Solve for x 3^(2x^2+12)=(1/9)^(-14x+34)

$3^{2 x^{2} + 12} = {(\frac{1}{9})}^{- 14 x + 34}$

Apply the product rule to $\frac{1}{9}$ .

$3^{2 x^{2} + 12} = \frac{1^{- 14 x + 34}}{9^{- 14 x + 34}}$

One to any power is one.

$3^{2 x^{2} + 12} = \frac{1}{9^{- 14 x + 34}}$

Move $9^{- 14 x + 34}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$3^{2 x^{2} + 12} = 9^{- (- 14 x + 34)}$

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

$3^{2 x^{2} + 12} = 3^{2 (- (- 14 x + 34))}$

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

$2 x^{2} + 12 = 2 (- (- 14 x + 34))$

Solve for $x$ .

Simplify $2 (- (- 14 x + 34))$ .

Rewrite.

$2 x^{2} + 12 = 0 + 0 + 2 (- (- 14 x + 34))$

Simplify by adding zeros.

$2 x^{2} + 12 = 2 (- (- 14 x + 34))$

Apply the distributive property.

$2 x^{2} + 12 = 2 (- (- 14 x) - 1 \cdot 34)$

Multiply.

Multiply $- 14$ by $- 1$ .

$2 x^{2} + 12 = 2 (14 x - 1 \cdot 34)$

Multiply $- 1$ by $34$ .

$2 x^{2} + 12 = 2 (14 x - 34)$

Apply the distributive property.

$2 x^{2} + 12 = 2 (14 x) + 2 \cdot - 34$

Multiply.

Multiply $14$ by $2$ .

$2 x^{2} + 12 = 28 x + 2 \cdot - 34$

Multiply $2$ by $- 34$ .

$2 x^{2} + 12 = 28 x - 68$

Subtract $28 x$ from both sides of the equation.

$2 x^{2} + 12 - 28 x = - 68$

Add $68$ to both sides of the equation.

$2 x^{2} + 12 - 28 x + 68 = 0$

Add $12$ and $68$ .

$2 x^{2} - 28 x + 80 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 28 x + 80$ .

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

$2 (x^{2}) - 28 x + 80 = 0$

Factor $2$ out of $- 28 x$ .

$2 (x^{2}) + 2 (- 14 x) + 80 = 0$

Factor $2$ out of $80$ .

$2 x^{2} + 2 (- 14 x) + 2 \cdot 40 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 14 x)$ .

$2 (x^{2} - 14 x) + 2 \cdot 40 = 0$

Factor $2$ out of $2 (x^{2} - 14 x) + 2 \cdot 40$ .

$2 (x^{2} - 14 x + 40) = 0$

Factor.

Factor $x^{2} - 14 x + 40$ 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 $40$ and whose sum is $- 14$ .

$- 10, - 4$

Write the factored form using these integers.

$2 ((x - 10) (x - 4)) = 0$

Remove unnecessary parentheses.

$2 (x - 10) (x - 4) = 0$

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

$x - 10 = 0$

$x - 4 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $2 (x - 10) (x - 4) = 0$ true.

$x = 10, 4$

Solve for x 3^(2x^2+12)=(1/9)^(-14x+34)

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