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: 78

Answer

$2x^{\frac{1}{6}}$

Work Step by Step

RECALL: (i) $a^{-m} = \dfrac{1}{a^m},a\ne0$ (ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ (iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$ Simplify the numerator to obtain: $=\dfrac{\sqrt[3]{2^3x^2}}{\sqrt{x}} \\=\dfrac{2\sqrt[3]{x^2}}{\sqrt{x}}$ Use rule (ii) above to obtain: $=\dfrac{2x^{\frac{2}{3}}}{x^{\frac{1}{2}}}$ Use rule (iii) above to to obtain: $=2x^{\frac{2}{3} - \frac{1}{2}} \\=2x^{\frac{4}{6}-\frac{3}{6}} \\=2x^{\frac{1}{6}}$
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