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EXAMPLE 1

5 1. 5. 5 1/3. 5 1/3. a. 7 1/4 7 1/2. b. (6 1/2 4 1/3 ) 2. = (6 1/2 ) 2 (4 1/3 ) 2. = 6( 1/2 2 ) 4( 1/3 2 ). = 6 4 2/3. = 6 1 4 2/3. 1. c. (4 5 3 5 ) –1/5. = [(4 3) 5 ] –1/5. = 12 [5 (–1/5)]. =. 12. d. =. 2. 42. 42 1/3 2. 1/3. e. =. = 7( 1/3 2 ).

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EXAMPLE 1

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  1. 51 5 51/3 51/3 a. 71/4 71/2 b. (61/2 41/3)2 = (61/2)2 (41/3)2 = 6(1/22) 4(1/32) = 6 42/3 = 61 42/3 1 c. (45 35)–1/5 = [(4 3)5]–1/5 = 12[5 (–1/5)] = 12 d. = 2 42 421/3 2 1/3 e. = = 7(1/3 2) 6 61/3 EXAMPLE 1 Use properties of exponents Use the properties of rational exponents to simplify the expression. = 73/4 = 7(1/4 + 1/2) = 12 –1 = (125)–1/5 = 5(1 – 1/3) = 52/3 = 72/3 = (71/3)2

  2. A mammal’s surface area S(in square centimeters) can be approximated by the model S = km2/3 where m is the mass (in grams) of the mammal and kis a constant. The values of kfor some mammals are shown below. Approximate the surface area of a rabbit that has a mass of 3.4kilograms (3.4 103grams). EXAMPLE 2 Apply properties of exponents Biology

  3. km2/3 S = =9.75(3.4 103)2/3 Substitute 9.75 for kand 3.4 103 for m. 9.75(2.26)(102) 2200 ANSWER The rabbit’s surface area is about 2200square centimeters. EXAMPLE 2 Apply properties of exponents SOLUTION Write model. = 9.75(3.4)2/3(103)2/3 Power of a product property. Power of a product property. Simplify.

  4. 3 31/4 2. 23/4 21/2 1. (51/3 71/4)3 5 73/4 25/4 33/4 3. 201/2 3 8 4. 51/2 for Examples 1 and 2 GUIDED PRACTICE Use the properties of rational exponents to simplify the expression. SOLUTION SOLUTION SOLUTION SOLUTION

  5. Use the information in Example 2 to approximate the surface area of a sheep that has a mass of 95kilograms(9.5 104grams). for Examples 1 and 2 GUIDED PRACTICE Biology 17,500cm2 SOLUTION

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