Answer
$x = 3$
Work Step by Step
$\log_6 (4x) + \log_6 x = 2$
$\log_6 (4x^{2}) = 2$
$6^{2} = 4x^{2}$
$36 = 4x^{2}$
$9 = x^{2}$
$x = ±\sqrt 9$
$x = ±3$
Since $x$ can't be negative, the only possible solution is $x = 3$. This is because $\log$ of a negative number does not exist and therefore we only accept the positive number of $x$.
Check:
$\log_6 (4(3)) + \log_6 3 \overset{?}{=} 2$
$\log_6 (12) + \log_6 3 \overset{?}{=} 2$
$\log_6 (12 \times 3) \overset{?}{=} 2$
$\log_6 (36) \overset{?}{=} 2$
$\log_6 (6^{2}) \overset{?}{=} 2$
$2\log_6 (6) \overset{?}{=} 2$
$2(1) \overset{?}{=} 2$
$2 = 2$
