EXAMPLE 1 Use properties of exponents

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Presentation transcript:

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

EXAMPLE 2 Apply properties of exponents Biology (3.4 103 grams). 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 k is a constant. The values of k for some mammals are shown below. Approximate the surface area of a rabbit that has a mass of 3.4 kilograms

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

GUIDED PRACTICE for Examples 1 and 2 Use the properties of rational exponents to simplify the expression. 1. (51/3 71/4)3 = (51/3)3 (71/4)3 51/3 3 71/4 3 = 51 73/4 = 5 73/4 = 2. 23/4 21/2 2(3/4 + 1/2) = 25/4 = 3. 3 31/4 31 31/4 = = 3(1 – 1/4) 33/4 = 201/2 3 4. 51/2 = 20 5 1/2 3 (41/2)3 = (22)3/2 = 8 =

GUIDED PRACTICE for Examples 1 and 2 Biology Use the information in Example 2 to approximate the surface area of a sheep that has a mass of 95 kilograms (9.5 104 grams). SOLUTION S = km2/3 Write model. = 9.75(9.5 104)2/3 Substitute 9.75 for k and 9.5 104 for m. = 9.75(9.5)2/3(104)2/3 Power of a product property.

The Sheep’s surface area is about 2200 square centimeters. ANSWER GUIDED PRACTICE for Examples 1 and 2 9.75(95)10/3 Power of a product property. 17,500 Simplify. The Sheep’s surface area is about 2200 square centimeters. ANSWER