1. MODELING REAL SITUATIONS USING EXPONENTIAL FUNCTIONS: PART 1 Patterns #4

2. Example 1 Remember the function A = 1000(1.08) n where $1000 was invested at 8% compounded annually? You were asked when the money doubled. Let’s solve it using logs instead of a graph!

3. Example 1 So A = 1000(1.08) n where $1000 was invested at 8% compounded annually, find out how long before the money doubles. 2000 = 1000(1.08) n

4. Example 1

5. Example 2 Coffee, tea, cola, and chocolate contain caffeine. When you consume caffeine, the percent, P, left in your body can be modeled as a function of the elapsed time, n hours, by the equation: P = 100 (0.87) n Determine how many hours it takes for the original amount of caffeine to drop by 50%.

6. Example 2 So to determine how many hours it takes for the original amount of caffeine to drop by 50% we use the equation: P = 100 (0.87) n. Solve:50 = 100 (0.87) n

7. Example 2

8. Example 3 The population, P million, of Alberta can be modeled by the equation P = 2.28 (1.014) n, where n is the number of years since 1981. Assume this pattern continues. Determine when the population of Alberta might become 4 million.

9. Example 3 The population, P million, of Alberta can be modeled by the equation P = 2.28 (1.014) n, where n is the number of years since 1981. Assume this pattern continues. Determine when the population of Alberta might become 4 million. To get started we need to solve the equation: 4 = 2.28 (1.014) n

10. Example 3

11. Example 4 In 1995, Canada’s population was 29.6 million, and was growing at about 1.24% per year. Estimate the doubling time for Canada’s population growth. We start with an equation: P = 29.6 x (1.0124) n So doubling would mean: 2 x 29.6 = 29.6 x (1.0124) n

12. Example 4

13. Example 5 Consider the equation P = 100 (0.87) n that models residual caffeine. Write this equation as an exponential function with ½ as the base instead of 0.87. To start: write 0.87 as a power of 0.5

14. Example 5

15. In Chemistry… Radioactive isotopes of certain elements decay with a characteristic half-life. You can use an equation of the form:

16. Example 6 In April 1986, there was a major nuclear accident at the Chernobyl plant in Ukraine. The atmosphere was contaminated with quantities of radioactive iodine-131, which has a half-life of 8.1 days. How long did it take for the level of radiation to reduce to 1 % of the level immediately after the accident?

17. Example 6 To begin: Solve for d, the number of days

18. Example 6

19. Example 7 In 1996, Kelowna’s population was approximately 89 000 and was growing at about 3.33% per year. Estimate the doubling time for Kelowna’s population growth.

20. Example 7

21. Example 8 Radioactive phosphorus-32 is used to study liver function. It has a half-life of 14.3 days. If a small amount of phosphorus-32 is injected into a person’s body, how long will it take for the level of radiation to drop to 5% of its original value?

22. Example 8

23. Textwork p. 92/1-12