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Exploring Exponential Models

1. 2 x for x = 3. 2. 4 x +1 for x = 1. 3. 2 3 x +4 for x = –1. 4. 3 x 3 x –2 for x = 2. 5. for x = 0 . 6. 2 x for x = –2. 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 2. x.

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Exploring Exponential Models

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  1. 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 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 2 x 8-1

  2. Exploring Exponential Models ALGEBRA 2 LESSON 8-1 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 = = = Solutions 1 2 1 2 0 x 1 22 1 2 • 2 1 4 8-1

  3. Step 1: Make a table of values. Step 2: Graph the coordinates. Connect the points with a smooth curve. x 3xy –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 1 9 1 3 Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Graph y = 3x. 8-1

  4. 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). 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 + 0.007Substitute 0.7%, or 0.007 for r. = 1.007 Simplify. 8-1

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

  6. 6 = ab1Substitute for x and y using (1, 6). = a Solve for a. 6 b 6 b 6 b 2 = • 1 Any number to the zero power equals 1. 2 = Simplify. b = 3 Solve for b. 6 b 2 = b0Substitute for x and y using (0, 2) and for a using . 6 b 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 = abxUse the general term. 8-1

  7. 6 b a = Use your equation for a. a = Substitute 3 for b. a = 2 Simplify. y = 2 • 3xSubstitute 2 for a and 3 for b in y = abx. 6 3 Exploring Exponential Models ALGEBRA 2 LESSON 8-1 (continued) The exponential for a graph that includes (1, 6) and (0, 2) is y = 2 • 3x. 8-1

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

  9. 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. Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Graph y = 36(0.5)x. Identify the horizontal asymptote. As x increases, y approaches 0. The horizontal asymptote is the x-axis, y = 0. 8-1

  10. Relate:  The value of the car decreases exponentially; b = 0.8. Define: Let x = number of years. Let y = value of the car. 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 + rUse r to find b. = 1 + (–0.20) = 0.80 Simplify. Write an equation, and then evaluate it for x = 6. 8-1

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

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

  13. 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 24–31.y = 0 is the horizontal asymptote. 27. 24. 1 3 x 25. 28. 29. 26. Exploring Exponential Models ALGEBRA 2 LESSON 8-1 8-1

  14. 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 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% 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 Exploring Exponential Models ALGEBRA 2 LESSON 8-1 8-1

  15. 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. 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 60. C 61. H 62. B 63.[2] [1] general form of graph correct, but not exact Exploring Exponential Models ALGEBRA 2 LESSON 8-1 8-1

  16. 3 2 3 2 2 3 2 x 3x 54 92 2 3 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 a = a = 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. Sample: y= x3 – 7x – 6 54 b2 1 2 54 b2 54 b 3 2 3 2 3 8-1

  17. 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 2 Exploring Exponential Models ALGEBRA 2 LESSON 8-1 8-1

  18. y = 0 y = 0 Exploring Exponential Models ALGEBRA 2 LESSON 8-1 Sketch the graph of each function. Identify the horizontal asymptote. 1.y = (0.8)x2.y = Without graphing, determine whether each equation represents exponential growth or decay. 3.y = 15(7)x4.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 growth decay y = 0.5(6)x y = 20(4)x 8-1

  19. 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 1 7 – – x –7 6 4 x 5 6 x Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 (For help, go to Lesson 2-6, 5-3, and 7-4.) 8-2

  20. Solutions 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 1 x5 5 4 4 – 5 4 1 7 – – 1 7 – –7 5 6 5 6 6 Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 8-2

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

  22. 1 2 1 2 1 2 Step 1: Graph y = 6. The horizontal asymptote is y = 0. x 1 2 x–3 Step 2: For y = 6 – 2, h = 3 and k = –2. So shift the graph of the present function 3 units right and 2 units down. Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 Graph y = 6 and y = 6 – 2. x x–3 The horizontal asymptote is y = –2. 8-2

  23. 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 Define: Let y = the amount of technetium-99m. Let x = the number of hours elapsed. Then x = the number of half-lives. 1 6 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. 8-2

  24. 1 6 1 2 x Write: y = 50 y = 50 Substitute 25 for x. 1 6 1 2 • 25 1 2 = 50 Simplify. 2.784 4.16 Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 (continued) After 25 hours, about 2.784 mg of technetium-99m remains. 8-2

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

  26. 115.49 Simplify. 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 • e0.144Simplify. = 100 • (1.154884) Evaluate e0.144. You will have about $115.49 in the account after three years. 8-2

  27. 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

  28. 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 14.3 1 2 x 1 2 1 8.14 x 1 2 1 5730 x Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 8-2

  29. 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. 1 2 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 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. 1 2 1 3 x 1 3 x + 5 1 2 x 1 2 x + 4 Properties of Exponential Functions ALGEBRA 2 LESSON 8-2 8-2

  30. 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

  31. 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 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 8000 e(0.06)(4) 3 4 4 4 2 + 8 3 + 5 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 8-2

  32. 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

  33. reflected over the x-axis and translated down 1 rises more steeply, is translated 2 to the right and up 1 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 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

  34. Solve each equation. 1. 8 = x32.x = 2 3. 27 = 3x4. 46 = 43x Graph each relation and its inverse on a coordinate plane. 5.y = 5x6.y = 2x27.y = –x38.y = x 1 4 1 2 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 (For help, go to Lessons 7-1 and 7-7.) 8-3

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

  36. E • 306 E • 303 Write a ratio. 306 303 = Simplify. 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. = 306–3Division Property of Exponents = 303Simplify. = 27,000 Use a calculator. The first earthquake released about 27,000 times as much energy as the second. 8-3

  37. Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Write: 32 = 25 in logarithmic form. If y = bx, then logby = x. Write the definition. If 32 = 25, then log232 = 5. Substitute. The logarithmic form of 32 = 25 is log2 32 = 5. 8-3

  38. Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Evaluate log3 81. Let log3 81 = x. Log381 = xWrite in logarithmic form. 81 = 3xConvert to exponential form. 34 = 3x Write each side using base 3. 4 = xSet the exponents equal to each other. So log3 81 = 4. 8-3

  39. Apple Banana pH = –log[H+] pH = –log[H+] 3.3 = –log[H+] 5.2 = –log[H+] log[H+] = –3.3 log[H+] = –5.2 [H+] = 10–3.3 [H+] = 10–5.2 5.0  10– 4 6.3  10– 6 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? The [H+] of the banana is about 6.3  10– 6. The [H+] of the apple is about 5.0  10– 4. The apple has a higher concentration of hydrogen ions, so it is more acidic. 8-3

  40. 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 = log4x. Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Graph y = log4x. By definition of logarithm, y = log4x is the inverse of y = 4x. 8-3

  41. Step 2: Graph the function by shifting the points from the table to the right 1 unit and up 2 units. x y = log5x –3 1 25 –2 1 125 –1 0 1 1 5 5 1 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. 8-3

  42. 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 more energy. 3. The earthquake in Missouri released about 8,759,310 times more energy. 4. The earthquake in Missouri released about 30 times more energy. 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 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 2 3 2 1 2 1 10 1 2 1 27 1 3 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 8-3

  43. 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

  44. 55. 16,807 = 75 56. 6 = 61 57. 1 = 40 58. = 3–2 59. = 2–1 60. 10 = 101 61. 8192 = 213 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 . 51. First rewrite y = log1x as 1y = x. For any real number y, x = 1. 52. Answers may vary. Sample: y = log3x; y = 3x 53. 128 = 27 54. 0.0001 = 10–4 1 9 1 2 1 3 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 8-3

  45. 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. 73. 74. 75. domain {x|x > 0}, range: all reals 76. domain {x|x > 0}, range: all reals x 2 1 x Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 8-3

  46. 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 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 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 8-3

  47. 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) 5 3 z8 t2 n x p 1 2 1 2 w3 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 98. y = 0 99. y = 9 8-3

  48. 1 25 Logarithmic Functions as Inverses ALGEBRA 2 LESSON 8-3 Evaluate each logarithm. 1. log232 2. log10(–100) 3. log5 –2 undefined 5 Write each equation in exponential form. 4. log 0.01 = –2 5. log327 = 3 6. Graph y = log6(x – 3) + 1. 33 = 27 10–2 = 0.01 8-3

  49. x6 x9 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 – x5.x5 • x26.7.x3 + x2 8-4

  50. Solutions 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 x = 2 y = 3 log39 – log327 = 2 – 3 = –1 3. log216 = x     log264 = y 2x = 16 2y = 64 x = 4 y = 6 log216 ÷ log264 = 4 ÷ 6 = = 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 1 33 1 27 x6 x9 36 99 4 6 2 3 Properties of Logarithms ALGEBRA 2 LESSON 8-4 8-4

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