# How do you solve log_4(v+8)=log_4(-4v-2)?

Sep 26, 2016

$v = - 2$

#### Explanation:

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

$\Leftrightarrow {\log}_{4} \left(v + 8\right) - {\log}_{4} \left(- 4 v - 2\right) = 0$

or ${\log}_{4} \left(\frac{v + 8}{- 4 v - 2}\right) = 0$

Therefore $\left(\frac{v + 8}{- 4 v - 2}\right) = {4}^{0} = 1$ or

$v + 8 = - 4 v - 2$ or

$v + 4 v = - 2 - 8$ or

$5 v = - 10$ or $v = - 2$