1. Copyright © Cengage Learning. All rights reserved. Exponential and Logarithmic Functions

2. Copyright © Cengage Learning. All rights reserved. 4.5 Exponential and Logarithmic Equations

3. 3 Objectives ► Exponential Equations ► Logarithmic Equations ► Compound Interest

4. 4 Exponential Equations

5. 5 An exponential equation is one in which the variable occurs in the exponent. For example, 2 x = 7 The variable x presents a difficulty because it is in the exponent. To deal with this difficulty, we take the logarithm of each side and then use the Laws of Logarithms to “bring down x” from the exponent.

6. 6 Exponential Equations 2 x = 7 ln 2 x = ln 7 x ln 2 = ln 7  2.807 Given Equation Law 3 (Bring down exponent) Take In of each side Solve for x Calculator

7. 7 Exponential Equations Recall that Law 3 of the Laws of Logarithms says that log a A C = C log a A. The method that we used to solve 2 x = 7 is typical of how we solve exponential equations in general.

8. 8 Example 1 – Solving an Exponential Equation Find the solution of the equation 3 x + 2 = 7, rounded to six decimal places. Solution: We take the common logarithm of each side and use Law 3. 3 x + 2 = 7 log(3 x + 2 ) = log 7 Given Equation Take log of each side

9. 9 Example 1 – Solution (x + 2)log 3 = log 7 x + 2 =  –0.228756 Calculator Subtract 2 Divide by log 3 Law 3 (bring down exponent) cont’d

10. 10 Example 1 – Solution Check Your Answer Substituting x = –0.228756 into the original equation and using a calculator, we get 3 (–0.228756) + 2  7 cont’d

11. 11 Example 4 – An Exponential Equation of Quadratic Type Solve the equation e 2x – e x – 6 = 0. Solution: To isolate the exponential term, we factor. e 2x – e x – 6 = 0 (e x ) 2 – e x – 6 = 0 Given Equation Law of Exponents

12. 12 Example 4 – Solution (e x – 3)(e x + 2) = 0 e x – 3 = 0 or e x + 2 = 0 e x = 3 e x = –2 The equation e x = 3 leads to x = ln 3. But the equation e x = –2 has no solution because e x > 0 for all x. Thus, x = ln 3  1.0986 is the only solution. Zero-Product Property Factor (a quadratic in e x ) cont’d

13. 13 Logarithmic Equations

14. 14 Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. For example, log 2 (x + 2) = 5 To solve for x, we write the equation in exponential form. x + 2 = 2 5 x = 32 – 2 = 30 Solve for x Exponential form

15. 15 Logarithmic Equations Another way of looking at the first step is to raise the base, 2, to each side of the equation. 2 log 2 (x + 2) = 2 5 x + 2 = 2 5 x = 32 – 2 = 30 Solve for x Property of logarithms Raise 2 to each side

16. 16 Logarithmic Equations The method used to solve this simple problem is typical. We summarize the steps as follows.

17. 17 Example 6 – Solving Logarithmic Equations Solve each equation for x. (a) ln x = 8 (b) log 2 (25 – x) = 3 Solution: (a) ln x = 8 x = e 8 Therefore, x = e 8  2981. Given equation Exponential form

18. 18 Example 6 – Solution We can also solve this problem another way: ln x = 8 e ln x = e 8 x = e 8 Property of ln Raise e to each side Given equation cont’d

19. 19 Example 6 – Solution (b) The first step is to rewrite the equation in exponential form. log 2 (25 – x) = 3 25 – x = 2 3 25 – x = 8 x = 25 – 8 = 17 Given equation Exponential form (or raise 2 to each side) cont’d

20. 20 Example 6 – Solution Check Your Answer If x = 17, we get log 2 (25 – 17) = log 2 8 = 3 cont’d

21. 21 Logarithmic Equations Logarithmic equations are used in determining the amount of light that reaches various depths in a lake. (This information helps biologists to determine the types of life a lake can support.) As light passes through water (or other transparent materials such as glass or plastic), some of the light is absorbed. It’s easy to see that the murkier the water, the more light is absorbed. The exact relationship between light absorption and the distance light travels in a material is described in the next example.

22. 22 Example 10 – Transparency of a Lake If I 0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer Lambert Law, where k is a constant depending on the type of material.

23. 23 Example 10 – Transparency of a Lake (a) Solve the equation for I. (b) For a certain lake k = 0.025, and the light intensity is I 0 = 14 lumens (lm). Find the light intensity at a depth of 20 ft. cont’d

24. 24 Example 10 – Solution (a) We first isolate the logarithmic term. I = I 0 e –kx Given equation Multiply by –k Exponential form Multiply by I 0

25. 25 Example 10 – Solution (b) We find I using the formula from part (a). I = I 0 e –kx = 14e (–0.025)(20)  8.49 The light intensity at a depth of 20 ft is about 8.5 lm. From part (a) I 0 = 14, k = 0.025, x = 20 Calculator cont’d

26. 26 Compound Interest

27. 27 Compound Interest If a principal P is invested at an interest rate r for a period of t years, then the amount A of the investment is given by A = P(1 + r) A(t) = Pe rt We can use logarithms to determine the time it takes for the principal to increase to a given amount. Simple interest (for one year) Interest compounded n times per year Interest compounded continuously

28. 28 Example 11 – Finding the Term for an Investment to Double A sum of $5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannually (b) Continuously

29. 29 Example 11 – Solution (a) We use the formula for compound interest with P = $5000, A(t) = $10,000, r = 0.05, and n = 2 and solve the resulting exponential equation for t. (1.025) 2t = 2 log 1.025 2t = log 2 2t log 1.025 = log 2 Divide by 5000 Take log of each side Law 3 (bring down the exponent)

30. 30 Example 11 – Solution t  14.04 The money will double in 14.04 years. Calculator Divide by 2 log 1.025 cont’d

31. 31 Example 11 – Solution (b) We use the formula for continuously compounded interest with P = $5000, A(t) = $10,000, and r = 0.05 and solve the resulting exponential equation for t. 5000e 0.05t = 10,000 e 0.05t = 2 ln e 0.05t = ln 2 0.05t = ln 2 Pe rt = A Divide by 5000 Take ln of each side Property of ln cont’d

32. 32 Example 11 – Solution t  13.86 The money will double in 13.86 years. Calculator Divide by 0.05 cont’d