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

$x=35$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x-3)^{2/5}=4 ,$ raise both sides to the exponent equal to $ \dfrac{5}{2} .$ 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 exponent equal to $ \dfrac{5}{2} ,$ the given equation becomes \begin{array}{l}\require{cancel} \left( (x-3)^{2/5} \right)^{5/2}=4^{5/2} .\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\frac{5}{2} }=4^{\frac{5}{2}} \\\\ x-3=4^{\frac{5}{2}} .\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-3=\left(\sqrt{4}\right)^{5} \\\\ x-3=\left(2\right)^{5} \\\\ x-3=32 \\\\ x=32+3 \\\\ x=35 .\end{array} Upon checking, $ x=35 $ satisfies the original equation.
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