# How do you solve log_12(v^2+35)=log_12(-12v-1)?

Jul 10, 2016

${\log}_{12} \left({v}^{2} + 35\right) - {\log}_{12} \left(- 12 v - 1\right) = 0$

${\log}_{12} \left(\frac{{v}^{2} + 35}{- 12 v - 1}\right) = 0$

$\frac{{v}^{2} + 35}{- 12 v - 1} = {12}^{0}$

${v}^{2} + 35 = 1 \left(- 12 v - 1\right)$

${v}^{2} + 35 = - 12 v - 1$

${v}^{2} + 12 v + 36 = 0$

${\left(v + 6\right)}^{2} = 0$

$\left(v + 6\right) \left(v + 6\right) = 0$

$v = - 6 \mathmr{and} - 6$

Checking in the original solution, we find this solution works.

This equation has a solution set $\left\{- 6\right\}$.

Hopefully this helps!