College Algebra 7th Edition

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

Chapter P, Prerequisites - Section P.8 - Solving Basic Equations - P.8 Exercises: 68

Answer

$\left\{-\dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{3}}{2}\right\}$

Work Step by Step

Divide 64 on both sides of the equation to obtain: $x^6=\dfrac{27}{64}$ Write $64$ as $2^6$ and 27 as $3^3$ to obtain: $x^6=\dfrac{3^3}{2^6}$ Take the sixth root of both sides to obtain: $x = \pm \sqrt[6]{\dfrac{3^3}{2^6}} \\x-\pm \dfrac{\sqrt[6]{3^3}}{2}$ Use the rule $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ to obtain: $x=\pm \dfrac{3^{\frac{3}{6}}}{2}$ Simplify the rational exponent to obtain: $x = \pm \dfrac{3^{\frac{1}{2}}}{2}$ Use the rule $a^{\frac{1}{2}}=\sqrt{a}$ to obtain: $x= \pm \dfrac{\sqrt{3}}{2}$ Thus, the solution set of the given equation is $\left\{-\dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{3}}{2}\right\}$.
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