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Lesson 4.6 Applying the Exponent Laws A 3. Use the product of powers law: When the bases are the same, add the exponents. a) x3 x 4 x3 4 x7 b) a 2 a 5 a 2 5 a 3 1 3 a Write with a positive exponent. c) b 3 b5 b 3 5 b2 d) m2 m 3 m 2 3 m1 1 m Write with a positive exponent. 4. a) 0.52 0.53 Use the product of powers law: When the bases are the same, add the exponents. 0.52 0.53 0.52 3 0.55 b) 0.52 0.53 Use the product of powers law: When the bases are the same, add the exponents. 0.52 0.53 0.52 ( 3) 0.51 Write with a positive exponent. = 1 0.5 Exercises (pages 241–243) c) 0.52 0.53 Use the quotient of powers law: When the bases are the same, subtract the exponents. 0.52 0.52 3 0.53 0.51 Write with a positive exponent. d) 1 0.5 0.52 0.53 Use the quotient of powers law: When the bases are the same, subtract the exponents. 0.52 0.52 ( 3) 3 0.5 0.55 5. Use the quotient of powers law: When the bases are the same, subtract the exponents. x4 x4 2 a) x2 x2 b) x2 x2 5 5 x x 3 Write with a positive exponent. = c) 1 x3 n 6 n5 n 6 5 n1 n d) a2 a2 6 a6 a 4 Write with a positive exponent. = 1 a4 6. Use the power of a power law: multiply the exponents a) n 2 3 n(2)(3) n6 b) z 2 3 z (2)( 3) z 6 1 z6 = c) n 4 3 Write with a positive exponent. n( 4)( 3) n12 d) c 2 2 c ( 2)(2) c 4 Write with a positive exponent. 1 c4 = 7. Use the power of a power law: multiply the exponents 4 (3)(4) 3 3 3 a) 5 5 12 3 5 3 3 b) 5 4 3 5 (3)( 4) 3 5 12 Write with a positive exponent. 12 5 = 3 c) 3 3 5 4 3 5 ( 3)( 4) 12 3 5 3 3 d) 5 4 3 5 ( 3)( 4) 12 3 5 8. a) Use the power of a quotient law. 2 a2 a b2 b b) Use the power of a quotient law. n2 n2 m3 m 3 3 Use the power of a power law. n(2)(3) m3 n6 3 m c2 2 d 4 d2 2 c d c 4 Writing with a positive exponent 2 4 2 4 d (2)(4) c (2)(4) d8 8 c c) Using the power of a quotient law Using the power of a power law d) Use the power of a quotient law. 2b 2b 2 5c 5c 2 2 22 b 2 52 c 2 4b 2 25c 2 Use the power of a product law. e) ab 2 a 2b 2 Using the power of a product law n m n m 2 3 2 3 3 Using the power of a product law n(2)(3) m3 6 =nm f) c d 3 2 4 Using the power of a power law 3 c3 d 2 4 4 Using the power of a product law c (3)( 4) d (2)( 4) Using the power of a power law = c 12 d 8 Write with positive exponents. = g) 1 c d8 12 xy x y 1 3 1 3 3 Using the power of a product law x3 y ( 1)(3) Using the power of a power law = x3 y 3 Write with a positive exponent. 3 = h) B 9. a) x y3 x 3 x 4 x 3 4 Using the product of powers law x1 x b) a 4 a 1 a 4 ( 1) a c) 5 Write with a positive exponent. 1 a5 b4 b3 b2 b4 ( 3) 2 b d) Using the product of powers law Using the product of powers law 3 m8 m2 m6 m8 ( 2) ( 6) m 1 0 Using the product of powers law e) x 5 x 5 2 x2 x 7 f) g) 1 x7 Using the quotient of powers law b 8 b 8 ( 3) b 3 b 5 Using the quotient of powers law Write with a positive exponent. 1 b5 t 4 t 4 ( 4) t 4 t0 =1 3 10. a) Write with a positive exponent. s5 s 5 ( 5) s 5 s10 = h) Using the quotient of powers law 1 Using the quotient of powers law 3 1.5 2 1.5 2 1.5 2 1.5 1 2 Using the product of powers law 4 2 1.52 2.25 3 3 4 b) 4 5 3 3 4 3 4 4 4 8 3 4 4 3 4 9 16 2 5 4 Using the product of powers law 1 c) 5 1 0.6 3 0.6 3 0.6 3 0.6 6 3 0.6 2 5 3 Using the product of powers law 0.36 4 4 3 d) 5 4 5 4 3 4 4 3 5 4 5 1 e) 0.6 0.6 1 2 3 2 1 0.6 2 0.6 3 2 4 3 Using the product of powers law 0 Using the quotient of powers law 2 2 0.61 Write with a positive exponent. 1 0.6 1 6 10 10 6 5 3 f) 2 3 3 2 1 8 3 3 3 1 8 3 3 8 3 3 3 8 1 3 8 3 8 Using the quotient of powers law. g) 5 5 4 0.49 2 2 0.49 4 0.49 5 3 2 0.49 2 0.49 8 2 Write with a positive exponent. 1 Using the quotient of powers law 3 0.49 2 1 0.49 3 1 0.73 1 0.343 1 343 1000 1000 343 5 h) 0.027 3 0.027 4 3 5 0.027 3 4 3 Using the quotient of powers law 1 0.027 3 3 0.027 0.3, or 11. a) x 1 y 2 3 3 10 x 1 y 2 3 3 = x ( 1)( 3) y ( 2)( 3) = x3 y 6 x3 y 6 Using the power of a product law Using the power of a power law b) c) 2a 2 2 2 b 4m n 2 3 3 21 a 2 b 2 2 2 2 Using the power of a product law = 2(1)( 2) a ( 2)( 2) b(2)( 2) Using the power of a power law = 22 a 4 b 4 Write with positive exponents. 4 a 2 b4 a4 4b 4 2 41 m 2 n3 3 3 3 = 4(1)( 3) m(2)( 3) n(3)( 3) 3 6 =4 m n 1 3 4 m6 n9 9 Using the power of a product law Using the power of a power law Write with positive exponents. 1 64m6 n9 d) 3 2 3 m n 2 4 4 31 4 4 m 2 n 3 2 3 = 2 Using the power of a product law (1)( 4) m( 2)( 4) n( 3)( 4) Using the power of a power law 4 3 = m8 n12 2 4 2 m8 n12 3 16 m8 n12 81 16m8 n12 81 Write with a positive exponent. 12. The volume of a cone with base radius r and height h is given by the formula: 1 V πr 2 h 3 The cone has equal height and radius. So, substitute: r = h 1 V πh 2 h 3 1 πh 2 h1 Use the product of powers law. 3 1 πh 2 1 3 1 3 πh 3 Substitute V = 1234, then solve for h. 1 1234 πh3 Multiply each side by 3. 3 1 3 1234 3 πh3 3 3 3702 πh Divide each side by π. 3702 πh3 π π 3702 h3 π To solve for h, take the cube root of each side by raising each side to the one-third power. 1 3 3702 3 3 h π 1 1 Use the power of a power law. 3702 3 Use a calculator. h π h 10.5623... The height of the cone is approximately 10.6 cm. 4 13. The volume, V, of a sphere with radius r is given by the formula: V πr 3 3 Substitute V = 375, then solve for r. 4 375 πr 3 Multiply each side by 3. 3 4 3 375 3 πr 3 3 3 1125 4πr Divide each side by 4π. 1125 4πr 3 4π 4π 1125 r3 4π To solve for r , take the cube root of each side by raising each side to the one-third power. 1 3 1125 3 3 r 4π 1 Use the power of a power law. 1 1125 3 r 4π r 4.4735... Use a calculator. The surface area, SA, of a sphere with radius r is given by the formula: Substitute: r 4.4735... SA 4πr 2 4π(4.4735...) 2 251.4808... The surface area of the sphere is approximately 251 square feet. a b a b 2 1 2 14. a) 3 3 a 2( 2) b( 1)( 2) a ( 3)(3) b(1)(3) Using the power of a power law = a 4 b 2 a 9 b3 Use the quotient of powers law. a 4 ( 9) b 2 3 a 5 b 1 5 a b Write with a positive exponent. b) c 3 d 1 c2 d 2 c ( 3)( 1) d (1)( 1) c2d c3 d 1 = 2 cd Using the power of a power law inside the large brackets 2 Use the quotient of powers law. c3 2 d 1 1 c1 d 2 2 2 2 Use the power of a power law. c1( 2) d ( 2)( 2) c 2 d 4 Write with a positive exponent. 4 = d c2 15. a) Simplify first. a3b2 a 2b3 a3 b2 a 2 b3 a 3 a 2 b 2 b3 =a 32 b Use the product of powers law. 23 a 5b5 Substitute: a = –2, b = 1 a 5b5 (2)5 1 5 (32)(1) 32 b) Simplify first. a 1b2 a 2b3 a1 b2 a2 b3 a 1 a 2 b 2 b 3 =a 1 ( 2) 3 5 a b 1 ab Substitute: a = –2, b = 1 1 1 3 5 ab (2)3 (1)5 1 (8)(1) 1 8 = c) Simplify first. 3 5 b Use the product of powers law. 2 ( 3) Write with positive exponents. a 4b5 a 4 1 b5 3 ab3 = a 5 b 2 Using the quotient of powers law Write with a positive exponent. 2 b a5 Substitute: a = –2, b = 1 b2 12 a 5 2 5 = 1 32 d) Simplify first. a 7 b7 9 10 a b 5 a 7 ( 9) b7 10 = a 2 b 3 a2 = 3 b 5 b3 = 2 a b3(5) = 2(5) a b15 10 a 5 5 5 Using the quotient of powers law inside the brackets Write the expression inside the brackets with a positive exponent. Write with a positive exponent. Use the power of a power law. Substitute: a = –2, b = 1 b15 115 10 10 a 2 2 1 1024 4 2 16. a) m 3 m 3 m 3 =m 6 3 m2 4 3 Using the product of powers law b) x 3 2 x 1 4 x =x x 3 1 2 4 Using the quotient of powers law 6 1 4 4 5 4 Write with a positive exponent. 1 5 x4 c) 3 9a 4b 4 2 3a b 1 4 3 9 a 4 b 4 1 3 a2 b4 Use the quotient of powers law. 3 = 3 a 4 2 b 4 1 4 2 = 3 a 6 b 4 1 3 a 6 b 2 Write with a positive exponent. 1 2 3b a6 d) 1 1 64c 6 3 3 64 1 1 c 6 1 9 12 a9 b 2 ab Simplify inside the brackets first. 1 1 3 = 64 a 9 b 2 c 6 1 3 (64) a 1 9 3 1 3 b 1 1 2 3 Use the power of a power law. c 1 6 3 1 6 (64) a 3 b c 2 1 4 a 3 b 6 c 2 1 4b 6 c 2 = a3 Write with a positive exponent. 17. a) In the second line, the exponents were multiplied instead of added. A correct solution is: 1 1 x2 y 3 x 2 y 1 x2 x 2 y 3 y 1 Use the product of powers law. x 2 1 2 4 1 2 x2 y 3 ( 1) y 3 ( 1) 5 2 x y 4 Write with a positive exponent. 5 2 = x y4 b) In the first line, –5 was multiplied by –2 instead of being raised to the power –2. A correct solution is: 5a 2 12 b 2 5 b 2 a 2( 2) 1 2 2 5 2 a 4 b 1 Using the power of a power law Write with positive exponents. b1 5 2 a4 b 25 a 4 18. I record the volume of water in the measuring cylinder, in millilitres. Then I carefully place the marble in the cylinder. The water level rises; this is the total volume of the water and the marble. I record this volume. I subtract the volume of the water alone from the total volume of the water and the marble to determine the volume of the marble in millilitres. Since 1 mL = 1 cm3, I can write the volume of the marble in cubic centimetres. 4 The volume, V, of a sphere with radius r is given by the formula: V πr 3 3 I substitute the volume of the marble in cubic centimetres for V. I then multiply both sides of the equation by 3, then divide both sides by 4π . To solve for r, 1 I raise each side to the one-third power. I then use the power of a power law to write r 3 3 as r. Once I have determined r, I multiply the radius by 2 to get the diameter of the marble. 19. a) There are two errors in the first line: the quotient of powers law was used before the power of a power law, and when using the quotient of powers law, the powers outside the brackets were subtracted. There is an error in the second line: when using the power of a power law, the product of 5 and –6 should have been –30. A correct solution is: m m 3 2 n2 n 4 3 2 m( 3)( 4) n 2( 4) m 2(2) n 3(2) Using the power of a power law m12 n 8 m 4 n 6 Use the quotient of powers law. = m12 4 n 8 ( 6) m8 n 2 Write with a positive exponent. m8 n2 b) There is an error in the first line: when using the power of a power law, the exponents added when they should have been multiplied. A correct solution is: 1 12 23 2 r s 1 1 1 1 1 r 4 s 2 r 2 2 s 2 2 r 4 s 2 1 1 1 3 r4 s r 1 1 4 4 s 1 r4 s 3 2 4 4 1 5 4 1 2 5 s4 1 1 2 1 Using the power of a power law Use the product of powers law. 3 1 4 2 r2 s r 3 4 2 4 r s 1 were Write with a positive exponent. 20. Width, in metres, of a piece of An paper: 2 2n 1 4 2n 1 Length, in metres, of a piece of An paper: 2 4 a) i) To determine the dimensions of a piece of A3 paper, substitute: n = 3 Width: 2 2n 1 4 2 2 2(3) 1 4 7 4 Write with a positive exponent. 1 7 24 This is the width in metres. Since 1 m = 1000 mm, to write the width in millimetres, multiply by 1000. 1 1000 1000 7 7 24 24 1000 So, the width of the paper is 7 mm. 24 1000 Use a calculator to evaluate: = 297.3017… 7 4 2 The width of a piece of A3 paper is about 297 mm. Length: 2 2n 1 4 2 2 2(3) 1 4 5 4 Write with a positive exponent. 1 5 24 This is the length in metres. Since 1 m = 1000 mm, to write the length in millimetres, multiply by 1000. 1 1000 1000 5 5 24 24 1000 mm. So, the length of the paper is 5 4 2 1000 Use a calculator to evaluate: = 420.4482… 5 4 2 The length of a piece of A3 paper is about 420 mm. 1000 1000 A piece of A3 paper has dimensions 5 mm by 7 mm, or 297 mm by 420 mm. 24 24 ii) To determine the dimensions of a piece of A4 paper, substitute: n = 4 Width: 2 2n 1 4 2 2 2(4) 1 4 9 4 Write with a positive exponent. 1 9 24 This is the width in metres. To write the width in millimetres, multiply by 1000. 1 1000 1000 9 9 24 24 1000 So, the width of the paper is 9 mm. 24 1000 Use a calculator to evaluate: = 210.2241… 9 4 2 The width of a piece of A4 paper is about 210 mm. Length: 2 2n 1 4 2 2 2(4) 1 4 7 4 Write with a positive exponent. 1 7 24 From part i, 1000 7 4 = 297.3017… 2 The length of a piece of A4 paper is about 297 mm. 1000 1000 A piece of A4 paper has dimensions 7 mm by 9 mm, or 210 mm by 297 mm. 24 24 iii) To determine the dimensions of a piece of A5 paper, substitute: n = 5 Width: 2 2n 1 4 2 2 2(5) 1 4 11 4 Write with a positive exponent. 1 11 24 This is the width in metres. To write the width in millimetres, multiply by 1000. 1 11 4 1000 1000 11 24 2 So, the width of the paper is Use a calculator to evaluate: 1000 11 4 2 1000 11 4 mm. = 148.6508… 2 The width of a piece of A5 paper is about 149 mm. Length: 2 2n 1 4 2 2 2(5) 1 4 9 4 Write with a positive exponent. 1 9 24 1000 From part ii, 9 = 210.2241… 24 The length of a piece of A5 paper is about 210 mm. 1000 1000 A piece of A5 paper has dimensions 9 mm by 11 mm, or 149 mm by 210 mm. 24 24 b) i) The paper is folded along a line perpendicular to its length. So, one dimension of a folded piece of A3 paper, in millimetres, half the original length: 1000 1 1000 5 5 2 24 2 4 21 1000 5 1 24 1000 5 4 24 4 1000 9 24 is one- The other dimension is the original width: 1000 2 7 4 So, a folded piece of A3 paper has dimensions mm 1000 2 7 4 mm by 1000 2 9 4 mm. ii) The paper is folded along a line perpendicular to its length. So, one dimension of a folded piece of A4 paper, in millimetres, is one-half the original length: 1000 1 1000 7 7 2 24 2 4 21 1000 7 1 24 1000 7 4 24 4 1000 11 24 1000 The other dimension is the original width: 9 mm 24 1000 1000 So, a folded piece of A4 paper has dimensions 9 mm by 11 mm. 24 24 iii) The paper is folded along a line perpendicular to its length. So, one dimension of a folded piece of A5 paper, in millimetres, is one-half the original length: 1000 1 1000 9 9 2 24 2 4 21 1000 9 1 24 1000 9 4 24 4 1000 13 24 1000 The other dimension is the original width: 11 mm 24 1000 1000 So, a folded piece of A5 paper has dimensions 11 mm by 13 mm. 24 24 c) I noticed that a piece of A4 paper has the same dimensions as a folded piece of A3 paper, and a piece of A5 paper has the same dimensions as a folded piece of A4 paper. C 21. a) 4 1 a 3 4 b1 4 a 3b c5 c5 1 2 4 3 2 4 4 1 3 1 c b c a b a a12 b 4 c 5 4 3 8 c a b a12b 4 c8 a 4b 3c 5 12 4 4 3 =a b c Using the power of a power law Rewrite without using fractions. Use the product of powers law. 85 a16b 7 c3 Write with a positive exponent. 16 3 = b) 2a b c 4a bc 1 4 3 2 2 4 2 a c b7 22 a ( 1) 2 b 4 2 c ( 3) 2 42 a 2 2 b 2 c ( 4) 2 Using the power of a power law 22 a 2 b 8 c 6 16 a 4 b 2 c 8 Rewrite 16 as a power of 2. 22 a 2 b 8 c 6 24 a 4 b 2 c 8 Use the quotient of powers law. 22 4 a 2 4 b 8 2 c 6 8 26 a 2 b 10 c14 14 c 2 a 2 b10 c14 64a 2b10 6 Write with positive exponents. 22. a) 2 2 1 12 23 2 2 3 2 x y x y 4 Using the power of a power law 2 Substitute: x = a 2 , y a 3 x y3 4 a 2 2 3 a3 Use the power of a power law. 2 4 3 a 2 a 3 8 a 2 a 9 a a 2 8 9 18 8 9 9 10 a 9 1 10 a9 Use the product of powers law. Write with a positive exponent. b) x y 3 4 1 2 3 x4 1 y2 3 3 x y Rewriting as a fraction 3 (3) 4 Using the power of a power law 1 3 2 9 2 x4 y Substitute: x = a 2 , y a 3 3 2 9 2 4 a a 2 3 a 2 Use the power of a power law. 3 2 9 4 2 3 2 a3 9 a 2 1 a a a a 9 1 2 9 2 2 2 7 2 Use the quotient of powers law. Write with a positive exponent. 1 7 a2 23. a) The product of powers law is: xm xn x m n 3 So, for the product to be x 2 , I need to find two exponents whose sum is For example: 1 1 x 2 x1 x 2 3 x2 2 2 1 5 1 x4 x4 x4 6 x4 3 x2 5 4 11 1 11 x 8 x8 x 8 12 x8 3 x2 1 8 3 . 2 b) The quotient of powers law is: xm xn xm n 3 So, for the quotient to be x 2 , I need to find two exponents whose difference is For example: 1 x2 x 2 x 2 1 2 5 5 1 5 x 2 x1 x 2 3 x2 x2 7 7 2 7 x 2 x2 x 2 2 2 x2 3 4 2 3 x2 x2 c) The power of a power law is: x m x mn n 3 So, for the result to be x 2 , I need to find two exponents whose product is For example: x 1 3 2 x 1 3 2 3 x2 9 13 2 3 2 x x 1 x x 9 1 x9 6 x 1 9 6 9 9 6 x6 3 2 x2 3 3 . 2 3 . 2 1 1 1 3 1 2 AC AD 2 2 1 2 24. In , substitute AB = , AC = , AE = , then solve for AD. 2 3 AB AE 3 3 1 3 1 2 2 3 2 3 1 2 1 3 1 2 1 2 3 1 3 2 2 3 1 2 1 1 1 2 Multiply both sides by . 3 AD 1 1 2 3 1 1 2 AD 1 3 1 2 3 1 3 1 2 1 2 2 3 3 1 AD Use the product of powers law. 2 2 3 1 3 1 2 2 3 1 2 AD 1 2 2 3 31 2 3 AD 1 2 2 3 1 2 AD 2 3 1 2 2 3 1 2 1 2 1 AD 1 3 2 AD 2 2 Use a calculator to evaluate: 1 1 3 2 AD 2 2 0.6123... Use a negative exponent to move the denominator to the numerator. Write with a positive exponent. 1 1 3 2 The length of AD is cm, or approximately 0.6 cm. 2 2