Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 (For help, go to Lessons 1-2 and 1-6.) Simplify each expression. 1 42 1. 23 2. 3. 42  22 4. (–3)3 5. –33 6. 62  12 Evaluate each expression for a = 2, b = –1, c = 0.5. a 2a bc c ab bc 7. 8. 9. 8-1

Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 1. 23 = 2 • 2 • 2 = 8 2. = = 3. 42  22 = = = = 4 4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27 5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27 6. 62  12 = 36  12 = 3 7. for a = 2: = 8. for b = –1, c = 0.5: = –1 9. for a = 2, b = –1, c = 0.5: = = 4 Solutions 1 42 1 4 • 4 1 16 42 22 4 • 4 2 • 2 16 4 a 2a 2 2 • 2 1 2 bc c –1 • 0.5 0.5 ab bc 2 • (–1) (–1) • 0.5 2 0.5 8-1

Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Simplify. = Use the definition of negative exponent. 1 32 a. 3–2 Simplify. 1 9 = b. (–22.4)0 Use the definition of zero as an exponent. = 1 8-1

Zero and Negative Exponents ALGEBRA 1 LESSON 8-1 Simplify 1 x –3 a. 3ab –2 1 b2 Use the definition of negative exponent. = 3a b. Rewrite using a division symbol. = 1  x –3 = 1  1 x 3 Use the definition of negative exponent. Simplify. 3a b 2 = = 1 • x 3 Multiply by the reciprocal of , which is x 3. 1 x3 = x 3 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 2y –3 = Use the definition of negative exponent. 4x 2 y 3 Substitute 3 for x and –2 for y. 4(3)2 (–2)3 = 36 –8 –4 1 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 Substitute 3 for x and –2 for y. 4(3)2 (–2)3 = Use the definition of negative exponent. 36 –8 –4 1 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   Substitute 0 for m. = 3000 • 1 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 • Simplify. = 750 1 4 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 Simplify each expression. 1. 3–4 2. (–6)0 3. –2a0b–2 4. 5. 8000 • 40 6. 4500 • 3–2 1 81 1 2 b2 – k m–3 km3 8000 500 8-1

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 • 5 • 5 • s • s • s = 53 • s3 = 53s3 5. –54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625 6. (–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625 7. (–5)0 = 1 8. (–5)–4 = (– )4 = (– )(– )(– )(– ) = ( )( ) = 1 5 1 5 1 5 1 5 1 5 1 25 1 25 1 625 8-3

Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Rewrite each expression using each base only once. Add exponents of powers with the same base. 73 + 2 = a. 73 • 72 = 75 Simplify the sum of the exponents. Think of 4 + 1 – 2 as 4 + 1 + (–2) to add the exponents. 44 + 1 – 2 = b. 44 • 41 • 4–2 = 43 Simplify the sum of the exponents. Add exponents of powers with the same base. 68 + (–8) = c. 68 • 6–8 = 60 Simplify the sum of the exponents. Use the definition of zero as an exponent. = 1 8-3

Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify each expression. a. p2 • p • p5 Add exponents of powers with the same base. p 2 + 1 + 5 = = p 8 Simplify. 4x6 • 5x–4 b. Commutative Property of Multiplication (4 • 5)(x 6 • x –4) = Add exponents of powers with the same base. = 20(x 6+(–4)) Simplify. = 20x 2 8-3

Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-3 Simplify each expression. a. a 2 • b –4 • a 5 Commutative Property of Multiplication a 2 • a 5 • b –4 = = a 2 + 5 • b –4 Add exponents of powers with the same base. Simplify. a 7 b 4 = Commutative and Associative Properties of Multiplication (2 • 3 • 4)(p 3)(q • q 4) = b. 2q • 3p3 • 4q4 = 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) = Commutative and Associative Properties of Multiplication (3 • 7)(10–3 • 10–5) = 21  10–8 Simplify. = 2.1  101 • 10–8 Write 21 in scientific notation. = 2.1  101 + (– 8) Add exponents of powers with the same base. = 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 seconds kilometers hour • Use dimensional analysis. = (3  108) • (1  10–3) • (3.6  103) m s km h Substitute. = (3 • 1 • 3.6)  (108 • 10–3 • 103) Commutative and Associative Properties of Multiplication = 10.8  (108 + (– 3) + 3) 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 Simplify each expression. 1. 34 • 35 2. 4x5 • 3x–2 3. (3  104)(5  102) 4. (7  10–4)(1.5  105) 5. (–2w –2)(–3w2b–2)(–5b–3) 6. What is 2 trillion times 3 billion written in scientific notation? 39 12x3 1.5  107 1.05  102 30 b5 – 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 = 8. n–3 • n–3 = n(–3 + (–3)) = n–6 = 1 y 6 1 n 6 8-4

More Multiplication Properties of Exponents ALGEBRA 1 LESSON 8-4 Simplify (a3)4. Multiply exponents when raising a power to a power. (a3)4 = a3 • 4 Simplify. = a12 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. Simplify. = b–4 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 Simplify each expression. 1. (x4)5 2. x(x5y–2)3 3. (5x4)3 4. (1.5  105)2 5. (2w–2)4(3w2b–2)3 6. (3  10–5)(4  104)2 x16 y6 x20 2.25  1010 125x12 432 b6w2 4.8  103 8-4

Division Properties of Exponents ALGEBRA 1 LESSON 8-5 (For help, go to Skills Handbook page 724.) Write each fraction in simplest form. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 5 20 125 25 60 100 124 4 6 15 8 30 10 35 18 63 5xy 15x 6y2 3x 3ac 12a 24m 6mn2 8-5

Division Properties of Exponents ALGEBRA 1 LESSON 8-5 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. = = 5 20 5 • 1 5 • 4 1 4 125 25 25 • 5 25 • 1 = = 5 60 100 20 • 3 20 • 5 3 = = 124 = = 31 4 • 31 4 • 1 6 15 2 3 • 2 3 • 5 = = 8 30 2 • 4 2 • 15 10 35 7 5 • 2 5 • 7 = = 18 63 9 • 2 9 • 7 5xy 15x y 5 • x • y 5 • 3 • x = = 6y2 3x 2y2 x 3 • 2 • y2 3 • x = = 3ac 12a 3 • a • c 3 • 4 • a c 24m 6mn2 6 • 4 • m 6 • m • n2 n2 = = Solutions 8-5

Division Properties of Exponents ALGEBRA 1 LESSON 8-5 Simplify each expression. x4 x9 = Subtract exponents when dividing powers with the same base. x4 – 9 a. Simplify the exponents. = x–5 Rewrite using positive exponents. 1 x5 = p3 j –4 p–3 j 6 = Subtract exponents when dividing powers with the same base. p3 – (–3)j –4 – 6 b. = p6 j –10 Simplify. Rewrite using positive exponents. p6 j10 = 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? 64 million beats 530 thousand min 6.4  107 beats 5.3  105 min = Write in scientific notation. 6.4 5.3  107–5 = Subtract exponents when dividing powers with the same base. 6.4 5.3  102 = Simplify the exponent. 1.21  102 Divide. Round to the nearest hundredth. = 121 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 y 3 4 Simplify . 3 y 3 4 34 (y 3)4 = 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 2 3 –3 a. Simplify . 2 3 –3 = Rewrite using the reciprocal of . 33 23 = Raise the numerator and the denominator to the third power. Simplify. 27 8 3 or = 8-5

Division Properties of Exponents ALGEBRA 1 LESSON 8-5 (continued) 4b c – b. Simplify . 4b c – –2 2 = Rewrite using the reciprocal of . Write the fraction with a negative numerator. c 4b – 2 = Raise the numerator and denominator to the second power. (–c)2 (4b)2 = Simplify. c2 16b2 = 8-5

Division Properties of Exponents ALGEBRA 1 LESSON 8-5 Simplify each expression. 1. 2. 3. 4. 5. 6. a8 a–2 w3 w7 1 w4 (3a)4(2a–2) 6a2 a10 27 1.6  103 4  10–2 24 5 2 –3 4  104 256 25 or 10 6 4x 3 3x 2 27 64x3 8-5

Find the common difference of each sequence. 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

Common difference: 2 17 – 19 = –2 Common difference: –2 Geometric Sequences ALGEBRA 1 LESSON 8-6 Solutions 1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ... 7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2 13 – 15 = –2, 15 – 17 = –2, Common difference: 2 17 – 19 = –2 Common difference: –2 3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ... –2.3 – (–1.1) = –1.2, –1.1 28.5 – 25 = 3.5, 25 – 21.5 = 3.5, – 0.1 = –1.2, 0.1 – 1.3 = –1.2 21.5 – 18 = 3.5 Common difference: –1.2 Common difference: 3.5 8-6

Solutions (continued) 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... Geometric Sequences ALGEBRA 1 LESSON 8-6 Solutions (continued) 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... 2(2) = 4, 4(2) = 8, 8(2) = 16, 4(3) = 12, 12(3) = 36, 16(2) = 32, 32(2) = 64 36(3) = 108, 108(3) = 324 Next two numbers: 32, 64 Next two numbers: 108, 324 7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ... (0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 200  2 = 100, 100  2 = 50, 1.6, 1.6(2) = 3.2, 3.2(2) = 6.4 50  2 = 25, 25  2 = 12.5, Next two numbers: 3.2, 6.4 12.5  2 = 6.25 8-6

(–5) (–5) (–5)  Geometric Sequences ALGEBRA 1 LESSON 8-6 Find the common ratio of each sequence. a. 3, –15, 75, –375, . . . 3 –15 75 –375 (–5) (–5) (–5) The common ratio is –5. b. 3, 3 2 4 8 , , ... 3 2 4 8  1 The common ratio is . 1 2 8-6

Find the next three terms of the sequence 5, –10, 20, –40, . . . Geometric Sequences ALGEBRA 1 LESSON 8-6 Find the next three terms of the sequence 5, –10, 20, –40, . . . 5 –10 20 –40 (–2) (–2) (–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 Determine whether each sequence is arithmetic ALGEBRA 1 LESSON 8-6 Determine whether each sequence is arithmetic or geometric. a. 162, 54, 18, 6, . . . 62 54 18 6 1 3  The sequence has a common ratio. The sequence is geometric. 8-6

The sequence has a common difference. Geometric Sequences ALGEBRA 1 LESSON 8-6 (continued) b. 98, 101, 104, 107, . . . 98 101 104 107 + 3 The sequence has a common difference. The sequence is arithmetic. 8-6

first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3 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

The first term is 2 meters, which is 200 cm. 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

A rule for the sequence is A(n) = 200 • 0.75n – 1. 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

2. Find the next three terms of the sequence 243, 81, 27, 9, . . . Geometric Sequences ALGEBRA 1 LESSON 8-6 1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . . 2. Find the next three terms of the sequence 243, 81, 27, 9, . . . 3. Determine whether each sequence is arithmetic or geometric. a. 37, 34, 31, 28, . . . b. 8, –4, 2, –1, . . . 4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1. 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%.) –2 3, 1, 1 3 arithmetic geometric 4, 2500, 1,562,500 A(n) = 6(1.2)n-1; about 14.9 in. 8-6