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: 78

Answer

$x=2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{1}{32}=x^{-5} ,$ use the laws of exponents and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{32}=\dfrac{1}{x^5} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} 1(x^5)=1(32) \\\\ x^5=32 .\end{array} Taking the fifth root of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt[5]{x^5}=\sqrt[5]{32} \\\\ x=\sqrt[5]{(2)^5} \\\\ x=2 .\end{array}
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