1. Lesson 10.6 Exponential Growth & Decay Value of Items (Appreciation) Ending amount = Starting amount (1 + rate) time Value of Items (Depreciation) Ending amount = Starting amount (1 - rate) time

2. Example 1: Appreciation A house was purchased for $156,000 and appreciates at a rate of 9.8%. What is the value of the house five years after the purchase? A = $156,000 (1 + 0.098) 5 Ending amount = Starting amount (1 + rate) time Rate as a decimal Move the decimal left two places A = $248,963.85

3. Example 2: Appreciation The Jones family purchased a house five years ago for $245,000. The house is now worth $328,000. Assuming a steady annual percentage growth rate, what was the approximate yearly appreciation rate? 328,000 = 245,000 (1 + r) 5 328,000 = (1 + r) 5 245,000 = (1 + r) = r

4. Example 3: Depreciation A company purchased a GPS for $12,500. The GPS depreciated by a fixed rate of 6% and is now worth $8600. How long ago did the company purchase the GPS? 8600 = 12500 ( 1 – 0.06) t 8600 = ( 1 – 0.06) t 12500 0.688 = ( 0.94) t Now, we need to apply the power property of logarithms Ending amount = Starting amount (1 - rate) time =

5. Your turn! Textbook pages 563-564 Numbers 10, 11, 20

6. Example 4: Population Growth How many days will it take a culture of bacteria to increase from 2000 to 50,000 if the growth rate per day is 93.2%

7. Example 5: Interest Continuously Compounded Alex invests $2000 in an account that has 6% annual rate of growth. To the nearest year, when will the investment be worth $3600?

8. Example 6: Population The population P in thousands of a city can be modeled by the equation, where t is the time in years. In how many years will the population of the city be 120,000?

9. Your turn! Textbook pages 557-558 Numbers 28, 58, 59

10. Example 7: Compounded Interest Suppose $500 is invested in an account at 6% annual interest compounded twice per year. How much will the investment be worth in 8 years?