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: 69

Answer

$x=\left\{-29, 35\right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x-3)^{2/5}=4 ,$ raise both sides to the 5th power. Then use the laws of exponents and the concepts of rational exponents to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the 5th power, the given equation becomes \begin{array}{l}\require{cancel} \left( (x-3)^{2/5} \right)^{5}=4^{5} .\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-3)^{\frac{2}{5}\cdot 5 }=1024 \\\\ (x-3)^2=1024 .\end{array} Taking the square root of both sides, the expression above is equivalent to \begin{array}{l}\require{cancel} x-3=\pm \sqrt{1024} \\\\ x-3=\pm \sqrt{32^2} \\\\ x-3 =\pm 32 \\\\ .\end{array} Adding 3 to both sides, the expression above is equivalent to: \begin{array}{l}\require{cancel} x=3\pm 32 \\\\ x_1=3+32 = 35 \\\\ x_2=3-32=-29 \end{array} Upon checking, both numbers satisfy the original equation.
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