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: 12

Answer

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

Work Step by Step

We know that $\log_a {x^n}=n\cdot \log_a {x}$, hence the equation $-2\log_4{x}=\log_4{9}$ becomes $\log_4{x^{-2}}=\log_4{9}.$ RECALL: $\log_a{b}=\log_a{c} \longrightarrow b=c$ Hence, $\log_4{(x^{-2})}=\log_4{9}\longrightarrow x^{-2}=9$ Solve the equation above to obtain \begin{align*} x^{-2}&=9\\ \left(x^{-2}\right)^{-\frac{1}{2}} &=9^{-\frac{1}{2}} \\ x&=\frac{1}{3}\end{align*}
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