## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{3}$
The domain of the variable requires that $x>0$. Recall: $\log_a M^r = r \log_a M$ Using the rule above gives: $-2 \log_4 x = \log_4 x^{-2}$ Thus, the given equation is equivalent to: $\log_4{\left(x^{-2}\right)} = \log_4 9$ Recall also that: $\text{If } \log_a M = \log_a N \text{, then } M=N$ Therefore, $x^{-2} = 9$ $\left(x^{-2}\right)^{-1} = (9)^{-1}$ $x^2 = \dfrac{1}{9}$ $x = \pm \sqrt{\dfrac{1}{9}}$ $x = \pm \dfrac{1}{3}$ $x$ cannot be negative so $x=-\frac{1}{3}$ will be rejected. Hence, $x= \boxed{\dfrac{1}{3}}$