# How do you solve 2 log_4 x =1+ log_4 (x + 8)?

May 7, 2016

Solve using the following properties:
$a \log n = \log {n}^{a}$
${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \left(\frac{n}{m}\right)$

#### Explanation:

${\log}_{4} \left({x}^{2}\right) - {\log}_{4} \left(x + 8\right) = 1$

${\log}_{4} \left(\frac{{x}^{2}}{x + 8}\right) = 1$

$\frac{{x}^{2}}{x + 8} = 4$

${x}^{2} = 4 \left(x + 8\right)$

${x}^{2} = 4 x + 32$

${x}^{2} - 4 x - 32 = 0$

$\left(x - 8\right) \left(x + 4\right) = 0$

$x = 8 \mathmr{and} x = - 4$

Checking in the original equation, we find that only $x = 8$ works, so the $x = - 4$ is extraneous and you therefore will not include it inside the solution set.

Hopefully this helps!