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Elementary Algebra

Lynn Marecek, MaryAnne Anthony-Smith

$Math Studyset 91影视 Explanations$ Math

2nd Edition 路 1240 pages

ISBN: 9780998625713

Chapter 9: Q 9.221. (page 1135)

Simplify:
a) $8^{\frac{- 5}{3}}$
b) $81^{\frac{- 3}{2}}$
c) $16^{\frac{- 3}{4}}$

Short Answer

Expert verified

a) $8^{\frac{- 5}{3}} = \frac{1}{32}$

b) $81^{\frac{- 3}{2}} = \frac{1}{729}$

c) $16^{\frac{- 3}{4}} = \frac{1}{8}$

Step by step solution

01

Part (a) Step 1: Given Information

The given expression is $8^{\frac{- 5}{3}}$ .

02

Part (a) Step 2: Simplification

Rewrite using $b^{- p} = \frac{1}{b^{p}}$

$\frac{1}{8^{\frac{5}{3}}}$

Change in radical form. The power of radical is numerator of exponent and the index of radical is denominator of exponent.

$= \frac{1}{{(\sqrt[3]{8})}^{5}}$

$= \frac{1}{{(2)}^{5}}$

$= \frac{1}{32}$

03

Part (b) Step 1: Given Information

The given expression is $81^{\frac{- 3}{2}}$ .

04

Part (b) Step 2: Simplification

Rewrite using $b^{- p} = \frac{1}{b^{p}}$

$\frac{1}{81^{\frac{3}{2}}}$

Change in radical form. The power of radical is numerator of exponent and the index of radical is denominator of exponent.

$= \frac{1}{{(\sqrt{81})}^{3}}$

localid="1650520802071" $= \frac{1}{{(9)}^{3}}$

$= \frac{1}{729}$

05

Part (c) Step 1: Given Information

The given expression is $16^{\frac{- 3}{4}}$ .

06

Part (c) Step 2: Simplification

Rewrite using $b^{- p} = \frac{1}{b^{p}}$

$\frac{1}{16^{\frac{3}{4}}}$

Change in radical form. The power of radical is numerator of exponent and the index of radical is denominator of exponent.

$= \frac{1}{{(\sqrt[4]{16})}^{3}}$

$= \frac{1}{{(2)}^{3}}$

$= \frac{1}{8}$

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