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Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 (For help, go to Lessons 1-2 and 1-6.) Simplify each expression. 1 1. 23 2. 42 3. 42 22 4. (–3)3 5. –33 6. 62 12 Evaluate each expression for a = 2, b = –1, c = 0.5. 7. a 2a 8. bc c 9. 8-1 ab bc Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Solutions 1. 23 = 2 • 2 • 2 = 8 1 = 1 = 1 42 4 • 4 16 2 3. 42 22 = 42 = 4 • 4 = 16 = 4 2•2 4 2 2. 4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27 5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27 6. 62 12 = 36 12 = 3 a for a = 2: 2 = 1 2•2 2 2a 8. bc for b = –1, c = 0.5: –1 • 0.5 = –1 c 0.5 9. ab for a = 2, b = –1, c = 0.5: 2 • (–1) = 2 = 4 bc (–1) • 0.5 0.5 7. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Simplify. a. 1 3–2 = 32 1 = 9 b. (–22.4)0 = 1 Use the definition of negative exponent. Simplify. Use the definition of zero as an exponent. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Simplify a. 3ab –2 = 3a 1 b2 b. Rewrite using a division symbol. Use the definition of negative exponent. = 3a b2 1 –3 = 1 x –3 x 1 =1 3 x Use the definition of negative exponent. Simplify. = 1 • x3 1 Multiply by the reciprocal of x3 , which is x 3. = x3 Identity Property of Multiplication 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Evaluate 4x 2y –3 for x = 3 and y = –2. Method 1: Write with positive exponents first. 4x 2 4x 2y –3 = y 3 Use the definition of negative exponent. 4(3)2 = (–2)3 36 Substitute 3 for x and –2 for y. 1 = –8 = –4 2 Simplify. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 (continued) Method 2: Substitute first. 4x 2y –3 = 4(3)2(–2)–3 4(3)2 = (–2)3 36 Substitute 3 for x and –2 for y. Use the definition of negative exponent. 1 = –8 = –4 2 Simplify. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 In the lab, the population of a certain bacteria doubles every month. The expression 3000 • 2m models a population of 3000 bacteria after m months of growth. Evaluate the expression for m = 0 and m = –2. Describe what the value of the expression represents in each situation. a. Evaluate the expression for m = 0. 3000 • 2m = 3000 • 20 = 3000 • 1 Substitute 0 for m. Simplify. = 3000 When m = 0, the value of the expression is 3000. This represents the initial population of the bacteria. This makes sense because when m = 0, no time has passed. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 (continued) b. Evaluate the expression for m = –2. 3000 • 2m = 3000 • 2–2 Substitute –2 for m. = 3000 • 1 4 = 750 Simplify. When m = –2, the value of the expression is 750. This represents the 750 bacteria in the population 2 months before the present population of 3000 bacteria. 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 pages 397–399 Exercises 1. –1 2. 1 16 3. 1 25 4. – 1 25 5. 1 16 6. – 1 81 7. 1 64 8. – 1 12 11. – 1 64 12. – 1 64 13. –2 14. 3; 4 15. 0; –3 16. –5 17. 3a 18. 54 x 19. x 7 9. 1 20. c 10. 1 21. 78 1 25p 22. 14 a 3 x2y 24. 7a 3b 2 w 25. 51 7 x y 7 26. y 5 x 23. 33. 1 25 34. 1 9 35. – 1 9 36. 1 37. 3 27. 4c3 38. 7st 3 28. 5 39. 29. 40. 6 ac 3 30. x 2 8z 7 31. y 7 t 11 32. 14 m 2t 5 8-1 41. 42. 25 1 100 25 81 81 25 – 25 27 1 25 43. –27 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 44. – 27 54. 10–4 63. 45 45. a. $20.48; $.32 b. No; the value of the allowance rapidly becomes very great. 55. 10–5 64. 6 400 56. 0.001 57. 0.000001 65. 40 58. 0.7 66. 1 59. 0.03 67. – 1 47. pos. 60. 0.0005 68. 16 48. pos. 61. a. 5–2, 5–1, 50, 51, 52 b. 54 n c. a 46. neg. 49. neg. 50. neg. 51. 10–1 52. 10–2 53. 10–3 4 1 62. In –30, 3 is raised to the zero power, and then the opposite is determined. In (–3)0, the number –3 is raised to the zero power. 8-1 243 69. 2 9 70. 1 8 71. 1 16 72. –1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 73. 74. a. b. 79. a. 1 correct, 0.4096; 2 correct, 0.1536; 3 correct, 0.0256; 4 correct, 0.0016 b. 0 or 1 1 They are reciprocals for a =/ 0; 1 = a –n and 1 = 1 = an. an a –n 1 an 75. A, B, D 76. Check students’ work. 77. No; • = 9 • = 9. The product of reciprocals should be 1. 3x–2 3x2 x0 78. The student multiplied b by zero instead of raising b to the zero power, which would equal 1. 80. about 4 students; about 16 students; about 29 students 81. 8 – 48m2 82. 21 83. 141 m2 84. 2.9375 4 nr 7y 2 86. –7 1 4 85. 87. 1 and –1 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 88. 2 3 89. 1 9 90. 1 64 95. c. Answers may vary slightly. Sample: y = 53x – 4328 d. Answers may vary slightly. Sample: $1,237,000,000 96. 98. y = –x + 4 91. 1 99. y = 5x – 2 92. 0.26 100. y = 2 x – 3 93. 258.4 94. 5 101. y = – 3 x – 17 11 97. a-b. 102. y = 5 x + 1 9 3 103. y = 1.25x – 3.79 8-1 Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Simplify each expression. 1. 3–4 1 81 2. (–6)0 k m–3 1 3. –2a0b–2 – 2 2 4. 5. 8000 • 40 6. 4500 • 3–2 b 8000 8-1 km3 500 Scientific Notation ALGEBRA 1 LESSON 8-2 (For help, go to Lessons8-1.) Simplify each expression. 1. 6 • 104 2. 7 • 10–2 3. 8.2 • 105 4. 3 • 10–3 5. 3.4 • 101 6. 5.24 • 102 7. Simplify 3 102 + 6 101 + 7 100 + 8 10–1. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 Solutions 1. 6 • 104 = 6 • 10,000 = 60,000 2. 7 • 10–2 = 7 • 0.01 = 0.07 3. 8.2 • 105 = 8.2 • 100,000 = 820,000 4. 5. 6. 7. 3 • 10–3 = 3 • 0.001 = 0.003 3.4 • 101 = 3.4 • 10 = 34 5.24 • 102 = 5.24 • 100 = 524 3 102 + 6 101 + 7 100 + 8 10–1 = (3 102) + (6 101) + (7 100) + (8 10–1) = (3 100) + (6 10) + (7 1) + (8 0.1) = 300 + 60 + 7 + 0.8 = 367.8 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 Is each number written in scientific notation? If not, explain. a. 0.46 104 b. 3.25 10–2 c. 13.2 106 No; 0.46 is less that 1. yes No; 13.2 is greater than 10. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 Write each number in scientific notation. a. 234,000,000 234,000,000 = 2.34 108 Move the decimal point 8 places to the left and use 8 as an exponent. Drop the zeros after the 4. b. 0.000063 0.000063 = 6.3 10–5 Move the decimal point 5 places to the right and use –5 as an exponent. Drop the zeros before the 6. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 Write each number in standard notation. a. elephant’s mass: 8.8 104 kg 8.8 104 = 8.8000. A positive exponent indicates a number greater than 10. Move the decimal point 4 places to the right. = 88,000 b. ant’s mass: 7.3 10–5 kg 7.3 10–5 = 0.00007.3 A negative exponent indicates a number between 0 and 1. Move the decimal point 5 places to the left. = 0.000073 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 List the planets in order from least to greatest distance from the sun. Planet Distance from the Sun Jupiter 4.84 108 mi Earth 9.3 107 mi Neptune 4.5 109 mi Mercury 3.8 107 mi Order the powers of 10. Arrange the decimals with the same power of 10 in order. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 (continued) 3.8 107 Mercury 9.3 107 Earth 4.84 108 Jupiter 4.5 109 Neptune From least to greatest distance from the sun, the order of the planets is Mercury, Earth, Jupiter, and Neptune. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 Order 0.0063 105, 6.03 104, 6103, and 63.1 103 from least to greatest. Write each number in scientific notation. 0.0063 105 6.3 102 6.03 104 63.1 103 6103 6.03 104 6.103 103 6.31 104 Order the powers of 10. Arrange the decimals with the same power of 10 in order. 6.3 102 6.103 103 6.03 104 Write the original numbers in order. 0.0063 105 6103 6.03 104 8-2 6.31 104 63.1 103 Scientific Notation ALGEBRA 1 LESSON 8-2 Simplify. Write each answer using scientific notation. a. 6(8 10–4) = (6 • 8) 10–4 b. 0.3(1.3 103) Use the Associative Property of Multiplication. = 48 10–4 Simplify inside the parentheses. = 4.8 10–3 Write the product in scientific notation. = (0.3 • 1.3) 103 Use the Associative Property of Multiplication. = 0.39 103 Simplify inside the parentheses. = 3.9 102 Write the product in scientific notation. 8-2 Scientific Notation ALGEBRA 1 LESSON 8-2 pages 402–404 Exercises 1. No; 55 > 10. 2. yes 11. 3.25 10–3 12. 8.003 106 3. No; 0.9 < 1. 13. 9.2 10–4 4. yes 14. 1.56 10–2 5. yes 15. 500 6. No; 46 > 10. 16. 0.05 7. 9.04 109 17. 2040 8. 2.0 10–2 18. 720,000 9. 9.3 19. 0.897 106 10. 2.17 104 22. 0.0048 23. 10–3, 10–1, 100, 101, 105 24. 6 10–10, 8 10–8, 9 10–7, 7 10–6 25. 0.52 10–3, 50.1 10–3, 4.8 10–1, 56 10–2 26. 5300 10–1, 5.3 105, 20. 1.3 21. 0.0000274 8-2 0.53 107, 530 108 27. C, A, B 28. 5.6 10–2 29. 2.4 1015 Scientific Notation ALGEBRA 1 LESSON 8-2 30. 6.0 101 31. 3.18 10–3 32. 2.46 10–3 33. 3.4 105 34. 5400 35. 7 101 36. 1 101 37. 4.6 10–2 38. 0.0005 39. 3 10–26 40. Answers may vary. 43. Sample: Yes, if you regard the 1 in 44. 1 105 as “understood” as happens when 1 is the coefficient of a term like x, then 105 is in scientific notation. 41. 48 million = 48 106. Write 48 in scientific notation; then add the powers of 10: 4.8 101 106 = 4.8 107. 48 millionths = 48 10–6. 45. So 4.8 101 10–6 = 4.8 10–5. 42. about $1.65 1012 8-2 2.796 1010 instructions; 1.6776 1012 instructions Answers may vary. Sample: Since the national debt is expressed in dollars, use standard notation, which most people will understand, rather than scientific notation, which is used mainly in science. a. 5 1014 b. about 1.6 108 years Scientific Notation ALGEBRA 1 LESSON 8-2 46. about 2.61 109 people 47. a. 6.08 1010 km3 b. 1.09 1012 km3 c. 9.17 1014 km3 48. 49. 50. 51. 3.3 10–3 D G C 53. 4 54. 3 55. 2 3 56. 64 49 57. 1 9 58. 52. [2] (8 10–4)(1000) = 0.0008 1000 = 0.8; the diameter is 0.8 mm [1] minor computational error 59. 8-2 60. Scientific Notation ALGEBRA 1 LESSON 8-2 1. Write each number in scientific notation. a. 0.00627 b. 3,486,000 6.27 10–3 3.486 106 2. Write each number in standard form. a. 9.4 104 b. 2.3 10–6 94,000 0.0000023 3. Order the following numbers from least to greatest. 0.98 10–1, 1.6 103, 2.4 10–1, 11 100 0.98 10–1, 2.4 10–1, 11 100, 1.6 103 4. Simplify. Write the answer in scientific notation. 7(6.1 10–2) 4.27 10–1 8-2 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 (For help, go to Lesson 1-6.) Rewrite each expression using exponents. 1. t•t•t•t•t•t•t 2. (6 – m)(6 – m)(6 – m) 3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) 4. 5 • 5 • 5 • s • s • s Simplify. 5. –54 6. (–5)4 7. (–5)0 8. (–5)–4 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Solutions 1. t • t • t • t • t • t • t = t7 2. (6 – m)(6 – m)(6 – m) = (6 – m)3 3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) = (r + 5)5 4. 5. 6. 7. 8. 5 • 5 • 5 • s • s • s = 53 • s3 = 53s3 –54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625 (–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625 (–5)0 = 1 (–5)–4 = (– 1 )4 5 = (– 1 )(– 1 )(– 1 )(– 1 ) 5 5 = ( 1 )( 1 ) 25 25 = 1 625 5 5 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Rewrite each expression using each base only once. a. 73 • 72 = 73 + 2 = 75 b. 44 • 41 • 4–2 = 44 + 1 – 2 = 43 Add exponents of powers with the same base. Simplify the sum of the exponents. Think of 4 + 1 – 2 as 4 + 1 + (–2) to add the exponents. Simplify the sum of the exponents. = 60 Add exponents of powers with the same base. Simplify the sum of the exponents. =1 Use the definition of zero as an exponent. c. 68 • 6–8 = 68 + (–8) 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify each expression. a. p2 • p • p5 = p 2 + 1 + 5 = p8 Add exponents of powers with the same base. Simplify. b. 4x6 • 5x–4 = (4 • 5)(x 6 • x –4) Commutative Property of Multiplication = 20(x 6+(–4)) Add exponents of powers with the same base. = 20x 2 Simplify. 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify each expression. a. a 2 • b –4 • a 5 = a 2 • a 5 • b –4 Commutative Property of Multiplication Add exponents of powers with the same base. = a 2 + 5 • b –4 a7 = 4 b Simplify. b. 2q • 3p3 • 4q4 = (2 • 3 • 4)(p 3)(q • q 4) Commutative and Associative Properties of Multiplication = 24(p 3)(q 1 • q 4) Multiply the coefficients. Write q as q 1. = 24(p 3)(q 1 + 4) Add exponents of powers with the same base. = 24p 3q 5 Simplify. 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify (3 10–3)(7 10–5). Write the answer in scientific notation. (3 10–3)(7 10–5) = (3 • 7)(10–3 • 10–5) Commutative and Associative Properties of Multiplication = 21 10–8 Simplify. = 2.1 101 • 10–8 Write 21 in scientific notation. = 2.1 Add exponents of powers with the same base. 101 + (– 8) = 2.1 10–7 Simplify. 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 The speed of light is 3 108 m/s. If there are1 10–3 km in 1 m, and 3.6 103 s in 1 h, find the speed of light in km/h. Speed of light = meters kilometers seconds • • seconds meters hour m km s = (3 108) s • (1 10–3) • (3.6 103) m h = (3 • 1 • 3.6) (108 • 10–3 • 103) = 10.8 (108 + (– 3) + 3) Use dimensional analysis. Substitute. Commutative and Associative Properties of Multiplication Simplify. 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 (continued) = 10.8 108 Add exponents. = 1.08 101 • 108 Write 10.8 in scientific notation. = 1.08 109 Add the exponents. The speed of light is about 1.08 109 km/h. 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 pages 407–410 Exercises 1. 210 12. –4.8n3 22. 6 105 5 2. 28 13. b3 23. 6 109 3. 1 14. –7 24. 4 103 4. (0.99)3 15. –45a4 25. 3.4 10–5 5. 69 3 16. y x 26. 5.6 10–7 6. 1 17. 45x 7y 6 27. 1.5 1022 7. c5 18. 12a6c8 28. about 2.5578 1013 mi 8. 3r 5 19. x10y2 29. 1.08 1021 dollars 9. 10t –7 20. a8b 10. 56x 6 3 21. – 240m 30. about 3.84 105 km 5 r 11. 3x4 8-3 31. 9 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 32. –4 44. –6x6 54. about 1.01 g 33. –3 45. 12a7 34. 11 46. x10 35. –5 47. 34 • 22 36. 5 48. 2.7 10–8 37. –4 49. 8.0 105 55. a. y1y7; y2y6; y3y5; y4y4 b. Answers may vary. Sample: y–1y9; y–2y10; y–3y11; y–4y12 c. An infinite number; there are an infinite number of integer pairs with a sum of 8. 38. 0 50. 2.1 10–5 39. 2, –3 40. 6x3 + 2x2 51. 1.2 10–4 41. 4x4 52. 8.0 10–8 42. 4y5 + 8y2 53. 1.5 108 56. a. about 10–7 m b. Longer; 1 < 4 < 7 so 1 10–7 < 4 10–7 < 7 10–7. 57. Answers may vary. Sample: The property of multiplying powers only applies when 2 terms have the same base. 43. 4c4 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 58. about 5.85 103 m 59. 7.65 1014 60. 4.0392 108 61. 7.039305 10–7 62. 1.7882786 10–12 63. about 6.7 1033 molecules 64. 1.428 1033 molecules 69. 8m5 + 56m3 78. 700 times 70. –8x5 + 36x4 79. D 71. 81 80. F 72. 22n+3 81. A 73. 2x+y • 3x+2 82. B 74. 1 a+b 75. (t + 3)2 65. x3 66. 1 76. 25 67. 5c3 77. a. 1.833 10–9 km3 a 68. 6a3 + 10a2 b. 1.833 m3 8-3 83. [2] 365 4.7 107 = 1715.5 107 = 1.7155 1010: about 1.7 1010 diapers [1] no work shown OR answer not written in scientific notation Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 84. [4] a. 4r 2 b. 4(5)2 = 4 25 = 100 in.2 c. 144 = 4r 2, 36 = r 2 so r = 6, d = 2r = 12, 12 in. [3] radius found in (c) but not the diameter [2] only two questions answered correctly [1] only one question answered correctly 89. 876,000,000 90. 0.001052 91. 910,000,000,000 92. 0.00029 93. 96. 18; 34; 46 97. –1; 7; 13 98. 4; –12; –24 85. 1.28 106 86. 3.5 10–3 95. 99. –6.8; –22.8; –34.8 94. 87. 9.0 10–5 88. 6.2 106 8-3 Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify each expression. 1. 34 • 35 2. 4x5 • 3x–2 39 3. (3 104)(5 102) 1.5 107 12x3 4. (7 10–4)(1.5 105) 1.05 102 5. (–2w –2)(–3w2b–2)(–5b–3) – 305 b 6. What is 2 trillion times 3 billion written in scientific notation? 6 1021 8-3 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 (For help, go to Lesson 8-3.) Rewrite each expression using each base only once. 1. 32 • 32 • 32 2. 23 • 23 • 23 • 23 3. 57 • 57 • 57 • 57 4. 7 • 7 • 7 Simplify. 5. x3 • x3 6. a2 • a2 • a2 7. y–2 • y–2 • y–2 8. n–3 • n–3 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Solutions 1. 32 • 32 • 32 = 3(2 + 2 + 2) = 36 2. 23 • 23 • 23 • 23 = 2(3 + 3 + 3 + 3) = 212 3. 57 • 57 • 57 • 57 = 5(7 + 7 + 7 + 7) = 528 4. 7 • 7 • 7 = 73 5. x3 • x3 = x(3 + 3) = x6 6. a2 • a2 • a2 = a(2 + 2 + 2) = a6 7. y–2 • y–2 • y–2 = y(–2 + (–2) + (–2)) = y–6 = 16 y 8. n–3 • n–3 = n(–3 + (–3)) = n–6 = 16 n 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify (a3)4. (a3)4 = a3 • 4 = a12 Multiply exponents when raising a power to a power. Simplify. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify b2(b3)–2. b2(b3)–2 = b2 • b3 • (–2) Multiply exponents in (b3)–2. = b2 • b–6 Simplify. = b2 + (–6) Add exponents when multiplying powers of the same base. = b–4 Simplify. 1 = b4 Write using only positive exponents. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify (4x3)2. (4x3)2 = 42(x3)2 Raise each factor to the second power. = 42x6 Multiply exponents of a power raised to a power. = 16x6 Simplify. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify (4xy3)2(x3)–3. (4xy3)2(x3)–3 = 42x2(y3)2 • (x3)–3 Raise the three factors to the second power. = 42 • x2 • y6 • x–9 Multiply exponents of a power raised to a power. = 42 • x2 • x–9 • y6 Use the Commutative Property of Multiplication. = 42 • x–7 • y6 Add exponents of powers with the same base. 16y6 = x7 Simplify. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 An object has a mass of 102 kg. The expression 102 • (3 108)2 describes the amount of resting energy in joules the object contains. Simplify the expression. 102 • (3 108)2 = 102 • 32 • (108)2 Raise each factor within parentheses to the second power. = 102 • 32 • 1016 Simplify (108)2. = 32 • 102 • 1016 Use the Commutative Property of Multiplication. = 32 • 102 + 16 Add exponents of powers with the same base. = 9 1018 Simplify. Write in scientific notation. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 pages 413–416 Exercises 12. 1 12g 4 24. 9 1010 1. c10 13. 36y4 25. 8 10–30 2. c10 14. 81n24 26. 8 10–9 3. n32 15. 4. q100 16. 1 5. c19 17. x16 6. d15 18. 8x5y3 7. 1 t 14 1 8y 12 19. 1 8. x 7 20. 118 9. 625y4 21. 9a6b8 10. 1024m5 22. 11. 49a2 23. 1.6 1011 c a32 32c 26 8-4 27. 4.9 109 28. 3.6 1025 29. 6.25 10–18 30. 4.2875 10–11 31. 32. 33. 34. 35. 36. 8.57375 10–10 m3 3 –4 4 –3 0 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 37. 8 48. 9 38. –2 49. 4.3 104 39. 0 50. –256x4y5 40. –3 51. a. 41. The student who wrote b. x5 + x5 = 2x5 is correct; c. x5 times x5 is x10. d. 24x2; 96x2 4 times 8x3; 64x3 8 times 42. 1 52. (mn)4 43. 243x3 53. (ab)5 44. b17 54. (7xyz)2 45. 30x2 46. 0.16 4 x 47. –8a9b6 57. a. b. c. d. 106 109 109 1018 58. a. 223 bits b. 230 bytes; 233 bits 59. a. about 5.15 1014 m2 b. about 3.60 1014 m2 c. about 1.37 1018 m3 60. C 61. Add exponents for products of powers as 55. (2xy)2 in a2a4. Multiply exponents 56. Check students’ work. for powers of powers, as in (a2)4. 8-4 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 62. 3 73. [2] No, (x2 + 3y)2 =/ x4 + 9y2; 63. 6 for x = 2, y = 4: (x2 + 3y)2 80. (7, 26) 64. 12 = (4 + 12)2 = 162 = 256; 81. (–9, –5) 65. 3 x4 + 9y2 = 24 + 9(4)2 = 82. – 3 16 + 9 • 16 = 160 (OR equivalent explanation) [1] appropriate conclusion but no work shown 66. 4 67. –5 68. x12; x81; no b2 c6 69. B 74. 70. I 75. a8b3 71. C 76. 54m5n4 72. H 77. –4t5 78. (1, 8) 8-4 79. (4 2 , 1 1 ) 3 4 83. 6 84. – 3 2 85. – 9 11 3 More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify each expression. 1. (x4)5 3. (5x4)3 5. x(x5y–2)3 x20 2. 125x12 4. (1.5 105)2 (2w–2)4(3w2b–2)3 432 b6w2 x16 y6 2.25 1010 6. (3 10–5)(4 104)2 8-4 4.8 103 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 (For help, go to Skills Handbook page 724.) Write each fraction in simplest form. 5 1. 20 5. 6 15 9. 5xy 15x 125 2. 25 6. 8 30 10. 6y2 60 124 4 3. 100 4. 7. 10 8. 18 11. 3ac 12. 24m 12a 6mn2 35 3x 8-5 63 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 Solutions 1. 5 = 5•1 = 1 20 5•4 4 2. 125 25 • 5 = =5 25 25 • 1 3. 60 20 • 3 3 = = 100 20 • 5 5 4. 124 4 • 31 = = 31 4 4•1 5. 6 3•2 2 = = 15 3•5 5 6. 8 2•4 4 = = 30 2 • 15 15 7. 10 5•2 2 = = 35 5•7 7 8. 18 9•2 2 = = 63 9•7 7 9. 5xy 5•x•y y = = 15x 5 • 3 • x 3 10. 6y2 = 3 • 2 • y2 = 2y2 3x 3•x x 11. 3ac = 3 • a • c = c 12a 3•4•a 4 12. 8-5 24m 6•4•m 4 = = 6mn2 6 • m • n2 n2 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 Simplify each expression. a. Subtract exponents when dividing powers with the same base. x4 4–9 x = 9 x = x–5 Simplify the exponents. 1 Rewrite using positive exponents. = x5 b. p3 j –4 p3 – (–3)j = –3 6 p j = p6 j –10 p6 = j10 –4 – 6 Subtract exponents when dividing powers with the same base. Simplify. Rewrite using positive exponents. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 A small dog’s heart beats about 64 million beats in a year. If there are about 530 thousand minutes in a year, what is its average heart rate in beats per minute? 6.4 107 beats 64 million beats = 530 thousand min 5.3 105 min Write in scientific notation. 6.4 107–5 Subtract exponents when dividing powers with the same base. 6.4 102 Simplify the exponent. = 5.3 = 5.3 1.21 102 = 121 Divide. Round to the nearest hundredth. Write in standard notation. The dog’s average heart rate is about 121 beats per minute. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 3 y3 Simplify 3 y3 4 4 . 34 = 34 (y ) Raise the numerator and the denominator to the fourth power. 34 = y 12 Multiply the exponent in the denominator. 81 = y 12 Simplify. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 a. Simplify 2 3 –3 = –3 2 3 3 2 . 3 2 Rewrite using the reciprocal of 3 . Raise the numerator and the denominator to the third power. 33 = 23 27 3 = 8 or 3 8 Simplify. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 (continued) 4b b. Simplify – c . – 4b c –2 = – c 4b 2 = – c 4b 2 Rewrite using the reciprocal of – 4b . c Write the fraction with a negative numerator. (–c)2 = (4b)2 Raise the numerator and denominator to the second power. c2 = 16b2 Simplify. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 20. 1 10–12 s t 11 pages 420–423 Exercises 12. 27m4 1. 7 13. 5 107 2. 1 14. 3 10–5 3. –3 4. 0 5. 1 4 6. 4 7. 1 c3 8. m3 2 9. s 2 10. y 21. 9 22. 15. 6 102 23. 16. 4.2 103 24. 17. 1.5 10–6 25. 18. 7 10–3 19. a. 3.86 1011 26. h; 2.65 108 people b. about 1457 h c. about 4.0 h 11. 1 2 cd 8-5 27. 28. 29. 25 1 x3 32x 5 y5 81a 4 16b 4 64 125 1 9 36 n12 8p 3 125 3 2 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 30. 9 40. x0 simplifies to 1. 4 31. 9 4 41. The base d should appear only once. 32. – 27 42. 9 8 33. x8 8 1 16m12 44. 1 25 25 34. 8 n3 35. 112 c 43. 45. a6 36. 1 6 46. t 27 47. 1 n28 48. 9 4k 10 37. 53 simplifies to 125. 38. y–2 contains a negative exponent. 39. Each term should be raised to the 4th power and simplified. 49. 49 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 50. a. 9.74 107 households; 5.44 1011 local calls; 9.7 1010 long-distance calls b. about 5585 local calls c. about 996 long-distance calls 51. a. Answers may vary. 4 Sample: c 6 can be written as c c4–6 or c–2; c–2 = 12 . c b. Check students’ work. 4 52. a 10 4b 5 5 53. a 3c b 54. 3 4 55. 5 56. r5 p 25q5 57. 20,736 6 58. y 81 59. 3b 5 5a 60. a. about 0.02% b. about 99.98% 61. Answers may vary. Sample: You can raise the numerator and denominator to the power and then simplify or simplify and then raise to the power. 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 70. 62. a. about $12,988 b. about $20,733 c. about 60% 71. 63. a. The student treated 54 as 5 b. 125 64. 3 5 65. m n 5 7 66. d 3 67. 1010 68. 3x 3 69. 2y 2 13m 2 5 5 4 . 7m 5n 5c 6 2 3 72. a. about 1.2 106 s b. about 13.9 days 73. a-b. Check students’ work. c. No, the power may remain the same or be one less. 74. def. of neg. exponent 75. dividing powers with the same base, def. of neg. exponent 76. raising a quotient to a power 77. mult. powers with the same base 78. raising a power to a power, dividing powers with the same base, def. of neg. exponents 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 79. n2 80. b. The closer the ratio is to 1, the more circular the orbit. c. Pluto, Venus n4x 81. x4 84. B 82. 1 n2 85. G 83. a. 86. D 87. B 88. C 89. B 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 90. [4] (7.43 108) ÷ (2.5 104) = 2.972 104, 29,720 h; 29,720 hours • 1 1 day • 1 year 24 hours 365 days 3.4 years about 3.4 years (OR equivalent explanation) [3] appropriate methods but one computational error [2] error in conversion factor used or a missing conversion factor [1] correct number of hours and years but no work shown 97. n4 91. 27y6 100. (–4, –7) 8 m 21 t 20 93. 8 r 6 94. 2s 27 95. 1 8c 2 92. 98. 1 99. (0, 0) 96. 9r6 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 101. (3, 5) 103. 102. no solution 8-5 Division Properties of Exponents ALGEBRA 1 LESSON 8-5 Simplify each expression. 1. a8 a–2 4. 1.6 103 2. a10 4 10–2 4 104 5. w3 w7 24 5 8-5 1 w4 2 256 6 or 10 25 25 4 –2 3. (3a) (2a ) 6a2 6. 4x 3 3x 2 27 –3 27 64x3 Geometric Sequences ALGEBRA 1 LESSON 8-6 (For help, go to Lesson 5-6.) Find the common difference of each sequence. 1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ... 3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ... Use inductive reasoning to find the next two numbers in each pattern. 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... 7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ... 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Solutions 1. 1, 3, 5, 7, ... 7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2 Common difference: 2 2. 19, 17, 15, 13, ... 13 – 15 = –2, 15 – 17 = –2, 17 – 19 = –2 Common difference: –2 3. 1.3, 0.1, –1.1, –2.3, ... –2.3 – (–1.1) = –1.2, –1.1 – 0.1 = –1.2, 0.1 – 1.3 = –1.2 Common difference: –1.2 4. 18, 21.5, 25, 28.5, ... 28.5 – 25 = 3.5, 25 – 21.5 = 3.5, 21.5 – 18 = 3.5 Common difference: 3.5 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Solutions (continued) 5. 2, 4, 8, 16, ... 2(2) = 4, 4(2) = 8, 8(2) = 16, 16(2) = 32, 32(2) = 64 Next two numbers: 32, 64 6. 4, 12, 36, ... 4(3) = 12, 12(3) = 36, 36(3) = 108, 108(3) = 324 Next two numbers: 108, 324 7. 0.2, 0.4, 0.8, 1.6, ... (0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 1.6, 1.6(2) = 3.2, 3.2(2) = 6.4 Next two numbers: 3.2, 6.4 8. 200, 100, 50, 25, ... 200 2 = 100, 100 2 = 50, 50 2 = 25, 25 2 = 12.5, 12.5 2 = 6.25 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Find the common ratio of each sequence. a. 3, –15, 75, –375, . . . –15 3 (–5) –375 75 (–5) (–5) The common ratio is –5. 3 3 3 b. 3, 2 , 4 , 8 , ... 3 2 3 1 2 3 4 3 8 1 2 1 2 1 The common ratio is 2 . 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Find the next three terms of the sequence 5, –10, 20, –40, . . . 5 –10 (–2) 20 (–2) –40 (–2) The common ratio is –2. The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320. 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Determine whether each sequence is arithmetic or geometric. a. 162, 54, 18, 6, . . . 62 54 1 3 18 1 3 6 1 3 The sequence has a common ratio. The sequence is geometric. 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 (continued) b. 98, 101, 104, 107, . . . 98 101 +3 104 +3 107 +3 The sequence has a common difference. The sequence is arithmetic. 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Find the first, fifth, and tenth terms of the sequence that has the rule A(n) = –3(2)n – 1. first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3 fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48 tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 Suppose you drop a tennis ball from a height of 2 meters. On each bounce, the ball reaches a height that is 75% of its previous height. Write a rule for the height the ball reaches on each bounce. In centimeters, what height will the ball reach on its third bounce? The first term is 2 meters, which is 200 cm. Draw a diagram to help understand the problem. 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 (continued) The ball drops from an initial height, for which there is no bounce. The initial height is 200 cm, when n = 1. The third bounce is n = 4. The common ratio is 75%, or 0.75. A rule for the sequence is A(n) = 200 • 0.75n – 1. A(n) = 200 • 0.75n – 1 Use the sequence to find the height of the third bounce. A(4) = 200 • 0.754 – 1 Substitute 4 for n to find the height of the third bounce. = 200 • 0.753 Simplify exponents. = 200 • 0.421875 Evaluate powers. = 84.375 Simplify. The height of the third bounce is 84.375 cm. 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 12. –48, 96, –192 24. –1.1; 70.4; 18,022.4 1. 4 13. geometric 25. A(n) = 6 • 0.5n–1; 0.375 2. 4 14. arithmetic 26. A(n) = –6 • 2n–1; –3072 3. 0.1 15. geometric 27. A(n) = 7 • (1.1)n–1; 9.317 4. 2.50 16. arithmetic 28. A(n) = 1 • (-4)n–1; 4096 5. –0.25 17. arithmetic 6. 2 18. geometric 29. a. A(n) = 100 • (0.64)n–1 b. about 10.74 cm 7. 40, 80, 160 19. 5; 135; 10,935 8. 48, 96, 192 20. –5; –135; –10,935 pages 427–429 Exercises 9. 20.25, 30.375, 45.5625 21. 5; –135; –10,935 10. –0.5, 0.25, –0.125 22. 0.5; 13.5, 1093.5 11. 0.36, 0.072, 0.0144 23. –2; –250; –156,250 8-6 30. 2 2 , 8 , 8 ; A(n) = 3 9 27 n–1 216 • 1 3 31. 1, 0.2, 0.04; A(n) = 625 • (0.2)n–1 Geometric Sequences ALGEBRA 1 LESSON 8-6 32. 656.1, 5904.9, 53,144.1; A(n) = 0.1 • 9n–1 41. a. A(n) = 36 • (0.9)n–1 b. 6; n = 1 corresponds to the 33. 1, –0.5, 0.25; A(n) = 16 • (–0.5)n–1 first swing, because A(1) = 36. 34. Check students’ work. c. 21.3 cm 35. If all consecutive terms have a common difference, the sequence is arithmetic. If all consecutive terms have a common ratio, the sequence is geometric. 42. a. 1 , 1 , 1 , 1 2 4 b. 2–1, 8 2–2, 16 2–3, 2–4 c. r = 2–n d. 2–10 or 110 36. a. 3 folds: 8; 4 folds: 16; 5 folds: 32 b. A(n) = 1 • 2n c. 1024 rectangles 2 43. No; if a term were 0, then all terms would be 0 because you multiply a term to get the next term. 37. arithmetic; 3, 1, –1 38. neither; –3, –8, –14 39. geometric; 1.125, 0.5625, 0.28125 40. arithmetic; 20, 22, 24 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 49. 6 10–2; 2.592 102, 44. a. 1, 3 , 9 , 27 4 16 64 n–1 b. r = 1 • 3 4 243 c. 1024 d. 0, 1 , 7 , 37 4 16 64 e. r = 1.5552 101, 9.3312 10–1 50. 5th 51. D 52. H 1–1• 3 n–1 4 f. 14,197 16,384 45. x; x5, x6, x7 46. 3x; 27x4, 81x5, 243x6 47. xy2; x5y9, x6y11, x7y13 48. ab; 2a4b2, 2a5b3, 2a6b4 53. B 54. [2] Day 1 350 Day 2 700 Day 3 1400 Day 4 2800 Day 5 5600 (OR equivalent explanation) [1] answer with no work shown 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 55. a4 66. y = – 2 x 56. 16 57. 101 20 x z 58. 1 59. 60. 61. 62. 67. y = 68. y = 69. y = 81 64 1 m14 1 p15 d5 c2 70. y = 71. y = 72. y = 5 – 7x 6 5x 3 7x 4 – 1x 4 5x 9 –1x 2 63. 2.467 10–3 64. 3.26 1011 gal 65. y = 8 x 3 8-6 Geometric Sequences ALGEBRA 1 LESSON 8-6 1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . . –2 2. Find the next three terms of the sequence 243, 81, 27, 9, . . . 1 3, 1, 3 3. Determine whether each sequence is arithmetic or geometric. a. 37, 34, 31, 28, . . . arithmetic b. 8, –4, 2, –1, . . . geometric 4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1. 4, 2500, 1,562,500 5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so that its dimensions are 20% larger than its original size. Write a rule for the length of the copies. What will be the length if you enlarge the photograph five times? (Hint: The common ratio is not just 0.2. You must add 20% to 100%.) A(n) = 6(1.2)n-1; about 14.9 in. 8-6 Exponential Functions ALGEBRA 1 LESSON 8-7 (For help, go to Lesson 5-6.) Graph each function. 1. y = 3x 2. y = 4x 3. y = –2x Simplify each expression. 4. 32 5. 5–3 6. 2 • 34 7. 2 • 3–2 8. 3 • 2–1 9. 10 • 32 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 Solutions 1. y = 3x 3. y = –2x 2. y = 4x 4. 32 = 3 • 3 = 9 5. 5–3 = 13 = 5 6. 2 • 34 = 2 • (3 • 3 • 3 • 3) = 2 • 81 = 162 7. 2 • 3–2 = 2 • 12 = 2 • 1 = 2 3 9 9 8. 3 • 2-1 = 3 • 1 = 3 • 1 = 3 or 1 1 21 2 2 2 9. 10 • 32 = 10 • (3 • 3) = 10 • 9 = 90 8-7 1 = 1 5 • 5 • 5 125 Exponential Functions ALGEBRA 1 LESSON 8-7 Evaluate each exponential function. a. y = 3x for x = 2, 3, 4 x 2 3 4 y = 3x 32 = 9 33 = 27 34 = 81 y 9 27 81 b. p(q) = 3 • 4q for the domain {–2, 3} q p(q) = 3 • 4q –2 3 • 4–2 = 3 • 16 = 16 3 3 • 43 = 3 • 64 = 192 192 1 p(q) 3 3 16 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 Suppose two mice live in a barn. If the number of mice quadruples every 3 months, how many mice will be in the barn after 2 years? ƒ(x) = 2 • 4x ƒ(x) = 2 • 48 In two years, there are 8 three-month time periods. ƒ(x) = 2 • 65,536 Simplify powers. ƒ(x) = 131,072 Simplify. 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 Graph y = 2 • 3x. x y = 2 • 3x –2 2 • 3–2 = 3 2 = –1 2 • 3–1 = (x, y) 2 2 31 = 2 9 (–2, 29 ) 2 3 (–1, 3 ) 2 0 2 • 30 = 2 • 1 = 2 (0, 2) 1 2 • 31 = 2 • 3 = 6 (1, 6) 2 2 • 32 = 2 • 9 = 18 (2, 18) 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 The function ƒ(x) = 1.25x models the increase in size of an image being copied over and over at 125% on a photocopier. Graph the function. x ƒ(x) = 1.25x (x, ƒ(x)) 1 1.251 = 1.25 2 1.252 = 1.5625 1.6 (2, 1.6) 3 1.253 = 1.9531 2.0 (3, 2.0) 4 1.254 = 2.4414 2.4 (4, 2.4) 5 1.255 = 3.0518 3.1 (5, 3.1) 1.3 (1, 1.3) 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 pages 432–435 Exercises 12. A 1. 216 13. C 2. 2 14. B 9 3. 2.5 4. 32 20. 15. B 16. D 21. 17. C 5. 4.5 18. A 6. 115.2 19. 22. 7. 1600 8. 0.576 9. $160,000; $320,000 10. $2000; $4000 11. $16,000, $32,000 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 23. 27. 100, 10, 1, 0.1, 0.01, 0.001; decrease 28. 0.3125, 1.25, 5, 20, 80, 320; increase 29. 4, 2, 1, 0.5, 0.25, 0.125; decrease 30. 9 , 3 , 1, 2 , 4 , 8 ; decrease 4 2 24. 3 9 27 31. 0.04, 0.4, 4, 40, 400, 4000; increase 32. 1111.1, 333.3, 100, 30, 9, 2.7; decrease 33. a. Answers may vary. Sample: y = 2(3)x b. Answers may vary. Sample: y = 2(0.3)x 25. 0.04, 0.2, 1, 5, 25, 125; increase 26. 0.16, 0.4, 1, 2.5, 6.25, 15.625; increase 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 34. a. b. y = 75 • 2x, where x is the number of 20-min time periods 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 35. a. b. (0, 1) c. No; there is no value of x for which y = 0. d. If the base is > 1, the graph gets steeper as the base increases. If the base is < 1, the graph gets steeper as the base decreases. 36. a. 1,000,000,000 plants b. 1,000,000,000,000 plants x y 1 –2 2 4 3 –8 4 16 5 –32 b. Every other value is negative. The absolute value of each term is double the previous term. c. No; in y = a • bx, b > 0. –2 < 0, so it is not exponential. 38. y = x5 37. a. 39. ƒ(t) = 200 • t 2 40. y = 3x 41. ƒ(x) = 100x2 42. {500}; b = 1 produces a linear graph. 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 43. a. 46. 6 55. [2] 47. 3 48. 4 49. 5 The graphs intersect between x = 1 and x = 2 (OR equivalent explanation). 50. a. 4 b. Between x = 1 and x = 3, the graph of y = x2 rises faster than the graph of y = 2x. The graphs intersect at x = 2. c. The graph of y = 6x is steeper than y = x2 and y = 2x. 44. 2 b. 3 c. y = 4 • 3x [1] answer with no work shown d. 4 , 324 9 51. B 52. G 53. A 54. H 45. –3 8-7 56. 5; 1250, 6250, 31,250 57. –3; 567, –1701, 5103 58. 2; –3.2, –6.4, –12.8 59. – 1 ; 1 , – 1 , 1 3 3 9 27 Exponential Functions ALGEBRA 1 LESSON 8-7 60. 0.1; 0.045, 0.0045, 0.00045 61. 0.25; 28, 7, 1.75 62. y = 5x 63. y = 3x + 1 64. y = –2x + 8 65. y = 0.4x – 3.8 8-7 Exponential Functions ALGEBRA 1 LESSON 8-7 1. Evaluate each function rule for the given value. a. y = 0.5x for x = 3 0.125 b. ƒ(x) = 4 • 3x for x = –2 4 9 2. Suppose an investment of $5000 doubles every 12 years. a. How much is the investment worth after 24 years? $20,000 b. After 48 years? 3. Graph y = 0.5 • 3x. $80,000 4. Graph y = –0.5 • 3x. 8-7 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 (For help, go to Lesson 4-3.) Use the formula I = prt to find the interest for principal p, interest rate r, and time t in years. 1. principal: $1000; interest rate: 5%; time: 2 years 2. principal: $360; interest rate: 6%; time: 3 years 3. principal: $2500; interest rate: 4.5%; time: 2 years 4. principal: $1680; interest rate: 5.25%; time: 4 years 5. principal: $1350; interest rate: 4.8%; time: 5 years 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 Solutions 1. l = prt = ($1000)(0.05)(2) = $50(2) = $100 2. l = prt = ($360)(0.06)(3) = $21.60(3) = $64.80 3. l = prt = ($2500)(0.045)(2) = $112.50(2) = $225 4. l = prt = ($1680)(0.0525)(4) = $88.20(4) = $352.80 5. l = prt = ($1350)(0.048)(5) = $64.80(5) = $324 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 In 1998, a certain town had a population of about 13,000 people. Since 1998, the population has increased about 1.4% a year. a. Write an equation to model the population increase. Relate: y = a • bx Define: Write: Use an exponential function. Let x = the number of years since 1998. Let y = the population of the town at various times. Let a = the initial population in 1998, 13,000 people. Let b = the growth factor, which is 100% + 1.4% = 101.4% = 1.014. y = 13,000 • 1.014x 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 (continued) b. Use your equation to find the approximate population in 2006. y = 13,000 • 1.014x y = 13,000 • 1.0148 14,529 2006 is 8 years after 1998, so substitute 8 for x. Use a calculator. Round to the nearest whole number. The approximate population of the town in 2006 is 14,529 people. 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 Suppose you deposit $1000 in a college fund that pays 7.2% interest compounded annually. Find the account balance after 5 years. Relate: y = a • bx Define: Write: Use an exponential function. Let x = the number of interest periods. Let y = the balance. Let a = the initial deposit, $1000 Let b = 100% + 7.2% = 107.2% = 1.072. y = 1000 • 1.072x = 1000 • 1.0725 1415.71 Once a year for 5 years is 5 interest periods. Substitute 5 for x. Use a calculator. Round to the nearest cent. The balance after 5 years will be $1415.71. 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 Suppose the account in the above problem paid interest compounded quarterly instead of annually. Find the account balance after 5 years. Relate: y = a • bx Define: Use an exponential function. Let x = the number of interest periods. Let y = the balance. Let a = the initial deposit, $1000 Let b = 100% + 7.2% There are 4 interest periods in 4 1 year, so divide the interest into 4 parts. = 1 + 0.018 = 1.018 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 (continued) Write: y = 1000 • 1.018x = 1000 • 1.01820 1428.75 Four interest periods a year for 5 years is 20 interest periods. Substitute 20 for x. Use a calculator. Round to the nearest cent. The balance after 5 years will be $1428.75. 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 Technetium-99 has a half-life of 6 hours. Suppose a lab has 80 mg of technetium-99. How much technetium-99 is left after 24 hours? In 24 hours there are four 6-hour half lives. After one half-life, there are 40 mg. After two half-lives, there are 20 mg. After three half-lives, there are 10 mg. After four half-lives, there are 5 mg. 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 Suppose the population of a certain endangered species has decreased 2.4% each year. Suppose there were 60 of these animals in a given area in 1999. a. Write an equation to model the number of animals in this species that remain alive in that area. Relate: y = a • bx Define: Write: Use an exponential function. Let x = the number of years since 1999 Let y = the number of animals that remain Let a = 60, the initial population in 1999 Let b = the decay factor, which is 100% - 2.4 % = 97.6% = 0.976 y = 60 • 0.976x 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 (continued) b. Use your equation to find the approximate number of animals remaining in 2005. y = 60 • 0.976x y = 60 • 0.9766 52 2005 is 6 years after 1999, so substitute 6 for x. Use a calculator. Round to the nearest whole number. The approximate number of animals of this endangered species remaining in the area in 2005 is 52. 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 pages 441–444 Exercises 10. 1.005 1. 20; 2 11. 0.75%, 0.25% 21. a. 3 half-lives b. 3.125 mCi 2. 200; 1.0875 12. 1%; 0.3% 22. 0.5 3. 10,000; 1.01 13. 1.125%; 0.375% 23. 0.1 4. 1; 1.5 14. 1.9%; 0.63% 24. 2 5. a. 50,000 b. 0.03; 1.03 c. 1.03 d. 50,000; 1.03; x e. about 104,689 people 6. 1.04 15. 1.5625%; 0.52083% 25. 0.9 26. exp. growth 16. $5352.90 7. 1.05 8. 1.037 9. 1.0875 3 17. $16,661.35 27. exp. decay 18. $634.87 28. exp. growth 19. $28,338.18 29. exp. decay 20. a. 4 half-lives b. 2.5 mCi 30. a. $22,000; 0.8 b. y = 22,000 • (0.8)x c. $5767.17 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 31. y = 130,000 • (1.01)x; about 142,179 people 40. linear function 32. y = 3,000,000 • (0.985)x; about 2,579,191 people 33. y = 2400 • (1.07)x; $4721.16 41. exponential function 34. y = 2400 • (1.00583333)x; $4823.19 35. a. y = 584 • (1.065)x; $2057.81 b. Check students’ work. 36. Linear function; it is a straight line. 37. Neither; it is not just one straight line. 38. Exponential function; it is a curve with y-values that increase as x-values increase. 39. Neither; it decreases and then increases, unlike an exponential function. 8-8 42. linear function Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 43. Answers may vary. Sample: $600; even after 10 years, there is more money in the account with an initial deposit of $600 ($977.34) than there is in the account with an initial deposit of $500 ($907.01). 44. 6 half-lives 48. 94% 49. 88% 50. 96.5% 51. 46.1% 52. a. 2 years b. 4 years 45. 4 half-lives 46. a. b. c. about 4 h 1 4 about 3.7 mg using the function and 53. a. b. c. d. e. f. $220.00 $3.96 $223.96 $193.96 9 months $18.07 15 mg 4 = 3.75 mg using the prediction 54. Check students’ work. x 47. a. y = 6,284,000 • (1.01) 55. 2003 b. 7,667,674 people 1 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 56. C 61. 57. H 58. A 59. [2] Using $1000 for deposit, quarterly: 1000(1 + 0.05 )20 1282.04; annually: 4 1000(1 + 0.055)5 1306.96. The account paying 5.5% will be greater. (OR equivalent explanation) [1] correct approach with minor computational error 60. 62. 63. 7.28 1011 gal 8-8 Exponential Growth and Decay ALGEBRA 1 LESSON 8-8 1. Identify the original amount a and the growth factor b in the exponential function y = 10 • 1.036x. a = 10, b = 1.036 2. A population of 24,500 people has been increasing at a rate of 1.8% a year. What will be the population in 15 years if it continues at that rate? about 32,017 people 3. Write an exponential function to represent $2000 principal earning 5.6% interest compounded annually. y = 2000 • 1.056x 4. Find the account balance on $3000 principal earning 6.4% interest compounded quarterly for 7 years. about $4678.91 5. The half-life of a certain substance is 4 days. If you have 100 mg of the substance, how much of it will remain after 12 days? 12.5 mg 6. The value of a $1200 computer decreases 27% annually. What will be the value of the computer after 3 years? about $466.82 8-8 Exponents and Exponential Functions ALGEBRA 1 CHAPTER 8 3 1. r12 t 4 2. m12 a 5 3. m8 t 4. c 2v 9 5. 6. 7. 8. h 2d 2 k3 9j 2 y5 p 35 w 14 h 26 32y 9 9. 1.728 10. –27 q 11. C 12. 4.4909 107 3 20. a. A(n) = 12 • 5 votes 13. 4.5 105 reptiles b. 2.592 ft 14. No; 76 > 10. 21. –192 15. yes 22. 15, 75, 375 16. no; two powers of 10 23. 2, 8, 32 17. No; 32.5 > 10. 18. a. about 6.7068 108 mi b. about 1.5 h 24. 3.8, 3.61, 3.4295 25. 3.75, 2.8125, 2.109375 26. 19. a. – 1 2 b. –2, 1, – 1 2 c. A(n) = –32 – 1 d. – 1 8 n–1 n–1 2 8-A Exponents and Exponential Functions ALGEBRA 1 CHAPTER 8 27. 30. growth for b > 1, decay for 0 < b < 1 31. $1082.86; $1220.19 28. 32. a. b. c. d. growth; 1.07 about 7.1 kw-h about 0.66 kw-h Check students’ work. 33. a. 0.85; the car’s value depreciates 15% annually. b. $17,000 c. $10,440 29. Answers may vary. Sample: A computer loses 20% of its value each year. How much will a $3500 computer be worth in 3 years? $1792 34. a. 8% b. $10,000; $12,597.12 c. Check students’ work. 8-A