Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 13

Answer

$x=\dfrac{1}{3}$

Work Step by Step

We know that $\log_a {x^n}=n\cdot \log_a {x}$, hence the equation $3\log_2{x}=-\log_2{27}$ becomes $\log_2{x^{3}}=\log_2{27^{-1}}.$ RECALL: $\log_a{b}=\log_a{c} \longrightarrow b=c$ Hence, $\log_2{x^{3}}=\log_2{27^{-1}}\longrightarrow x^{3}=27^{-1}$. Solve the equation above to obtain \begin{align*} x^{3}&=27^{-1}\\ x^{3}&=\frac{1}{27} \\ \sqrt[3]{x^{3}}&=\sqrt[3]{\frac{1}{27}}\\ x &=\frac{1}{3}\end{align*}
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