Answer
$x \approx 14.434$
Work Step by Step
$\log_5 (3x) + \log_5 x = 4$
$\log_5 (3x)(x) = 4$
$\log_5 (3x^{2}) = 4$
$5^{4} = 3x^{2}$
$625 = 3x^{2}$
$x^{2} = \frac{625}{3}$
$x = ±\sqrt {\frac{625}{3}}$
$x = ±14.433756...$
$x \approx ±14.434$
We can't have $x = -14.434$, so the only possible solution is $x \approx 14.434$. This is because $\log$ of a negative number does not exist and therefore we only accept the positive number of $x$.
Check:
$\log_5 (3\sqrt {\frac{625}{3}}) + \log_5 \sqrt {\frac{625}{3}}\overset{?}{=}4$
$\log_5 (3(\sqrt {\frac{625}{3}})( \sqrt {\frac{625}{3}})) \overset{?}{=}4$
$\log_5 (3(\frac{625}{3}) \overset{?}{=}4$
$\log_5 (625) \overset{?}{=}4$
$\log_5 (5^{4}) \overset{?}{=}4$
$4\log_5 (5) \overset{?}{=}4$
$4(1) \overset{?}{=}4$
$4 = 4$
