Exploring Exponential Models ALGEBRA 2 LESSON 8-1 (For help, go to Lesson 1-2.) Evaluate each expression for the given value of x. 1. 2x for x = 3 2. 4x+1 for x = 1 3. 23x+4 for x = –1 4. 3x3x–2 for x = 2 5. for x = 0 6. 2x for x = –2 1 2 x 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Solutions 1. 2x for x = 3: 23 = 2 • 2 • 2 = 8 2. 4x+1 for x = 1: 41+1 = 42 = 4 • 4 = 16 3. 43x+4 for x = –1: 23(–1)+4 = 2–3+4 = 21 = 2 4. 3x3x–2 for x = 2: 32 • 32 – 2 = 32 • 30 = 3 • 3 • 1 = 9 5. for x = 0: = 1 6. 2x for x = –2: 2–2 = = = 1 2 1 2 x 1 22 1 2 • 2 1 4 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Graph y = 3x. x 3x y –3 3–3 = .037 –2 3–2 = .1 –1 3–1 = .3 0 30 1 1 31 3 2 32 9 3 33 27 1 27 9 3 Step 1: Make a table of values. Step 2: Graph the coordinates. Connect the points with a smooth curve. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 The population of the United States in 1994 was almost 260 million with an average annual rate of increase of about 0.7%. a. Find the growth factor for that year. b = 1 + r = 1 + 0.007 Substitute 0.7%, or 0.007 for r. = 1.007 Simplify. b. Suppose the rate of growth had continued to be 0.7%. Write a function to model this population growth. Relate: The population increases exponentially, so y = abx Define: Let x = number of years after 1994. Let y = the population (in millions). 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 (continued) Write: y = a(1.007)x 260 = a(1.007)0 To find a, substitute the 1994 values: y = 260, x = 0. 260 = a • 1 Any number to the zero power equals 1. 260 = a Simplify. y = 260(1.007)x Substitute a and b into y = abx. y = 260(1.007)x models U.S. population growth. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Write an exponential function y = abx for a graph that includes (1, 6) and (0, 2). y = abx Use the general term. 6 = ab1 Substitute for x and y using (1, 6). = a Solve for a. 6 b 2 = b0 Substitute for x and y using (0, 2) and for a using . 6 b 2 = • 1 Any number to the zero power equals 1. 2 = Simplify. b = 3 Solve for b. 6 b 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 (continued) a = Use your equation for a. 6 b a = Substitute 3 for b. a = 2 Simplify. y = 2 • 3x Substitute 2 for a and 3 for b in y = abx. 6 3 The exponential for a graph that includes (1, 6) and (0, 2) is y = 2 • 3x. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Without graphing, determine whether the function y = 3 represents exponential growth or decay. 2 3 x In y = , b = . 3 2 x Since b < 1, the function represents exponential decay. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Graph y = 36(0.5)x. Identify the horizontal asymptote. Step 1: Make a table of values. x –3 –2 –1 0 1 2 3 y 288 144 72 36 18 9 4 1 2 Step 2: Graph the coordinates. Connect the points with a smooth curve. As x increases, y approaches 0. The horizontal asymptote is the x-axis, y = 0. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Suppose you want to buy a used car that costs $11,800. The expected depreciation of the car is 20% per year. Estimate the depreciated value of the car after 6 years. The decay factor b = 1 + r, where r is the annual rate of change. b = 1 + r Use r to find b. = 1 + (–0.20) = 0.80 Simplify. Write an equation, and then evaluate it for x = 6. Define: Let x = number of years. Let y = value of the car. Relate:  The value of the car decreases exponentially; b = 0.8. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 (continued) Write: y = ab x     11,800 = a(0.8)0 Substitute using (0, 11,800).      11,800 = a Solve for a. y = 11,800(0.8)x Substitute a and b into y = abx. y = 11,800(0.8)6   Evaluate for x = 6.          3090 Simplify. The car’s depreciated value after 6 years will be about $3,090. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 1. 2. 3. 4. 6. 7. 5. 8. 9. a. 1.0126 b. y = 6.08(1.0126)x, where x = 0 corresponds to 2000 pages 426–429 Exercises 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 10. y = 0.5(2)x 11. y = 2.5(7)x 12. y = 8(1.5)x 13. y = 5(0.6)x 14. y = 3(0.5)x 15. y = 24 16. exponential growth 17. exponential decay 18. exponential growth 19. exponential decay 20. exponential decay 21. exponential growth 22. exponential growth 23. exponential decay 1 3 24–31. y = 0 is the horizontal asymptote. 24. 25. 26. 27. 28. 29. x 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 44. y = 30,000(0.7)x for car 1; y =15,000(0.8)x for car 2; car 2 will be worth more. 45. a. y = 80(0.965)x b. 56 animals c. about 47 years 46. 1.70 47. 6 48. 0.25 49. 0.45 30. 31. 32. y = 100(0.5)x; 1.5625 33. y = 12,000(0.9)x; 6377 34. y = 12,000(0.1)x; 0.012 35. a. y = 6500(0.857)x b. about $4091.25 36. a. Tokyo: 26,444,000, Mexico City: 18,850,649, Bombay: 23,148,579, São Paulo: 19,496,367 b. yes; Tokyo, Bombay, São Paulo, Mexico City 37. 63% increase 38. 30% increase 39. 35% decrease 40. 70% increase 41. 87.5% decrease 42. 75% decrease 43. a. 5.6% b. 0.0017% 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 60. C 61. H 62. B 63. [2] [1] general form of graph correct, but not exact 59. a. A negative growth rate would be represented by adding the negative rate to 1. b. Armenia: y = 9.2(1.06)x, Canada: y = 688.3(1.03)x, Oman: y = 18.6(.915)x, Paraguay: y = 19.8(.995)x, x in each equation represents the number of years since 1998. c. Armenia: $13.8 billion, Canada: $846.5 billion, Oman: $10.0 billion, Paraguay: $19.1 billion 50. 1.125 51. 0.999 52. 1.001 53. 2 54. y = 34(1.26)x, where x represents the number of years since 1995. 55. Check students’ work. 56. about $42,140 57. C 58. B; the graph shows a decreasing function, so b < 1, which eliminates A. The graph is always positive, which eliminates C. 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 64. [4] y = abx 54 = ab2 a = 2 =  b 2 = 2b = 54 b = 27 b = 27 b = 9 y =  9x [3] appropriate methods with one computational error [2] starts problem correctly, but fails to finish it properly [1] correct answer, without work shown 65. 66. 67. 68. 6n2 5n 69. 84r 2 9r 2 70. 71. Answers may vary. Sample: y = x3 – 5x2 + 4x 72. Answers may vary. y = x3 – 7x – 6 54 b2 54 b2 1 2 54 b 3 2 3 3 2 3 2 2 x 3x 2 3 54 92 2 3 2 3 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 73. Answers may vary. Sample: y = x3 – 7x2 + 10x 74. y = x2 + 2 75. y = –5x2 + 2 76. y = 2x2 + 2 77. y = 10x2 + 2 78. y = x2 + 2 79. y = 3x2 + 2 80. y = – x2 + 2 81. y = –x2 + 2 3 2 1 8-1

Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Sketch the graph of each function. Identify the horizontal asymptote. 1. y = (0.8)x 2. y = Without graphing, determine whether each equation represents exponential growth or decay. 3. y = 15(7)x 4. y = 1285(0.5)x Write an exponential function for a graph that includes the given points. 5. (0, 0.5), (1, 3) 6. (–1, 5), (0.5, 40) 1 4 x y = 0 y = 0 growth decay y = 0.5(6)x y = 20(4)x 8-1

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 (For help, go to Lesson 2-6, 5-3, and 7-4.) Write an equation for each translation. 1. y = | x |, 1 unit up, 2 units left 2. y = –| x |, 2 units down 3. y = x2, 2 units down, 1 unit right 4. y = –x2, 3 units up, 1 unit left Write each equation in simplest form. Assume that all variables are positive. 5. y = 6. y = 7. y = 8. Use the formula for simple interest I = Prt. Find the interest for a principal of $550 at a rate of 3% for 2 years. 5 4 – x 1 7 6 –7 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 1. y = | x |, 1 unit up and 2 units left: y = | x + 2 | + 1 2. y = –| x |, 2 units down: y = –| x | – 2 3. y = x2, 2 units down and 1 unit right: y = (x – 1)2 – 2 4. y = –x2, 3 units up and 1 unit left: y = –(x + 1)2 + 3 5. y = x ; y = x (4); y = x–5 or y = 6. y = x ; y = x (–7); y = x1; y = x 7. y = x ; y = x (6); y = x5 8. I = Prt = $550(0.03)(2) = $16.50(2) = $33 5 4 – 1 x5 7 6 Solutions –7 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Graph y = 3 • 2x and y = –3 • 2x. Label the asymptote of each graph. x y = 3 · 2x y = –3 · 2x –3 –2 –1 0 3 –3 1 6 –6 2 12 –12 3 24 –24 3 8 4 1 2 – Step 1: Make a table of values Step 2: Graph each function. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Graph y = 6 and y = 6 – 2. 1 2 x 1 2 x–3 Step 1: Graph y = 6 . The horizontal asymptote is y = 0. 1 2 x So shift the graph of the present function 3 units right and 2 units down. Step 2: For y = 6 – 2, h = 3 and k = –2. 1 2 x–3 The horizontal asymptote is y = –2. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Since a 100-mg supply of technetium-99m has a half-life of 6 hours, find the amount of technetium-99m that remains from a 50-mg supply after 25 hours. Relate: The amount of technetium-99m is an exponential function of the number of half-lives. The initial amount is 50 mg. The decay factor is . One half-life equals 6 h. 1 2 1 6 Define: Let y = the amount of technetium-99m. Let x = the number of hours elapsed. Then x = the number of half-lives. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 (continued) Write: y = 50 y = 50 Substitute 25 for x. 1 6 • 25 2 x = 50 Simplify. 2.784 1 2 4.16 After 25 hours, about 2.784 mg of technetium-99m remains. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Graph y = ex. Evaluate e3 to four decimal places. Step 1: Graph y = ex. Step 2: Find y when x = 3. The value of e3 is about 20.0855. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Suppose you invest $100 at an annual interest rate of 4.8% compounded continuously. How much will you have in the account after 3 years? A = Pert = 100 • e0.048(3) Substitute 100 for P, 0.048 for r, and 3 for t. = 100 • e0.144 Simplify. = 100 • (1.154884) Evaluate e0.144. 115.49 Simplify. You will have about $115.49 in the account after three years. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 pages 434–436  Exercises 1–8. Asymptote is y = 0. 1. 2. 6. 7. 8. 3. 4. 5. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 9. 10. 11. 12. 13. 14. 15. y = 50 ; 0.85 mg 16. y = 200 ; 0.43 mg 17. y = 24 ; 0.64 mg 18. 20.0855 19. 403.4288 20. 0.1353 21. 1 22. 12.1825 23. 15.1543 24. $2330.65 25. $448.30 26. $1819.76 27. 0 1 2 14.3 8.14 5730 x 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 28. 1 29. If c < 0, the graph models exponential decay. If c = 0, the graph is a horizontal line. If c > 0, the graph models exponential growth. 30. $6168.41 31. a. Answers may vary. Sample: y = –2(1.3)x b. Answers may vary. Sample: I am in debt for $2 and my debt is growing at a rate of 30% per year. c. The graph of exponential decay approaches the asymptote y = 0 as x increases. The graph of negative exponential growth approaches the asymptote y = 0 as x decreases. 32. y = 4 ; y = 4 + 3 33. y = –3x; y = –3x–8 + 2 1 2 34. y = (2)x; y = (2)x–6 – 7 35. y = –3 ; y = –3 – 1 36. 75.0 pascals 37. 8.7 yr 38. A deficit that is growing exponentially is modeled by y = abcx, where a < 0, and either b > 1 and c > 0 or 0 < b < 1 and c < 0. 3 x x + 4 x + 5 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 39. a. GDP = 8.511(1.038)t where t = 0 corresponds to 1998 and 8.511 is trillions. b. It almost doubles. (about 195.7%) c. In 9 years, the growth is about 40%. 40. a. $2501.50 b. $3.15 more 41. $399.97 42. exponential growth 43. exponential growth 44. exponential decay 45. exponential growth 46. exponential decay 47. exponential growth 48. a. y = 8001 – 3x, where y is the number of uninfected people and x represents days. b. 5814 people c. about 9 days 49. a. about 10 names; about 24 names b. Graphically, it will never happen; the graph has y = 30 as an asymptote. (In reality, you would be close to knowing all the names in about 21 days.) c. Answers may vary. Sample: I learn names pretty quickly; my learning rate might be 0.4. 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 71. x2 – x – 6 72. x2 – 1 73. x2 + x + 1 74. 13x + 1 75. 3x2 – 7x + 2 76. x – y = 5, x + z = 5, –y + z = 5 77. x + y = –2, x + 4z = –2, y + 4z = –2 78. x + y = 8, x – 2z = 8, y – 2z = 8 58. y = –5(2)x 59. y = 8(0.5)x 60. y = 70(10)x 61. y = 10,000(0.8)x 62. 6 5 63. – 64. 5( ) 65. 2( ) 66. 3 67. 11 7 68. 15 – 6 6 69. 3 + 5 70. x – 3, R –1 3 4 3 + 5 2 + 8 50. a. 2928 m3 b. V = 2928 – 15(2x – 1) c. ninth weekend 51. A 52. I 53. C 54. F 55. [2] A = Pert 8000 = Pe(0.06)(4) = P P = $6293.02 [1] answer only, without work shown 56. y = 3x 57. y = –2(4)x 8000 e(0.06)(4) 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 79. x – y = 8, x + 2z = 8, –y + 2z = 8 80. 3x + y = –18, x + 3z = –6, y + 9z = –18 81. –2x + y = 10, –2x – 5z = 10, y – 5z = 10 8-2

Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Describe how the graph of each function relates to its parent function. Then graph the function. 1. y = –5x – 1 2. y = 3(5)x – 2 + 1 reflected over the x-axis and translated down 1 rises more steeply, is translated 2 to the right and up 1 3. Use the formula A = Pert to find the amount in a continuously compounded account where the principal is $2000 at an annual interest rate of 5% for 3 years. 4. Use the graph of y = ex to evaluate e to four decimal places. about $2324 1 3 1.3956 8-2

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 (For help, go to Lessons 7-1 and 7-7.) Solve each equation. 1. 8 = x3 2. x = 2 3. 27 = 3x 4. 46 = 43x Graph each relation and its inverse on a coordinate plane. 5. y = 5x 6. y = 2x2 7. y = –x3 8. y = x 1 4 2 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 2. x = 2; since 16 = = 2, x = 16 6. y = 2x2 Inverse: x = 2y2 y2 = y2 = ± 8. y = x x = y y = 2x x 2 1. 8 = x3; since 23 = 8, x = 2 3. 27 = 3x; since 33 = 27, x = 3 5. y = 5x x = 5y y = x 7. y = –x3 x = –y3 y3 = –x y = 1 5 3 –x 4 16 4. 46 = 43x; since 6 = 3(2), x = 2 Solutions 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Compare the amount of energy released in an earthquake that registers 6 on the Richter scale with one that registers 3. Write a ratio. E • 306 E • 303 = Simplify. 306 303 = 306–3 Division Property of Exponents = 303 Simplify. = 27,000 Use a calculator. The first earthquake released about 27,000 times as much energy as the second. 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Write: 32 = 25 in logarithmic form. If y = bx, then logb y = x. Write the definition. If 32 = 25, then log2 32 = 5. Substitute. The logarithmic form of 32 = 25 is log2 32 = 5. 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Evaluate log3 81. Let log3 81 = x. Log3 81 = x Write in logarithmic form. 81 = 3x Convert to exponential form. 34 = 3x Write each side using base 3. 4 = x Set the exponents equal to each other. So log3 81 = 4. 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 The pH of an apple is about 3.3 and that of a banana is about 5.2. Recall that the pH of a substance equals –log[H+], where [H+] is the concentration of hydrogen ions in each fruit. Which is more acidic? Apple pH = –log[H+] 3.3 = –log[H+] log[H+] = –3.3 [H+] = 10–3.3 5.0  10– 4 [H+] = 10–5.2 pH = –log[H+] 5.2 = –log[H+] log[H+] = –5.2 Banana 6.3  10– 6 The [H+] of the apple is about 5.0  10– 4. The [H+] of the banana is about 6.3  10– 6. The apple has a higher concentration of hydrogen ions, so it is more acidic. 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Graph y = log4 x. By definition of logarithm, y = log4 x is the inverse of y = 4x. Step 1: Graph y = 4x. Step 2: Draw y = x. Step 3: Choose a few points on 4x. Reverse the coordinates and plot the points of y = log4 x. 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Graph y = log5 (x – 1) + 2. Step 1: Make a table of values for the parent function. Step 2: Graph the function by shifting the points from the table to the right 1 unit and up 2 units. 1 125 25 5 –3 –2 –1 x y = log5 x 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 pages 441–444  Exercises 1. The earthquake in Missouri released about 1.97 times more energy. 2. The earthquake in Chile released about 231 times 3. The earthquake in Missouri released about 8,759,310 times more energy. 4. The earthquake in Missouri released about 30 times 5. The earthquake in Alaska released about 83 times more energy. 6. log7 49 = 2 7. 3 = log 1000 8. log5 625 = 4 9. log = –1 10. 2 = log8 64 11. log 4 = –2 12. 3 = log ( ) 1 10 2 3 27 13. –2 = log 0.01 14. 4 15. 16. 1 17. 18. 3 19. 20. undefined 21. 2 22. 5 23. 1 24. 4 25. 3 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 26. 6.3  10–6 27. 6.3  10–3 28. 1.0  10–8 29. 7.9  10–4 30. 5.0  10–7 31. 1.3  10–5 32. 1.0  10–4 33. 4.0  10–6 34. 2.5  10–4 35. 36. 37. 38. 39. 40. 41. 0.6990; 0 42. –4.2147; –5 43. –1.0969; –2 44. 2.3010; 2 45. –0.7782; –1 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 46. 1.2435; 1 47. 7.1139; 7 48. 0.5119; 0 49. apple juice: 3.5, acidic; buttermilk: 4.6, acidic; cream: 6.6, acidic; ketchup: 3.9, acidic; shrimp sauce: 7.1, basic; strained peas: 6, acidic 50. The error is in the second line. It should read 3 = 27x; the correct answer is . 1 3 51. First rewrite y = log1 x as 1y = x. For any real number y, x = 1. 52. Answers may vary. Sample: y = log3 x; y = 3x 53. 128 = 27 54. 0.0001 = 10–4 55. 16,807 = 75 56. 6 = 61 57. 1 = 40 58. = 3–2 59. = 2–1 60. 10 = 101 61. 8192 = 213 9 2 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 62. a. buffalo bone: 9826 to 10,128 years old, bone fragment: 9776 to 10,180 years old, pottery shard: 0 to 183 years old, charcoal: 9718 to 10,242 years old, spear shaft: 9776 to 10,263 years old b. The pottery shard; answers may vary. Samples: the pottery may be from a later civilization, or the mass or the beta radiation emissions may have been measured incorrectly. 63. y = 4x 64. y = 0.5x 65. y = 10x 66. y = 2x–1 67. y = 10x – 1 68. y = 10x–1 69. y = 10x + 2 70. y = 5 71. y = 3 72. x 2 1 73. 74. 75. domain {x|x > 0}, range: all reals 76. domain {x|x > 0}, range: all reals 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 77. domain {x|x > 3}, range: all reals 78. domain {x|x > 0}, range: all reals 79. domain {x|x > 2}, range: all reals 80. domain {x|x > –1}, range: all reals 81. domain {x|x > 0}, range: all reals 82. domain {x|x > 0}, range: all reals 83. domain {x|x > t }, range: all reals 84. 100 85. 70 86. 60 87. 20 88. 10 89. a. III b. I c. IV d. II 90. a. II b. III c. I 91. D 92. I 93. C 94. A 95. B 96. C 97. y = –100 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 w3 3 t 2 5 z8 n x p 1 2 100. 101. 4 102. 103. 104. –2, –4 105. –0.162, 6.162 106. –1.217, 0, 8.217 107. (2x – 3)(2x – 1) 108. ( b – 2)( b + 2) 109. (5x – 2)(x + 3) 98. y = 0 99. y = 9 8-3

Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Evaluate each logarithm. 1. log232 2. log10(–100) 3. log5 1 25 5 undefined –2 Write each equation in exponential form. 4. log 0.01 = –2 5. log327 = 3 6. Graph y = log6(x – 3) + 1. 10–2 = 0.01 33 = 27 8-3

Properties of Logarithms ALGEBRA 2 LESSON 8-4 (For help, go to Lessons 8-3 and 1-2.) Simplify each expression. 1. log24 + log28 2. log39 – log327 3. log216 ÷ log264 Evaluate each expression for x = 3. 4. x3 – x 5. x5 • x2 6. 7. x3 + x2 x6 x9 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 1. log24 = x     log28 = y 2x = 4 2y = 8 x = 2 y = 3 log24 + log28 = 2 + 3 = 5 2. log39 = x     log327 = y 3x = 9 3y = 27 log39 – log327 = 2 – 3 = –1 3. log216 = x     log264 = y 2x = 16 2y = 64 x = 4 y = 6 log216 ÷ log264 = 4 ÷ 6 = = 4 6 2 3 4. x3 – x for x = 3: 33 – 3 = 27 – 3 = 24 5. x5 • x2 for x = 3: 35 • 32 = 3(5+2) = 37 = 2187 6. for x = 3: = 3(6–9) = 3–3 = = 7. x3 + x2 for x = 3: 33 + 32 = 27 + 9 = 36 x6 x9 36 99 1 33 27 Solutions 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 State the property or properties used to rewrite each expression. a. log 6 = log 2 + log 3 Product Property: log 6 = log (2•3) = log 2 + log 3 b. logb = 2 logb x – logb y x2 y Quotient Property: logb = logb x2 – logb y x2 y Power Property: logb x2 – logb y = 2 logb x – logb y 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 Write each logarithmic expression as a single logarithm. a. log4 64 – log4 16 log4 64 – log4 16 = log4 Quotient Property 64 16 = log4 4 or 1 Simplify. b. 6 log5 x + log5 y 6 log5 x + log5 y = log5 x6 + log5 y Power Property = log5 (x6y) Product Property So log4 64 – log4 16 = log4 4, and 6 log5 x + log2 y = log5 (x6y). 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 Expand each logarithm. a. log7 t u log7 = log7 t – log7 u Quotient Property t u b. log(4p3) log(4p3) = log 4 + log p3 Product Property = log 4 + 3 log p Power Property 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 Manufacturers of a vacuum cleaner want to reduce its sound intensity to 40% of the original intensity. By how many decibels would the loudness be reduced? Relate: The reduced intensity is 40% of the present intensity. Define: Let l1 = present intensity. Let l2 = reduced intensity. Let L1 = present loudness. Let L2 = reduced loudness. Write: l2 = 0.04 l1 L1 = 10 log L2 = 10 log l1 l0 l2 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 (continued) L1 – L2 = 10 log l1 l0 l2 – 10 log Find the decrease in loudness L1 – L2. = 10 log l1 l0 0.40l1 – 10 log Substitute l2 = 0.40l1. – 10 log 0.40 • Product Property = 10 log l1 l0 – 10 ( log 0.40 + log ) = 10 log l1 l0 – 10 log 0.40 – 10 log Distributive Property = –10 log 0.40 Combine like terms. 4.0 Use a calculator. The decrease in loudness would be about 4 decibels. 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 pages 449–451  Exercises 1. Product Property 2. Quotient Property 3. Power Property 4. Power Property 5. Power Property, Quotient Property 6. Power Property 7. Power Property, Quotient Property 8. Power Property, Product Property 9. Power Property, Quotient Property 10. Power Property, Product Property 11. log 14 12. log2 3 13. log 972 14. log 15. log 16. log 17. log6 5x 18. log7 19. 3 log x + 5 log y 2 3 m4 n 5 2k xy z 20. log7 22 + log7 x + log7 y + log7 z 21. log4 5 + log4 x 22. log 3 + 4 log m – 2 log n 23. log5 r – log5 s 24. 2 log3 2 + 2 log3 x 25. log3 7 + 2 log (2x – 3) 26. 2 log a + 3 log b – 4 log c 27. log 2 + log x – log y 28. 1 + log8 3 + log8 a 29. log s + log 7 – 2 log t 1 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 t s 30. –logb x 31. 9 dB 32. 13 dB 33. –2 34. 1 35. 6 36. 2 37. 2 38. 1 39. 1 40. –2 41. 1 1 2 51. –1.398 52. 1.398 53. –0.7782 54. 1.5564 55. 0.3495 56. 12 dB 57. 0.00001 58. True; log2 4 = 2 and log2 8 = 3. 59. False; log3 3 = log3 3 , not log3 . 3 42. The coefficient is missing in log4 s; log 4 = log4 = (log4 t – log4 s) = log4 t – log4 s. 43. Answers may vary. Sample: log 150 = log 15 + log 10 44. 1.3803 45. 1.4772 46. 1.2042 47. 2.097 48. 0.1761 49. –0.0969 50. –0.6021 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 60. True; it is an example of the Power Property since 8 = 23. 61. False; the two logs have different bases. 62. False; this is not an example of the Quotient Property. log (x – 2) log x – log 2. 63. False; logb = logb x – logb y. 64. False; the exponent on the left means the log x, quantity squared, not the log of x2. x y 65. False; log4 7 – log4 3 = log4 , not log4 4. 66. True; log x + log (x2 + 2) = log x(x2 + 2), which equals log (x3 + 2x). 67. False; the three logs have different bases. 68. True; the power and quotient properties are used correctly. 69. True; the left side equals logb ( • 43), which equals logb 8. 70. 102 dB 7 3 1 8 71. No; the expression (2x + 1) is a sum, so it is not covered by the Product, Quotient, or Power properties. 72. The log of a product is equal to the sum of the logs. log (MN) = log M + log N. So log (5 • 2) = log 10 = 1, log 5 • log 2 (0.7)(0.3) = 0.21, which is not equal to 1. = / 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 81. log 2 + log 4 + log r – log s 82. logb x + logb y – logb z 83. log4 x + log4 y – log4 z – 4 log4 w 84. log (x2 – 4) – 2 log (x + 3) 85. log x + log 2 – log y 1 2 3 5 7 4 4 2x 2 y z3 27 2 mxn p x2 y3 z5 3 1 y z 3 x5 73. log3 74. logx 75. log 76. log4 77. logb 78. log 79. 3 log 2 + log x – 3 log 5 80. 3 log m – 4 log n + 2 log p 86. log3 x + log3 y – 6 log3 z 87. log7 (r + 9) – 2 log7 s – log7 t 88. v = logb N bv = N MN = bu • bv = bu+v logb MN = u + v logb MN = logb M + logb N 1 2 3 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 89. 1. u = logb M (given) 2. bu = M (Rewrite in exponential form.) 3. (bu)x = Mx (Raise each side to x power.) 4. bux = Mx (Power Property of exponents) 5. logb bux = logb Mx (Take the log of each side.) 6. ux = logb Mx (Simplify.) 7. logb Mx = x • logb M (substitution) 90. 1. u = logb M (given) 2. bu = M (Rewrite in exponential form.) 3. v = logb N (given) 4. bv = N (Rewrite in exponential form.) 5. = = bu–v (Quotient Property of Exponents) 6. logb = logb bu–v (Take the log of each side.) 7. logb = u – v (Simplify.) 8. logb = logb M – logb N (substitution) M N bu bv 91. B 92. G 93. D 94. [2] By the Quotient Property, log5 = log5 1.4307 – 1.8614 = –0.4307. [1] correct answer, without work shown 1 2 10 20 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 101. 102. 2 103. – 104. i, –4i 105. –2i, –4 – i 106. – 2,– i – 1 107. 108. 2i + 3, –i 64 7 3, 5 7, 11 95. [4] 1) log = log 24 – log 2; Quotient Property 2) log 2 • 6 = log 2 + log 6; Product Property 3) log 144 = log 144; Power Property 4) log 3 • 22 = log 3 + 2 log 2; Product and Power Properties [3] log 12 written three ways 24 2 1 [2] log 12 written two ways OR written 4 ways but properties not named [1] log 12 written only 2 ways and properties not named 96. log7 49 = 2 97. 3 = log5 125 98. log8 = – 99. –3 = log5 100. 8, –8 1 4 2 3 125 8-4

Properties of Logarithms ALGEBRA 2 LESSON 8-4 Write each expression as a single logarithm. State the property you used. 1. log 12 – log 3 2. 3 log115 + log117 Expand each logarithm. 3. logc 4. log3x4 Use the properties of logarithms to evaluate each expression. 5. log 0.001 + log 100 6. logyy log 4; Quotient Property log11(53 • 7); Power Property and Product Property a b 4 log3x logca – logcb 1 2 –1 1 2 8-4

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 (For help, go to Lessons 8-3 and 7-4.) Evaluate each logarithm. 1. log981 • log93 2. log10 • log39 3. log216 ÷ log28 4. Simplify 125 . 2 3 – 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 4. 125 = = = = = 1 3 (125 ) 2 1. log981 = x     log93 = y 9x = 81 9y = 3 x = 2 y = log981 • log93 = 2 • = 1 3. log216 = x     log28 = y 2x = 16 2y = 8 x = 4 y = 3 log216 ÷ log28 = 4 ÷ 3 = or 1 4 2. log10 = x     log39 = y 10x = 10 3y = 9 x = 1 y = 2 log10 • log39 = 1 • 2 = 2 125 52 25 ( ) Solutions – 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve log 52x = 16. 52x = 16 52x = log 16 Take the common logarithm of each side. 2x log 5 = log 16 Use the power property of logarithms. x = Divide each side by 2 log 5. log 16 2 log 5 0.8614 Use a calculator. Check: 52x 16 52(0.8614) 16 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Use the Change of Base Formula to evaluate log6 12. Then convert log6 12 to a logarithm in base 3. log6 12 = Use the Change of Base Formula. log 12 log 6 1.387 Use a calculator. 1.0792 0.7782 log6 12 = log3 x Write an equation. 1.387 log3x Substitute log6 12 = 1.3868 1.387 Use the Change of Base Formula. log x log 3 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 (continued) 1.387 • log 3 log x Multiply each side by log 3. 1.387 • 0.4771 log x Use a calculator. 0.6617 log x Simplify. x 100.6617 Write in exponential form. 4.589 Use a calculator. The expression log6 12 is approximately equal to 1.3869, or log3 4.589. 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 52x = 120. 52x = 120 log5 52x = log5 120 Take the base-5 logarithm of each side. 2x = log5 120 Simplify. 2x = Use the Change of Base Formula. log 120 log 5 x 1.487 Use a calculator to solve for x. 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 43x = 1100 by graphing. Graph the equations y = 43x and y = 1100. Find the point of intersection. The solution is x 1.684 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 The population of trout in a certain stretch of the Platte River is shown for five consecutive years in the table, where 0 represents the year 1997. If the decay rate remains constant, in the beginning of which year might at most 100 trout remain in this stretch of river? Time t 0 1 2 3 4 Pop. P(t) 5000 4000 3201 2561 2049 Step 1: Enter the data into your calculator. Step 2: Use the Exp Reg feature to find the exponential function that fits the data. 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 (continued) Step 3: Graph the function and the line y = 100. Step 4: Find the point of intersection. The solution is x 18, so there may be only 100 trout remaining in the beginning of the year 2015. 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve log (2x – 2) = 4. log (2x – 2) = 4 2x – 2 = 104 Write in exponential form. 2x – 2 = 10000 x = 5001 Solve for x. log 104 = 4 log 10,000 4 log (2 • 5001 – 2) 4 Check: log (2x – 2) 4 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 3 log x – log 2 = 5. 3 log x – log 2 = 5 x3 2 Log ( ) = 5 Write as a single logarithm. x3 2 = 105 Write in exponential form. x3 = 2(100,000) Multiply each side by 2. x = 10 200, or about 58.48. 3 The solution is 10 200, or about 58.48. 3 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 pages 456–460 Exercises 1. 1.5850 2. 2.1240 3. 2.7320 4. 3.0101 5. 3 6. 3.4650 7. 0.9534 8. 0.3579 9. 3.2056 10. 0.2720 11. 3.1699; log8 729 12. 1.5; log8 22.627 13. 3.6309; log8 1901.3 14. 2.5643; log8 206.93 15. 3.1827; log8 748.56 16. 2.8074; log8 343 17. 3.8737; log8 3149.6 18. 0.0792; log8 1.1790 19. 2.3219 20. 0.8496 21. 0.0499 22. 3.0101 23. 1.0219 24. 0.9746 25. 0.2009 26. 5.2379 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 27. 0.5690 28. 1.2871 29. 4.7027 30. 14.4894 31. about 7.1 years 32. about 2018 33. 0.05 34. , or 0.3162 35. 33 36. 10,000 37. , or 0.0167 38. 12 39. 10, or 3.1623 40. 100 10 – 1, or 315.2 41. 2 42. 3  108 43. 100,000 5, or 223,606.8 44. 5 45. 46. 1357.2 47. 7 1 60 1 4 10 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 59. 3 60. – 61. a. Florida growth factor = 1.0213, y = 15,982,378 • (1.0213)x; New York growth factor = 1.0054, y = 18,976,457 • (1.0054)x b. 2011 48. a. 18.9658 b. 18.9658 c. Answers may vary. Sample: You don’t have to use the Change of Base formula with the base-10 method, but there is less algebra with the base-2 method. 49. 5.1 50. 20 = 8(1.2)x, 5 years 51. 2 = 10( ) , 2.7 min 52. 75 = 125(0.88)x, 4 years 53. –1 54. 3 55. 56. 3 57. 58. –2 1 2 x 1.17 3 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 65. Answers may vary. Sample: log x = 1.6 101.6 = x, x 39.81 66. Answers may vary. Sample: A possible model is y = 1465(1.0838)x where x = 0 represents 1991; the growth is probably exponential and 1465(1.0838)10 3276; using this model, there will be 10,000 manatees in about 2015. 62. a. Texas growth factor = 1.0208, y = 20,851,820 (1.0208)x; California growth factor = 1.013, y = 33,871,648 • (1.013)x b. 2063 63. Since Florida’s growth rate is larger than Texas’s growth rate, in theory, given constant conditions, Florida would exceed Texas in about 543 years. However, since no state has unlimited capacity for growth, it is unrealistic to predict over a long period of time. 64. 2 – 1 log102 log101 = / 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 76. a. 100 (or 1) W/m2, 104 W/m2 b. 10,000 times more intense 77. a. top up: 10–5 W/m2, top down: 10–2.5 W/m2 b. 99.68% 78. a. 10–3 W/m2, 106 W/m2 b. 109 times more intense 79. 2.9315 67. a. x = b. x = logab = c. Substituting the result from part (a) into the results from part (b), or vice versa, yields logab = . This justifies the Change of Base Formula for c = 10. log b log a 68. 69. 70. 71. 72. 73. 74. 75. log 2 log 7 log 8 log 3 log 140 log 5 log 3.3 log 9 log 3x log 4 log (1 – x) log 6 log x log (x + 1) 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 80. 0.2098 81. 0.6225 82. 625 83. 2.3094 84. 10 85. 0.8505 86. 1.5 87. 7.4168 88. 200.8 89. 2.9615 90. 2.7944 91. 1 92. 500 93. 1.0451 94. 114.3 95. 1.3063 96. 3.0417 97. a. bassoon, guitar, harp, violin, viola, cello b. bassoon, guitar, harp, cello, bass c. harp, violin d. harp, violin 98. 478,630 times 99. no; solving 0.65 = for x, the age in years of the sample, yields an age of about 3561 yrs. 100. 5 x (0.5)5430 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 110. 27.7 111. 249.5 112. 103 113. log 2 + 3 log x – 2 log y 114. log3 x – log3 y 115. 3 log2 3 + 3 log2 x 116. log3 7 + 2 log3 (2x – 3) 117. log4 5 + log4 x 118. log2 5 + log2 a – 2 log2 b 119. x2 + 3x – 1 120. x2 – 3x – 1 121. 3x3 – 3x 122. 1, ±i 101. –4, 2 102. –9, 9 103. 1 104. 20,031 m above sea level 105. a. 91 hours or 4 days b. 0.928 mg or 1.061 mg c. Estimate in hours is more accurate; the days have a larger rounding error. 106. 1.11 107. 7.30 108. 2.19 109. 138.8 1 2 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 123. ±2, ±2i 124. ± 2, ± 3 125. –2, –1, 3 126. 2(x) + 2(x + 12) = 128; 26 ft 127. = 133; 119 679 + x 6 8-5

Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Use mental math to solve each equation. 1. 2x = 2. log42 = x 3. 106x = 1 4. Solve 52x = 125. 1 8 1 2 –3 3 2 8-5

Natural Logarithms ALGEBRA 2 LESSON 8-6 (For help, go to Lessons 8-2 and 8-5.) Use your calculator to evaluate each expression to the nearest thousandth. 1. e5 2. 2e3 3. e–2 4. 5. 4.2e 1 e Solve. 6. log3x = 4 7. log164 = x 8. log16x = 4 8-6

Natural Logarithms 7. log164 = x 16x = 4 x = ALGEBRA 2 LESSON 8-6 7. log164 = x 16x = 4 x = 1 2 1. e5 148.413 2. 2e3 40.171 3. e–2 0.135 4. 0.368 5. 4.2e 11.417 e 6. log3x = 4 34 = x 81 = x 8. log16x = 4 164 = x 65,536 = x Solutions 8-6

Write 2 ln 12 – ln 9 as a single natural logarithm. Natural Logarithms ALGEBRA 2 LESSON 8-6 Write 2 ln 12 – ln 9 as a single natural logarithm. 2 ln 12 – ln 9 = ln 122 – ln 9 Power Property = ln Quotient Property 122 9 = ln 16 Simplify. 8-6

v = –0.0098t + c ln R Use the formula. Natural Logarithms ALGEBRA 2 LESSON 8-6 Find the velocity of a spacecraft whose booster rocket has a mass ratio 22, an exhaust velocity of 2.3 km/s, and a firing time of 50 s. Can the spacecraft achieve a stable orbit 300 km above Earth? Let R = 22, c = 2.3, and t = 50. Find v. v = –0.0098t + c ln R Use the formula. = –0.0098(50) + 2.3 ln 22 Substitute. –0.49 + 2.3(3.091) Use a calculator. 6.62 Simplify. The velocity is 6.6 km/s is less than the 7.7 km/s needed for a stable orbit. Therefore, the spacecraft cannot achieve a stable orbit at 300 km above Earth. 8-6

3 ln (2x – 4) = 6 Power Property Natural Logarithms ALGEBRA 2 LESSON 8-6 Solve ln (2x – 4)3 = 6. ln (2x – 4)3 = 6 3 ln (2x – 4) = 6 Power Property ln (2x – 4) = 2 Divide each side by 3. 2x – 4 = e2 Rewrite in exponential form. x = Solve for x. e2 + 4 2 x 5.69 Use a calculator. Check: ln (2 • 5.69 – 4)3 6 ln 401.95 6 5.996 6 8-6

Use natural logarithms to solve 4e3x + 1.2 = 14. ALGEBRA 2 LESSON 8-6 Use natural logarithms to solve 4e3x + 1.2 = 14. 4e3x + 1.2 = 14 4e3x = 12.8 Subtract 1.2 from each side. e3x = 3.2 Divide each side by 4. ln e3x = ln 3.2 Take the natural logarithm of each side. 3x = ln 3.2 Simplify. x = Solve for x. ln 3.2 3 x 0.388 8-6

A = Pert Continuously compounded interest formula. Natural Logarithms ALGEBRA 2 LESSON 8-6 An initial investment of $200 is now valued at $254.25. The interest rate is 6%, compounded continuously. How long has the money been invested? A = Pert Continuously compounded interest formula. 254.25 = 200e0.06t Substitute 254.25 for A, 200 for P, and 0.06 for r. 1.27125 = e0.06t Divide each side by 200. ln 1.27125 = ln e0.06t Take the natural logarithm of each side. ln 1.27125 = 0.06t Simplify. = t Solve for t. ln 1.27125 0.06 4 t Use a calculator. The money has been invested for 4 years. 8-6

Natural Logarithms pages 464–467 Exercises 1. In 125 2. In 18 3. In 4 ALGEBRA 2 LESSON 8-6 pages 464–467 Exercises 1. In 125 2. In 18 3. In 4 4. In 40,960 5. In 6. In 1 7. In 8. In 9. In 10. 20.92 1 81 m5 n3 xy z4 3 a c b2 23. 2.890 24. 1.151 25. 2.401 26. 5.493 27. 1.242 28. 23.752 29. 6 years 30. 78% 31. 1 32. 2 33. 10 34. 10 11. 24.13 12. 7.79 km/s; yes 13. 25 seconds 14. 134.476 15. 0.135 16. 1.078  1015 17. 1488.979 18. 5.482, –3.482 19. ±11.588 20. 110.196 21. ±2.241 22. ±0.908 8-6

63. No; using the Change of Base Formula would result in one of Natural Logarithms ALGEBRA 2 LESSON 8-6 35. 0 36. 37. 1 38. 83 39. 10.8 40. 301 days; 26 days 41. sometimes 42. never 43. always 44. about 5.8% per hour 45. 19.8 hours 46. about 40,000 bacteria 1 4 47. 3.6 48. 6.7 49. 9.4 50. 11.8 51. 13.9 52. 15.8 53. 17.5 54. 19.1 55. 542.31 56. 1 57. 0.0794 58. 81.286 59. 1.2639 60. no solution 61. 27,347.9 62. 78.342 63. No; using the Change of Base Formula would result in one of the log expressions being written as a quotient of logs, which couldn’t then be combined with the other expression to form a single logarithm. 8-6

( ) Natural Logarithms 64. a. y = 300e0.241t 65. a. about 43 min ALGEBRA 2 LESSON 8-6 64. a. y = 300e0.241t b. 2002 c. 2006 d. t = , where y is the number of Internet users in million and t is time in years. e. Substitute the number of users found in (b) and (c) into the equation in (d). Determine whether your answers in years are the same as t for each. y 300 ln ( ) 0.241 65. a. about 43 min b. t  ln 1.7, 6.0, 11.3, 18.1, 27.6, 43.1, 97.6 66. Check students’ work. 67. C 68. H 1 0.041 T  72 164 8-6

[3] correct model computation error in (b) or (c) Natural Logarithms ALGEBRA 2 LESSON 8-6 69. B 70. [4] a. 8% b. A = Pert = 500e0.08t c. 1800 = 500e0.08t 3.6 = e0.08t ln 3.6 = 0.08t = t 16 t about 16 years [3] correct model computation error in (b) or (c) [2] incorrect model, solved correctly [1] correct model, but without work shown in (c) ln 3.6 0.08 71. 4 72. 2.846 73. 0.272 74. 3333.3 75. 1.002 76. 9.0  10–5 77. y = ; yes 78. y = ; yes 79. y = ± 5 – x ; no 80. 4060 possible combinations x – 7 5 x – 10 2 3 8-6

1. Write 4 ln 6 – 2 ln 3 as a single natural logarithm. Natural Logarithms ALGEBRA 2 LESSON 8-6 1. Write 4 ln 6 – 2 ln 3 as a single natural logarithm. 2. Solve e3x = 15. 3. Simplify ln e7. ln 144 about 0.903 7 8-6

Exponential and Logarithmic Functions ALGEBRA 2 CHAPTER 8 Page 472 1. 2. 3. 8-A

Exponential and Logarithmic Functions ALGEBRA 2 CHAPTER 8 4. 5. Answers may vary. Both answers should be of the form y = abx. For the growth model, b > 1. For the decay model, 0 < b < 1. 6. y = (3)x 7. y = 2(5)x 8. y = 4(4)x 9. 4.11 years 10. y = 3x translated up 2 units 11. y = 3( )x translated left 1 unit 1 3 12. y = 5x translated right 2 units and down 1 unit 13. y = 2x translated left 2 units and reflected over the x-axis 14. 3 15. 1 16. 2 17. 3 18. 0 19. 4 2 8-A

Exponential and Logarithmic Functions ALGEBRA 2 CHAPTER 8 20. 21. 22. 23. 24. log2 2916 25. log 26. log7 a – log7 b 27. log 3 + 3 log x + 2 log y 28. 1 29. 13 30. –1 31. 3 32. Answers may vary. Sample: Taking common logarithms of both sides gives 2x = . Taking logarithms to the base 3 of both sides gives 2x = log3 4, which by the Change of Base Formula is equivalent to 2x = . 33. –15.28 34. 3.89 35. 250 36. 0.01 a3 b2 log 4 log 3 log 4 log 3 8-A

Exponential and Logarithmic Functions ALGEBRA 2 CHAPTER 8 37. 38. 39. 40. 1.359 41. 7.456 42. 17.058, –16.058 43. 4.780 44. 99 cm log 16 log 3 1 log 2 log 8 log 7 8-A