1. Slide 10- 1 Copyright © 2012 Pearson Education, Inc.

2. 9.7 Applications of Exponential and Logarithmic Functions ■ Applications of Logarithmic Functions ■ Applications of Exponential Functions

3. Slide 9- 3 Copyright © 2012 Pearson Education, Inc. Applications of Logarithmic Functions

4. Slide 9- 4 Copyright © 2012 Pearson Education, Inc. Sound Levels To measure the volume, or “loudness”, of a sound, the decibel scale is used. The loudness L, in decibels (dB), of a sound is given by where I is the intensity of the sound, in watts per square meter (W/m 2 ), and I 0 = 10 –12 W/m 2. (I 0 is approximately the intensity of the softest sound that can be heard by the human ear.)

5. Slide 9- 5 Copyright © 2012 Pearson Education, Inc. Solution Example a) If the intensity of sound is 10 –2.2 W/m 2, find its decibel level. b) What is the intensity level of sounds at a decibel level of 75 dB? a) The volume is 98 decibels.

6. Slide 9- 6 Copyright © 2012 Pearson Education, Inc. b) The sound intensity is 10 –4.5 W/m 2. Solution continued

7. Slide 9- 7 Copyright © 2012 Pearson Education, Inc. Chemistry: pH of Liquids. In chemistry, the pH of a liquid is a measure of its acidity. We calculate pH as follows where [H + ] is the hydrogen ion concentration in moles per liter.

8. Slide 9- 8 Copyright © 2012 Pearson Education, Inc. Applications of Exponential Functions

9. Slide 9- 9 Copyright © 2012 Pearson Education, Inc. Example Interest compounded annually. The amount of money A that a principal P will be worth after t years at interest rate i, compounded annually, is given by the formula a) How long will it take to accumulate $15,000 in the account? b) Find the amount of time it takes for the $9,000 to double itself. Suppose that $9,000 is invested at 5% interest, compounded annually.

10. Slide 9- 10 Copyright © 2012 Pearson Education, Inc. log(15/9) = log(1.05) t log(15/9) = t log(1.05) log(15/9)/log(1.05) = t It will take about 10.5 yr for $9,000 to grow to $15,000. Solution a) A = P(1 + i) t 15,000 = 9,000(1 +.05) t 15/9 = (1.05) t

11. Slide 9- 11 Copyright © 2012 Pearson Education, Inc. log(2) = log(1.05) t log(2) = t log(1.05) log(2)/log(1.05) = t At an interest rate of 5% per year, the doubling time is about 14.2 yr. Solution continued b) To find the doubling time, we replace A(t) with 18,000 and solve for t. 18,000 = 9,000(1 +.05) t 2 = (1.05) t

12. Slide 9- 12 Copyright © 2012 Pearson Education, Inc. Exponential Growth An exponential growth model is a function of the form where P 0 is the population at time 0, P(t) is the population at time t, and k is the exponential growth rate for the situation. The doubling time is the amount of time necessary for the population to double in size.

13. Slide 9- 13 Copyright © 2012 Pearson Education, Inc. Interest compounded continuously. When the amount of money P 0 is invested at interest rate k, compounded continuously, interest is computed every “instant” and added to the original amount. The balance P(t), after t years, is given by the exponential growth model

14. Slide 9- 14 Copyright © 2012 Pearson Education, Inc. Example Suppose that $45,000 is invested and grows to $60,743.65 in 5 yr. Find the exponential growth function. We have P(0) = 45,000. Thus the exponential growth function is P(t) = 45,000e kt, where k must still be determined. Knowing that for t = 5 we have P(5) = 60,743.65, it is possible to solve for k: Solution

15. Slide 9- 15 Copyright © 2012 Pearson Education, Inc. 60,743.65 = 45,000e k(5) 60,743.65/45,000 = e k(5) ln(1.349858889) = ln(e k(5) ) 1.349858889 = e k(5) ln(1.349858889) = 5k ln(1.349858889)/5 = k The interest rate is about 0.06, or 6%, compounded continuously. The exponential growth function is P(t) = 45,000e 0.06t.

16. Slide 9- 16 Copyright © 2012 Pearson Education, Inc. Exponential Decay An exponential decay model is a function of the form where P 0 is the quantity present at time 0, P(t) is the amount present at time t, and k is the decay rate. The half-life is the amount of time necessary for half of the quantity to decay.

17. Slide 9- 17 Copyright © 2012 Pearson Education, Inc. The equation can be used for any subsequent carbon-dating problem.

18. Slide 9- 18 Copyright © 2012 Pearson Education, Inc. Example For each of the following graphs, determine whether an exponential function might fit the data.

19. Slide 9- 19 Copyright © 2012 Pearson Education, Inc. Solution a. As the sound intensity increases, the safe exposure time decreases. The amount of decrease gets smaller as the intensity increases. It appears that an exponential decay function might fit the data.

20. Slide 9- 20 Copyright © 2012 Pearson Education, Inc. Solution b. The number of health-care providers and social workers increased between 2004 and 2008 at approximately the same rate each year. It does not appear that an exponential function models the data; instead; a linear function might be appropriate.

21. Slide 9- 21 Copyright © 2012 Pearson Education, Inc. Solution c. The number hurricanes first fell, then rose, and then fell again. This does not fit an exponential model.

22. Slide 9- 22 Copyright © 2012 Pearson Education, Inc. Solution d. The number of text messages increased from 2003 to 2008, and the amount of yearly growth also increased during that time. This suggests that an exponential growth function could be used to model this situation.