# How do you solve (\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}?

Jan 20, 2018

$x = - 4$ or $x = - 2$

#### Explanation:

Start by working on the left side:

Since $81 = {9}^{2}$ we know $\frac{1}{81} = \frac{1}{9} ^ 2$ and by the rules $\frac{1}{a} = {a}^{-} 1$, we know that $\frac{1}{9} ^ 2 = {9}^{-} 2$

By the rule ${\left({a}^{b}\right)}^{c} = {a}^{b c}$, we know

${\left({9}^{-} 2\right)}^{6 x + 2} = {9}^{- 12 x - 4}$.

So our equation now looks like:

${9}^{- 12 x - 4} = {9}^{2 {x}^{2} + 12}$

If ${a}^{b} = {a}^{c}$, then $b = c$, so:

$- 12 x - 4 = 2 {x}^{2} + 12$

moving everything to one side and collecting like terms we have:

$2 {x}^{2} + 12 x + 16 = 0$

Dividing thruogh by 2:

${x}^{2} + 6 x + 8 = 0$

factoring:

$\left(x + 4\right) \left(x + 2\right) = 0$

Solving:

$x = - 4$ or $x = - 2$