Answer
$x = \sqrt {312.5}$
Work Step by Step
$\log_5 2x + \log_5 x = 4$
$\log_5 2x^{2} = 4$
$5^{4} = 2x^{2}$
$625 = 2x^{2}$
$x^{2} = 312.5$
$x = ±\sqrt {312.5}$. The negative value is not in the domain of $\log_5 x$.
$x = \sqrt {312.5}$
Check:
$\log_5 2x + \log_5 x \overset{?}{=} 4$
$\log_5 2(\sqrt {312.5}) + \log_5 (\sqrt {312.5}) \overset{?}{=} 4$
$\log_5 (2\sqrt {312.5}) + \log_5 (\sqrt {312.5}) \overset{?}{=} 4$
$\log_5 (2\sqrt {312.5})(\sqrt {312.5}) \overset{?}{=} 4$
$\log_5 625 \overset{?}{=} 4$
$5^{4}\overset{?}{=} 625$
$625 = 625$
