College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.4 - Rational Exponents and Radicals - P.4 Exercises - Page 30: 68

Answer

(a) $\frac{2y^{4/3}}{x^{2}}$ (b) $-\frac{y^{3/4_{Z}2}}{2}$

Work Step by Step

(a) $(\frac{x^{8}y^{-4}}{16y^{4/3}})^{-1/4}=16^{-1*-1/4}x^{8*-1/4}*y^{(-4)(-1/4)-(4/3)*(-1/4)}=\frac{2y^{4/3}}{x^{2}}$ (b) $(\frac{-8y^{3/4}}{y^{3}z^{6}})^{-1/3}=(-8)^{-1/3}y^{3/4(-1/3)-3(-1/3)}z^{-6*-1/3}=-\frac{y^{3/4_{Z}2}}{2}$
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