College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

1. College Algebra Fifth Edition James StewartLothar RedlinSaleem Watson

2. Exponential and Logarithmic Functions 5

3. Exponential and Logarithmic Equations 5.4

4. Exponential and Logarithmic Equations • In this section, we solve equations that involve exponential or logarithmic functions. • The techniques we develop here will be used in the next section for solving applied problems.

5. Exponential Equations

6. Exponential Equations • An exponential equationis one in which the variable occurs in the exponent. • For example, 2x = 7

7. Solving Exponential Equations • 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.

8. Solving Exponential Equations • Law 3 of the Laws of Logarithms says that: logaAC =C logaA

9. Solving Exponential Equations • The method we used to solve 2x = 7 is typical of how we solve exponential equations in general.

10. Guidelines for Solving Exponential Equations • Isolate the exponential expression on one side of the equation. • Take the logarithm of each side, and then use the Laws of Logarithms to “bring down the exponent.” • Solve for the variable.

11. E.g. 1—Solving an Exponential Equation • Find the solution of 3x + 2 =7 correct to six decimal places. • We take the common logarithm of each side and use Law 3.

12. E.g. 1—Solving an Exponential Equation

13. Check Your Answer • Substituting x = –0.228756 into the original equation and using a calculator, we get: 3(–0.228756)+2≈ 7

14. E.g. 2—Solving an Exponential Equation • Solve the equation 8e2x = 20. • We first divide by 8 to isolate the exponential term on one side.

15. Check Your Answer • Substituting x = 0.458 into the original equation and using a calculator, we get: 8e2(0.458) = 20

16. E.g. 3—Solving Algebraically and Graphically • Solve the equation e3 – 2x = 4 algebraically and graphically

17. Solution 1 E.g. 3—Solving Algebraically • The base of the exponential term is e. • So, we use natural logarithms to solve. • You should check that this satisfies the original equation.

18. Solution 2 E.g. 3—Solving Graphically • We graph the equations y =e3–2x and y = 4 in the same viewing rectangle as shown. • The solutions occur where the graphs intersect. • Zooming in on the point of intersection, we see: x ≈ 0.81

19. E.g. 4—Exponential Equation of Quadratic Type • Solve the equation e2x –ex – 6 = 0. • To isolate the exponential term, we factor.

20. E.g. 4—Exponential Equation of Quadratic Type • The equation ex = 3 leads to x = ln 3. • However, the equation ex = –2 has no solution because ex > 0 for all x. • Thus, x =ln 3 ≈ 1.0986 is the only solution. • You should check that this satisfies the original equation.

21. E.g. 5—Solving an Exponential Equation • Solve the equation 3xex +x2ex = 0. • First, we factor the left side of the equation. • Thus, the solutions are x = 0 and x = –3.

22. Logarithmic Equations

23. Logarithmic Equations • A logarithmic equationis one in which a logarithm of the variable occurs. • For example, log2(x + 2) = 5

24. Solving Logarithmic Equations • To solve for x, we write the equation in exponential form.x + 2 = 25x = 32 – 2 = 30

25. Solving Logarithmic Equations • Another way of looking at the first step is to raise the base, 2, to each side. 2log2(x +2) =25x + 2 = 25x = 32 – 2 = 30 • The method used to solve this simple problem is typical.

26. Guidelines for Solving Logarithmic Equations • Isolate the logarithmic term on one side of the equation. • You may first need to combine the logarithmic terms. • Write the equation in exponential form (or raise the base to each side). • Solve for the variable.

27. E.g. 6—Solving Logarithmic Equations • Solve each equation for x. • ln x = 8 • log2(25 – x) = 3

28. Example (a) E.g. 6—Solving Logarithmic Eqns. • ln x = 8x =e8Therefore, x =e8≈ 2981. • We can also solve this problem another way:

29. Example (b) E.g. 6—Solving Logarithmic Eqns. • The first step is to rewrite the equation in exponential form.

30. E.g. 7—Solving a Logarithmic Equation • Solve the equation 4 + 3 log(2x) = 16 • We first isolate the logarithmic term. • This allows us to write the equation in exponential form.

31. E.g. 7—Solving a Logarithmic Equation

32. E.g. 8—Solving Algebraically and Graphically • Solve the equation log(x + 2) + log(x – 1) = 1algebraically and graphically.

33. Solution 1 E.g. 8—Solving Algebraically • We first combine the logarithmic terms using the Laws of Logarithms.

34. Solution 1 E.g. 8—Solving Algebraically • We check these potential solutions in the original equation. • We find that x = –4 is not a solution. • This is because logarithms of negative numbers are undefined. • x = 3 is a solution, though.

35. Solution 2 E.g. 8—Solving Graphically • We first move all terms to one side of the equation: log(x + 2) + log(x – 1) – 1 = 0

36. Solution 2 E.g. 8—Solving Graphically • Then, we graph y = log(x + 2) + log(x – 1) – 1 • The solutions are the x-intercepts. • So, the only solution is x≈ 3.

37. E.g. 9—Solving a Logarithmic Equation Graphically • Solve the equation x2 = 2 ln(x + 2) • We first move all terms to one side of the equationx2 – 2 ln(x + 2) = 0

38. E.g. 9—Solving a Logarithmic Equation Graphically • Then, we graph y = x2 – 2 ln(x + 2) • The solutions are the x-intercepts. • Zooming in on them, we see that there are two solutions: x ≈ 0.71 x ≈ 1.60

39. Application • Logarithmic equations are used in determining the amount of light that reaches various depths in a lake. • This information helps biologists determine the types of life a lake can support.

40. Application • 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.

41. Application • The exact relationship between light absorption and the distance light travels in a material is described in the next example.

42. E.g. 10—Transparency of a Lake • Let: • I0 and Idenote the intensity of light before and after going through a material. • x be the distance (in feet) the light travels in the material.

43. E.g. 10—Transparency of a Lake • Then, according to the Beer-Lambert Law, where k is a constant depending on the type of material.

44. E.g. 10—Transparency of a Lake • Solve the equation for I. • For a certain lake k = 0.025 and the light intensity is I0 = 14 lumens (lm). Find the light intensity at a depth of 20 ft.

45. Example (a) E.g. 10—Transparency of a Lake • We first isolate the logarithmic term.

46. Example (b) E.g. 10—Transparency of a Lake • We find Iusing the formula from part (a). • The light intensity at a depth of 20 ft is about 8.5 lm.

47. Compound Interest

48. Types of Interest • If a principal P is invested at an interest rate r for a period of t years, the amount A of the investment is given by:

49. Compound Interest • We can use logarithms to determine the time it takes for the principal to increase to a given amount.

50. E.g. 11—Finding 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. • Semiannual • Continuous