Pleasantville and Ghosttown examples are a new type of function, because the input variable t is part of the exponent in the formula. Definition 17. An exponential function is a function of the form f(t) = bt, where b > 0 and b 6= 1.