2. Copyright © 2009 Pearson Education, Inc. CHAPTER 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions and Graphs 5.3 Logarithmic Functions and Graphs 5.4 Properties of Logarithmic Functions 5.5 Solving Exponential and Logarithmic Equations 5.6 Applications and Models: Growth and Decay; and Compound Interest

3. Copyright © 2009 Pearson Education, Inc. 5.6 Applications and Models: Growth and Decay; and Compound Interest  Solve applied problems involving exponential growth and decay.  Solve applied problems involving compound interest.  Find models involving exponential functions and logarithmic functions.

4. Slide 5.6 - 4 Copyright © 2009 Pearson Education, Inc. Population Growth The function P(t) = P 0 e kt, k > 0 can model many kinds of population growths. In this function: P 0 = population at time 0, P(t) = population after time t, t = amount of time, k = exponential growth rate. The growth rate unit must be the same as the time unit.

5. Slide 5.6 - 5 Copyright © 2009 Pearson Education, Inc. Population Growth - Graph

6. Slide 5.6 - 6 Copyright © 2009 Pearson Education, Inc. Example In 2006, the population of China was about 1.314 billon, and the exponential growth rate was 0.6% per year. a)Find the exponential growth function. b)Graph the exponential growth function. c)Estimate the population in 2010. d)After how long will the population be double what it was in 2006?

7. Slide 5.6 - 7 Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: a)At t = 0 (2006), the population was about 1.314 billion. We substitute 1.314 for P 0 and 0.006 for k to obtain the exponential growth function. b) P(t) = 1.314e 0.006t

8. Slide 5.6 - 8 Copyright © 2009 Pearson Education, Inc. Example (continued) c)In 2010, t = 4. To find the population in 2010 we substitute 4 for t: The population will be approximately 1.346 billion in 2010. The graph also displays this value. P(4) = 1.314e 0.006(4) = 1.314e 0.024  1.346

9. Slide 5.6 - 9 Copyright © 2009 Pearson Education, Inc. Example (continued) d)We are looking for the doubling time; T such that P(T) = 2 1.314 = 2.628. Solve The population of China will be double what it was in 2006 about 115.5 years after 2006.

10. Slide 5.6 - 10 Copyright © 2009 Pearson Education, Inc. Example (continued) d)Using the Intersect method we graph The population of China will be double that of 2006 about 115.5 years after 2006. and find the first coordinate of their point of intersection.

11. Slide 5.6 - 11 Copyright © 2009 Pearson Education, Inc. Interest Compound Continuously The function P(t) = P 0 e kt can be used to calculate interest that is compounded continuously. In this function: P 0 = amount of money invested, P(t) = balance of the account after t years, t = years, k = interest rate compounded continuously.

12. Slide 5.6 - 12 Copyright © 2009 Pearson Education, Inc. Example Suppose that $2000 is invested at interest rate k, compounded continuously, and grows to $2504.65 after 5 years. a. What is the interest rate? b. Find the exponential growth function. c. What will the balance be after 10 years? d. After how long will the $2000 have doubled?

13. Slide 5.6 - 13 Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: a. At t = 0, P(0) = P 0 = $2000. Thus the exponential growth function is P(t) = 2000e kt. We know that P(5) = $2504.65. Substitute and solve for k: The interest rate is about 0.045 or 4.5%.

14. Slide 5.6 - 14 Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: b.The exponential growth function is P(t) = 2000e 0.045t. c.The balance after 10 years is

15. Slide 5.6 - 15 Copyright © 2009 Pearson Education, Inc. Example (continued) d.To find the doubling time T, we set P(T) = 2 P 0 = 2 $2000 = $4000 and solve for T. Thus the orginal investment of $2000 will double in about 15.4 yr.

16. Slide 5.6 - 16 Copyright © 2009 Pearson Education, Inc. Growth Rate and Doubling Time The growth rate k and doubling time T are related by kT = ln 2 or or Note that the relationship between k and T does not depend on P 0.

17. Slide 5.6 - 17 Copyright © 2009 Pearson Education, Inc. Example The population of the world is now doubling every 60.8 yr. What is the exponential growth rate? Solution: The growth rate of the world population is about 1.14% per year.

18. Slide 5.6 - 18 Copyright © 2009 Pearson Education, Inc. Models of Limited Growth In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value. One model of such growth is which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

19. Slide 5.6 - 19 Copyright © 2009 Pearson Education, Inc. Models of Limited Growth - Graph

20. Slide 5.6 - 20 Copyright © 2009 Pearson Education, Inc. Exponential Decay Decay, or decline, of a population is represented by the function P(t) = P 0 e  kt, k > 0. In this function: P 0 = initial amount of the substance (at time t = 0), P(t) = amount of the substance left after time, t = time, k = decay rate. The half-life is the amount of time it takes for a substance to decay to half of the original amount.

21. Slide 5.6 - 21 Copyright © 2009 Pearson Education, Inc. Graphs

22. Slide 5.6 - 22 Copyright © 2009 Pearson Education, Inc. Decay Rate and Half-Life The decay rate k and the half-life T are related by kT = ln 2 or or Note that the relationship between decay rate and half- life is the same as that between growth rate and doubling time.

23. Slide 5.6 - 23 Copyright © 2009 Pearson Education, Inc. Example Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. The percentage of carbon-14 present in the remains of organic matter can be used to determine the age of that organic matter. Archaeologists discovered that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon-14 at the time it was found. How old was the linen wrapping?

24. Slide 5.6 - 24 Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: First find k when the half-life T is 5750 yr: Now we have the function

25. Slide 5.6 - 25 Copyright © 2009 Pearson Education, Inc. Example (continued) If the linen wrapping lost 22.3% of its carbon-14 from the initial amount P 0, then 77.7% is the amount present. To find the age t of the wrapping, solve for t: The linen wrapping on the Dead Sea Scrolls was about 2103 years old when it was found.

26. Slide 5.6 - 26 Copyright © 2009 Pearson Education, Inc. Exponential Curve Fitting We have added several new functions that can be considered when we fit curves to data.

27. Slide 5.6 - 27 Copyright © 2009 Pearson Education, Inc. Logarithmic Curve Fitting Logarithmic Logistic

28. Slide 5.6 - 28 Copyright © 2009 Pearson Education, Inc. Example The number of U.S. communities using surveillance cameras at intersections has greatly increased in recent years, as show in the table.

29. Slide 5.6 - 29 Copyright © 2009 Pearson Education, Inc. Example (continued) a. Use a graphing calculator to fit an exponential function to the data. b.Graph the function with the scatter plot of the data. c.Estimate the number of U.S. communities using surveillance cameras at intersections in 2010.

30. Slide 5.6 - 30 Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: a.Fit an equation of the type y = a b x, where x is the number of years since 1999. Enter the data... The equation is The correlation coefficient is close to 1, indicating the exponential function fits the data well.

31. Slide 5.6 - 31 Copyright © 2009 Pearson Education, Inc. Example (continued) b.Here’s the graph of the function with the scatter plot.

32. Slide 5.6 - 32 Copyright © 2009 Pearson Education, Inc. Example (continued) c.Using the VALUE feature in the CALC menu, we evaluate the function for x = 11 (2010 – 1999 = 11), and estimate the number of communities using surveillance cameras at intersections in 2010 to be about 651.