1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 4 Inverse, Exponential, and Logarithmic Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2. 2 4.5 Exponential and Logarithmic Equations Exponential Equations Logarithmic Equations Applications and Models

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Property of Logarithms If x > 0, y > 0, a > 0, and a ≠ 1, then the following holds. x = y is equivalent to log a x = log a y.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 SOLVING AN EXPONENTIAL EQUATION Solve 7 x = 12. Give the solution to the nearest thousandth. Solution The properties of exponents given in Section 4.2 cannot be used to solve this equation, so we apply the preceding property of logarithms. While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 SOLVING AN EXPONENTIAL EQUATION Solve 7 x = 12. Give the solution to the nearest thousandth. Solution Property of logarithms Power property Divide by In 7. Use a calculator. The solution set is {1.277}.

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Caution When evaluating a quotient like in Example 1, do not confuse this quotient with which can be written as In 12 – In 7. We cannot change the quotient of two logarithms to a difference of logarithms.

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Example 2 SOLVING AN EXPONENTIAL EQUATION Solve 3 2x – 1 = 0.4 x + 2. Give the solution to the nearest thousandth. Solution Take the natural logarithm on each side. Power property Distributive property

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 2 SOLVING AN EXPONENTIAL EQUATION Solve 3 2x – 1 = 0.4 x + 2. Give the solution to the nearest thousandth. Solution Write the terms with x on one side Factor out x. Divide by 2 In 3 – In 0.4. Power property

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 2 SOLVING AN EXPONENTIAL EQUATION Solution Apply the exponents. Product and quotient properties. This is approximate. This is exact. The solution set is { –0.236}. Solve 3 2x – 1 = 0.4 x + 2. Give the solution to the nearest thousandth.

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Example 3 SOLVING BASE e EXPONENTIAL EQUATIONS Solve each equation. Give solutions to the nearest thousandth. Solution (a) Take the natural logarithm on each side. In = x 2

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Example 3 SOLVING BASE e EXPONENTIAL EQUATIONS Solution Square root property Remember both roots. Use a calculator. The solution set is {  2.302}. Solve each equation. Give solutions to the nearest thousandth. (a)

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 3 SOLVING BASE e EXPONENTIAL EQUATIONS Solution (b) Take the natural logarithm on each side. Power property Solve each equation. Give solutions to the nearest thousandth. Divide by e;

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 3 SOLVING BASE e EXPONENTIAL EQUATIONS Solution In e = 1 Multiply by – ½. The solution set is {– 0.549}. Solve each equation. Give solutions to the nearest thousandth. (b) Use a calculator.

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 4 SOLVING AN EXPONENTIAL EQUATION QUADRATIC IN FORM Solve e 2x – 4 e x + 3 = 0. Give exact value(s) for x. Solution a mn = (a n ) m Zero-factor property This is an equation that is quadratic in form, because it can be rewritten with u = e x. Let u = e x. Factor.

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 4 SOLVING AN EXPONENTIAL EQUATION QUADRATIC IN FORM Solve e 2x – 4 e x + 3 = 0. Give exact value(s) for x. Solution Take the natural logarithm on each side. Substitute e x for u. ln e x = x Solve for u. Both values check, so the solution set is {0,ln 3}.

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 5 SOLVING LOGARITHMIC EQUATIONS Solve each equation. Give exact values. Solution Divide by 7. Write the natural logarithm in exponential form. (a) 7ln x = 28 The solution set is { e 4 }.

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Example 5 SOLVING LOGARITHMIC EQUATIONS Solve each equation. Give exact values. Solution Write in exponential form. Add 19. (b) log 2 (x 3 – 19) = 3 The solution set is {3}. Apply the exponent. Take cube roots.

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 6 SOLVING A LOGARITHMIC EQUATION Solve log(x + 6) – log(x + 2) = log x. Give exact value(s). Solution Quotient property Property of logarithms Keep in mind that logarithms are defined only for nonnegative numbers. Multiply by x + 2.

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Example 6 SOLVING A LOGARITHMIC EQUATION Solution Distributive property Standard form Factor. Zero-factor property The proposed negative solution (– 3) is not in the domain of log x in the original equation, so the only valid solution is the positive number 2. The solution set is {2}. Solve log(x + 6) – log(x + 2) = log x. Give exact value(s). Solve for x.

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Caution Recall that the domain of y = log a x is (0,  ). For this reason, it is always necessary to check that proposed solutions of a logarithmic equation result in logarithms of positive numbers in the original equation.

21. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 7 SOLVING A LOGARITHMIC EQUATION Solve log 2 [(3x – 7)(x – 4)] = 3. Give exact value(s). Solution Write in exponential form. Multiply. Standard form Factor.

22. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 7 SOLVING A LOGARITHMIC EQUATION Solve log 2 [(3x – 7)(x – 4) = 3. Give exact value(s). Solution Zero-factor property Solve for x. A check is necessary to be sure that the argument of the logarithm in the given equation is positive. In both cases, the product (3x – 7)(x – 4) leads to 8, and log 2 8 = 3 is true. The solution set is {4/3, 5}.

23. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 8 SOLVING A LOGARITHMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give exact value(s). Solution Substitute. Product property Property of logarithms The notation log x is an abbreviation for log 10 x, and 1 = log 10 10.

24. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 8 SOLVING A LOGARITMIC EQUATION Multiply. Subtract 10. Quadratic formula Solve log(3x + 2) + log(x – 1 ) = 1. Give exact value(s). Solution

25. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 8 SOLVING A LOGARITMIC EQUATION The first of these proposed solutions, is negative and when substituted for x in log(x – 1) results in a negative argument, which is not allowed. Therefore, this solution must be rejected. Solve log(3x + 2) + log(x – 1 ) = 1. Give exact value(s). Solution The two proposed solutions are and

26. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 8 SOLVING A LOGARITMIC EQUATION The second proposed solution, is positive. Substituting it for x in log(3x + 2) results in a positive argument, and substituting it for x in log(x + 1) also results in a positive argument, both of which are necessary conditions. Therefore, the solution set is Solve log(3x + 2) + log(x – 1 ) = 1. Give exact value(s). Solution The two proposed solutions are and

27. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Note We could have used the definition of logarithm in Example 8 by first writing Product property Definition of logarithm and then continuing as shown in the solution. Equation from Example 8

28. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 9 SOLVING A BASE e LOGARITHMIC EQUATION Solve In e In x – In(x – 3) = In 2. Give exact value(s). Solution Quotient property Property of logarithms

29. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 9 SOLVING A BASE e LOGARITHMIC EQUATION Solution Distributive property Solve for x. Check that the solution set is {6}. Solve In e In x – In(x – 3) = In 2. Give exact value(s). Multiply by x – 3.

30. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Solving Exponential or Logarithmic Equations To solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines. 1. a  (x) = b Solve by taking logarithms on both sides. 2. log a  (x) = b Solve by changing to exponential form a b =  (x). 3. log a  (x) = log a g(x) The given equation is equivalent to the equation  (x) = g(x). Solve algebraically.

31. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Solving Exponential or Logarithmic Equations 4. In a more complicated equation, such as in Example 3(b), it may be necessary to first solve for a  (x) or log a  (x) and then solve the resulting equation using one of the methods given above. 5. Check that each proposed solution is in the domain.

32. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Example 10 APPLYING AN EXPONENTIAL EQUATION TO THE STRENGTH OF A HABIT The strength of a habit is a function of the number of times the habit is repeated. If N is the number of repetitions and H is the strength of the habit, then, according to psychologist C. L. Hull, where k is a constant. Solve this equation for k.

33. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 10 APPLYING AN EXPONENTIAL EQUATION TO THE STRENGTH OF A HABIT Solution First solve the equation for e – kN. Divide by 1000. Subtract 1. Multiply by – 1 and rewrite

34. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Example 10 APPLYING AN EXPONENTIAL EQUATION TO THE STRENGTH OF A HABIT Solution Now solve for k. Take the natural logarithm on each side. In e x = x Multiply by.

35. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 10 APPLYING AN EXPONENTIAL EQUATION TO THE STRENGTH OF A HABIT Solution With the final equation, if one pair of values for H and N is known, k can be found, and the equation can then be used to find either H or N for given values of the other variable.

36. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Example 11 MODELING COAL CONSUMPTION IN THE U.S. Year Coal Consumption (in quads) 198015.42 198517.48 199019.17 199520.09 200022.58 200522.80 200822.39 The table gives U.S. coal consumption (in quadrillions of British thermal units, or quads) for several years. The data can be modeled by the function where t is the number of years after 1900, and  (t) is in quads.

37. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Example 11 MODELING COAL CONSUMPTION IN THE U.S. Solution (a) Approximately what amount of coal was consumed in the United States in 2003? How does this figure compare to the actual figure of 22.32 quads? The year 2003 is represented by t = 2003 – 1900 = 103. Use a calculator. Based on this model, 22.19 quads were used in 2003. This figure is very close to the actual amount of 22.32 quads. Let t = 103.

38. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 11 MODELING COAL CONSUMPTION IN THE U.S. Solution (b) Add 93.31. Replace  (t) with 25, and solve for t. Divide by 24.92. Rewrite. Write in exponential form. If this trend continues, approximately when will annual consumption reach 25 quads? f(t) = 25 in the given model.

39. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 11 MODELING COAL CONSUMPTION IN THE U.S. Use a calculator. Add 115 to 1900 to get 2015. Based on this model, annual consumption will reach 25 quads in approximately 2015. Solution (b) If this trend continues, approximately when will annual consumption reach 25 quads?