College Algebra (11th Edition)

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

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 70


$x=\{-416,16 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x+200)^{2/3}=36 ,$ raise both sides to the exponent equal to $3$ to get rid of the denominator. Then use the laws of exponents and the Square Root Principle to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the exponent equal to $ 3,$ the given equation becomes \begin{array}{l}\require{cancel} \left( (x+200)^{2/3} \right)^{3}=36^{3} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x+200)^{\frac{2}{3}\cdot3 }=36^{3} \\\\ (x+200)^2=36^{3} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} x+200=\pm\sqrt{36^{3}} \\\\ x+200=\pm(\sqrt{36})^3 \\\\ x+200=\pm(6)^3 \\\\ x+200=\pm216 \\\\ x=-200\pm216 .\end{array} The solutions are \begin{array}{l}\require{cancel} x=-200-216 \\\\ x=-416 \\\\\text{OR}\\\\ x=-200+216 \\\\ x=16 .\end{array} Upon checking, $ x=\{-416,16 \} $ satisfy the original equation.
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