Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set: 85

Answer

$\sqrt[15] (y^{11})$

Work Step by Step

$\sqrt[3] y\times\sqrt[5] (y^{2})=y^{\frac{1}{3}}\times y^{\frac{2}{5}}$ We can further simplify by using the product rule, which holds that $a^{m}\times a^{n}=a^{m+n}$. $y^{\frac{1}{3}}\times y^{\frac{2}{5}}=y^{(\frac{1}{3}+\frac{2}{5})}=y^{(\frac{5}{15}+\frac{6}{15})}=y^{\frac{11}{15}}=\sqrt[15] (y^{11})$
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