Solve for x (1/81)^(6x+2)=9^(2x^2+12)

${(\frac{1}{81})}^{6 x + 2} = 9^{2 x^{2} + 12}$

Apply the product rule to $\frac{1}{81}$ .

$\frac{1^{6 x + 2}}{81^{6 x + 2}} = 9^{2 x^{2} + 12}$

One to any power is one.

$\frac{1}{81^{6 x + 2}} = 9^{2 x^{2} + 12}$

Move $81^{6 x + 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$81^{- (6 x + 2)} = 9^{2 x^{2} + 12}$

Create equivalent expressions in the equation that all have equal bases.

$9^{2 (- (6 x + 2))} = 9^{2 x^{2} + 12}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 (- (6 x + 2)) = 2 x^{2} + 12$

Solve for $x$ .

Simplify $2 (- (6 x + 2))$ .

Rewrite.

$0 + 0 + 2 (- (6 x + 2)) = 2 x^{2} + 12$

Simplify by adding zeros.

$2 (- (6 x + 2)) = 2 x^{2} + 12$

Apply the distributive property.

$2 (- (6 x) - 1 \cdot 2) = 2 x^{2} + 12$

Multiply.

Multiply $6$ by $- 1$ .

$2 (- 6 x - 1 \cdot 2) = 2 x^{2} + 12$

Multiply $- 1$ by $2$ .

$2 (- 6 x - 2) = 2 x^{2} + 12$

Apply the distributive property.

$2 (- 6 x) + 2 \cdot - 2 = 2 x^{2} + 12$

Multiply.

Multiply $- 6$ by $2$ .

$- 12 x + 2 \cdot - 2 = 2 x^{2} + 12$

Multiply $2$ by $- 2$ .

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

Subtract $2 x^{2}$ from both sides of the equation.

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

Subtract $12$ from both sides of the equation.

$- 12 x - 4 - 2 x^{2} - 12 = 0$

Subtract $12$ from $- 4$ .

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

Factor the left side of the equation.

Factor $- 2$ out of $- 12 x - 2 x^{2} - 16$ .

Reorder $- 12 x$ and $- 2 x^{2}$ .

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

Factor $- 2$ out of $- 2 x^{2}$ .

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

Factor $- 2$ out of $- 12 x$ .

$- 2 x^{2} - 2 (6 x) - 16 = 0$

Factor $- 2$ out of $- 16$ .

$- 2 x^{2} - 2 (6 x) - 2 \cdot 8 = 0$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (6 x)$ .

$- 2 (x^{2} + 6 x) - 2 \cdot 8 = 0$

Factor $- 2$ out of $- 2 (x^{2} + 6 x) - 2 (8)$ .

$- 2 (x^{2} + 6 x + 8) = 0$

Factor.

Factor $x^{2} + 6 x + 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $6$ .

$2, 4$

Write the factored form using these integers.

$- 2 ((x + 2) (x + 4)) = 0$

Remove unnecessary parentheses.

$- 2 (x + 2) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$x + 4 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $- 2 (x + 2) (x + 4) = 0$ true.

$x = - 2, - 4$

Solve for x (1/81)^(6x+2)=9^(2x^2+12)

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