Solve for v log base 12 of v^2+35 = log base 12 of -12v-1

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

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

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

Solve for $v$ .

Add $12 v$ to both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Add $35$ and $1$ .

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

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$v^{2} + 12 v + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 v = 2 \cdot v \cdot 6$

Rewrite the polynomial.

$v^{2} + 2 \cdot v \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = v$ and $b = 6$ .

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

Set the $v + 6$ equal to $0$ .

$v + 6 = 0$

Subtract $6$ from both sides of the equation.

$v = - 6$