## Intermediate Algebra: Connecting Concepts through Application

$x \approx 22.627$
$\log_4 (6x^{3}) - \log_4 (3x) = 5$ $\log_4 \frac{6x^{3}}{3x} = 5$ $\log_4 \frac{6x^{2}}{3} = 5$ $\log_4 2x^{2} = 5$ $4^{5} = 2x^{2}$ $1024 = 2x^{2}$ $512 = x^{2}$ $x = ±\sqrt {512}$ $x \approx 22.627$ Since we can't take the log of a negative number, $x = \sqrt {512}$ is the only possible solution. Check: $\log_4 (6(\sqrt {512})^{3}) - \log_4 (3(\sqrt {512})) \overset{?}{=}5$ $\log_4 (69511.42504) - \log_4 (67.8822...) \overset{?}{=}5$ $\log_4 \frac{(69511.42504)}{(67.8822...) } \overset{?}{=}5$ $\log_4 (1024)\overset{?}{=}5$ $\log_4 (4^{5})\overset{?}{=}5$ $5\log_4 (4)\overset{?}{=}5$ $5(1)\overset{?}{=}5$ $5 = 5$