College Algebra (11th Edition)

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

Chapter 4 - Section 4.3 - Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions: 36

Answer

$x=243$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_{1/3} x=-5 ,$ change the expression to exponential form. Then use the laws of exponents. $\bf{\text{Solution Details:}}$ Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \left(\dfrac{1}{3}\right)^{-5}=x .\end{array} Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^m}{z^p} \right)^q=\dfrac{x^{mq}}{z^{pq}},$ the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1^{-5}}{3^{-5}}=x .\end{array} 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{3^{5}}{1^{5}}=x \\\\ \dfrac{243}{1}=x \\\\ x=243 .\end{array}
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