Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises - Page 23: 66

Answer

$a)$ $\dfrac{x^{3}}{\sqrt[5]{y}}$ $b)$ $r^{4}$

Work Step by Step

$a)$ $(x^{-5}y^{1/3})^{-3/5}$ Rewrite this expression as a fraction to change the sign of the exponent: $(x^{-5}y^{1/3})^{-3/5}=\dfrac{1}{(x^{-5}y^{1/3})^{3/5}}=...$ Evaluate the power in the denominator and simplify: $...=\dfrac{1}{x^{-3}y^{1/5}}=\dfrac{x^{3}}{y^{1/5}}=\dfrac{x^{3}}{\sqrt[5]{y}}$ $b)$ $(4r^{8}s^{-1/2})^{1/2}(32s^{-5/4})^{-1/5}$ Rewrite this expression as a fraction, to change the sign of the exponent of the second parentheses: $(4r^{8}s^{-1/2})^{1/2}(32s^{-5/4})^{-1/5}=\dfrac{(4r^{8}s^{-1/2})^{1/2}}{(32s^{-5/4})^{1/5}}=...$ Evaluate the powers in the numerator and the denominator and simplify: $...=\dfrac{(4^{1/2})r^{4}s^{-1/4}}{(32^{1/5})s^{-1/4}}=\dfrac{(\sqrt{4})r^{4}}{\sqrt[5]{32}}=\dfrac{2r^{4}}{2}=r^{4}$
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