Answer
$8$
Work Step by Step
The domain of the variable requires that $x>0$.
Recall:
$\log_a M^r = r \log_a M$
This means that:
$2\log_5 x = \log_5 x^2\\$
$3 \log_5 4 = \log_5 4^3$
Thus, the given equation is equivalent to:
$\log_5 x^2 = \log_5 4^3$
Recall also that:
$\text{If } \log_a M = \log_a N \text{, then } M=N$.
Therefore,
$x^2 = 4^3$
$x^2 = 64$
$x = \pm \sqrt{64}$
$x = \pm 8$
$x$ cannot be negative so $x=-8$ is not a solution.
Hence,
$x = \boxed{8}$
