Precalculus (10th Edition)

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