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 - Mid-Chapter Check Point - Page 541: 13

Answer

${x^{\frac{7}{12}}}$.

Work Step by Step

The given expression is $=\frac{\left (x^{\frac{2}{3}}\right )^2}{\left (x^{\frac{1}{4}}\right )^3}$ Use power rule. $=\frac{x^{\frac{2\cdot 2}{3}}}{x^{\frac{1\cdot 3}{4}}}$ Simplify. $=\frac{x^{\frac{4}{3}}}{x^{\frac{3}{4}}}$ Divide factors: $={x^{\frac{4}{3}-\frac{3}{4}}}$ Simplify. $={x^{\frac{16}{12}-\frac{9}{12}}}$ $={x^{\frac{16-9}{12}}}$ $={x^{\frac{7}{12}}}$.
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