## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises - Page 410: 77

#### Answer

$x=3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{1}{27}=x^{-3} ,$ 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}{27}=\dfrac{1}{x^3} .\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^3)=1(27) \\\\ x^3=27 .\end{array} Taking the cube root of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{x^3}=\sqrt[3]{27} \\\\ x=\sqrt[3]{(3)^3} \\\\ x=3 .\end{array}

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