1. 10.2 Exponential and Logarithmic Functions

2. Exponential Functions These functions model rapid growth or decay: # of users on the Internet 16 million (1995)  957 million (late 2005) Compound interest Population growth or decline

3. Comparison Linear Functions Rate of change is constant Exponential Functions Change at a constant PERCENT rate of change.   

4. The Exponential Function y = ab x b is the base: It must be greater than 0 It cannot equal 1 (Why?). x is the exponent: x can be any real number

5. Graph Exponential Functions (b > 1) Graph y = 2 x for x = -3 to 3 x y -3 -2 0 1 2 3 1/8 1/4 1/2 1 2 4 8

6. Graph Exponential Functions (0< b < 1) Graph y = (1/2) x for x = -3 to 3 x y -3 -2 0 1 2 3 8 4 2 1 1/2 1/4 1/8

7. What is the solution to the inequality according to the graph? X>0 or (0, ∞ )

8. All of the transformations that you learned apply to all functions, so what would the graph of look like? up 3 up 1 Reflected over x axis down 1right 2

9. Which function has a constant rate of change? Why? Blue or Red ? RED

10. How many solutions has the equation f(x)=g(x)? Why? Two Solutions

11. COMPOUND INTEREST Consider an initial principal P deposited in an account that pays interest at an annual rate r (expressed as a decimal), compounded n times per year. The amount A in the account after t years can be modeled by this equation: Although interest earned is expressed as an annual percent, the interest is usually compounded more frequently than once per year. Therefore, the formula y = a(1 + r) t must be modified for compound interest problems. Using Exponential Growth Models COMPOUND INTEREST Exponential growth functions are used in real-life situations involving compound interest. Compound interest is interest paid on the initial investment, called the principal, and on previously earned interest. (Interest paid only on the principal is called simple interest.) A = P 1 + rnrn nt ( )

12. annually SOLUTION With interest compounded annually, the balance at the end of 1 year is: FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. = 1000(1.08) 1 = 1080 P = 1000, r = 0.08, n = 1, t = 1 ( ) A = 1000 1 + 0.08 1 Use a calculator Finding the Balance in an Account The balance at the end of 1 year is $1080. Write compound interest model A = P 1 + rnrn nt ( ) Simplify

13. quarterly SOLUTION With interest compounded quarterly, the balance at the end of 1 year is: FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. = 1000(1.02) 4  1082.43 P = 1000, r = 0.08, n = 4, t = 1 Use a calculator Finding the Balance in an Account The balance at the end of 1 year is $1082.43. ( ) A = 1000 1 + 0.08 4 4 1 Write compound interest model A = P 1 + rnrn nt ( ) Simplify

14. daily SOLUTION With interest compounded daily, the balance at the end of 1 year is: FINANCE You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency.  1000(1.000219) 365  1083.28 P = 1000, r = 0.08, n = 365, t = 1 Simplify Use a calculator Finding the Balance in an Account The balance at the end of 1 year is $1083.28. ( ) A = 1000 1 + 0.08 365 365 1 Write compound interest model A = P 1 + rnrn nt ( )

15. Using Exponential Decay Models In this model, a is the initial amount and r is the percent decrease expressed as a decimal. When a real-life quantity decreases by a fixed percent each year (or other time period), the amount y of quantity after t years can be modeled by the equation: y = a(1 – r) t The quantity 1 – r is called the decay factor.

16. Write an exponential decay model for the value of the car. Use the model to estimate the value after 2 years. SOLUTION Let t be the number of years since you bought the car. The exponential decay model is: You buy a new car for $24,000. The value y of the car decreases by 16% each year. y = a(1 – r) t = 24,000(1 – 0.16) t = 24,000(0.84) t Write exponential decay model Substitute for a and r Simplify Modeling Exponential Decay When t = 2, the value is y = 24,000(0.84) 2  $16,934.

17. SOLUTION You buy a new car for $24,000. The value y of the car decreases by 16% each year. Graph the model. Modeling Exponential Decay The graph of the model is shown at the right. Notice that it passes through the points (0, 24,000) and (1, 20,160). The asymptote of the graph is the line y = 0.

18. Use the graph to estimate when the car will have a value of $12,000. SOLUTION You buy a new car for $24,000. The value y of the car decreases by 16% each year. Modeling Exponential Decay Using the graph, you can see that the value of the car will drop to $12,000 after about 4 years.

19. Modeling Exponential Decay In the previous example the percent decrease, 16%, tells you how much value the car loses from one year to the next. The decay factor, 0.84, tells you what fraction of the car’s value remains from one year to the next. The closer the percent decrease for some quantity is to 0%, the more the quantity is conserved or retained over time. The closer the percent decrease is to 100%, the more the quantity is used or lost over time.

20. Graphing Logarithmic Functions

27. Now it’s your time… Let’s practice