## Intermediate Algebra: Connecting Concepts through Application

$x \approx 14.434$
$\log_5 (3x) + \log_5 x = 4$ $\log_5 (3x)(x) = 4$ $\log_5 (3x^{2}) = 4$ $5^{4} = 3x^{2}$ $625 = 3x^{2}$ $x^{2} = \frac{625}{3}$ $x = ±\sqrt {\frac{625}{3}}$ $x = ±14.433756...$ $x \approx ±14.434$ 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: $\log_5 (3\sqrt {\frac{625}{3}}) + \log_5 \sqrt {\frac{625}{3}}\overset{?}{=}4$ $\log_5 (3(\sqrt {\frac{625}{3}})( \sqrt {\frac{625}{3}})) \overset{?}{=}4$ $\log_5 (3(\frac{625}{3}) \overset{?}{=}4$ $\log_5 (625) \overset{?}{=}4$ $\log_5 (5^{4}) \overset{?}{=}4$ $4\log_5 (5) \overset{?}{=}4$ $4(1) \overset{?}{=}4$ $4 = 4$