Answer
$\dfrac{1}{3}$
Work Step by Step
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,
$x= \boxed{\dfrac{1}{3}}$
