College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises: 85

Answer

$x=-27$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^{5/3}=-243 ,$ raise both sides to the power $\dfrac{3}{5}$ to make the exponent of the variable equal to $1.$ Then use the definition of rational exponents and the concepts of radicals to solve for the variable. $\bf{\text{Solution Details:}}$ Raising both sides of the equation to the power $\dfrac{3}{5},$ the equation above is equivalent to \begin{array}{l}\require{cancel} \left( x^{\frac{5}{3}} \right)^{\frac{3}{5}}=(-243)^{\frac{3}{5}} \\\\ x=(-243)^{\frac{3}{5}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x=\left(\sqrt[5]{-243}\right)^{3} \\\\ x=\left(\sqrt[5]{(-3)^5}\right)^{3} \\\\ x=\left( -3 \right)^{3} \\\\ x=-27 .\end{array}
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