The log is literally gone

Recall that if $\log_ a b = c$ , then $a^c = b$ .

$\log_x4=2$
$x^2$ = 4
$x^4$ = $4^2$
$x^4$ = 16

Given, $log_ x 4 = 2$

∴ By base changing property

( $\frac{log 4}{log x}$ = 2)

$log 4 = 2 log x$

$log 4 = log x^{2}$

$4 = x^{2}$

squaring both sides,

$4^{2} = x^{4}$

$16 = x^{4}$

$x^{4} = 16$

$\log_x 4 = 2$

$\Rightarrow x^2 = 4$

$\Rightarrow x = \sqrt 4$

$\Rightarrow x = 2$

Hence $x^4 = 2^4 = \boxed{16}$