Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set: 77

Answer

$27 y^{\frac{2}{3}}$

Work Step by Step

RECALL: (1) $a^m \cdot a^n = a^{m+n}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $(ab)^m =a^mb^m$ Use rule (3) above to obtain: $=\dfrac{3^3(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}} \\=\dfrac{3(3)(3)(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}} \\=\dfrac{27(y^{\frac{1}{4}})^3}{y^{\frac{1}{12}}}$ Use rule (3) above to obtain: $=\dfrac{27(y^{\frac{1}{4}(3)})}{y^{\frac{1}{12}}} \\=\dfrac{27y^{\frac{3}{4}}}{y^{\frac{1}{12}}}$ Use rule (2) above to obtain: $=27 y^{\frac{3}{4} - \frac{1}{12}}$ Make the fractional exponents similar using their LCD of $12$ to obtain: $=27 y^{\frac{9}{12} - \frac{1}{12}} \\=27 y^{\frac{9-1}{12}} \\=27 y^{\frac{8}{12}} \\=27 y^{\frac{2}{3}}$
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