Solve for x (1/9)^(-2x-42)=(1/3)^(-2x^2-14)

${(\frac{1}{9})}^{- 2 x - 42} = {(\frac{1}{3})}^{- 2 x^{2} - 14}$

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

${(\frac{1}{3})}^{2 (- 2 x - 42)} = {(\frac{1}{3})}^{- 2 x^{2} - 14}$

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

$2 (- 2 x - 42) = - 2 x^{2} - 14$

Solve for $x$ .

Simplify $2 (- 2 x - 42)$ .

Rewrite.

$0 + 0 + 2 (- 2 x - 42) = - 2 x^{2} - 14$

Simplify by adding zeros.

$2 (- 2 x - 42) = - 2 x^{2} - 14$

Apply the distributive property.

$2 (- 2 x) + 2 \cdot - 42 = - 2 x^{2} - 14$

Multiply.

Multiply $- 2$ by $2$ .

$- 4 x + 2 \cdot - 42 = - 2 x^{2} - 14$

Multiply $2$ by $- 42$ .

$- 4 x - 84 = - 2 x^{2} - 14$

Add $2 x^{2}$ to both sides of the equation.

$- 4 x - 84 + 2 x^{2} = - 14$

Add $14$ to both sides of the equation.

$- 4 x - 84 + 2 x^{2} + 14 = 0$

Add $- 84$ and $14$ .

$- 4 x + 2 x^{2} - 70 = 0$

Factor the left side of the equation.

Factor $2$ out of $- 4 x + 2 x^{2} - 70$ .

Factor $2$ out of $- 4 x$ .

$2 (- 2 x) + 2 x^{2} - 70 = 0$

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

$2 (- 2 x) + 2 (x^{2}) - 70 = 0$

Factor $2$ out of $- 70$ .

$2 (- 2 x) + 2 x^{2} + 2 \cdot - 35 = 0$

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

$2 (- 2 x + x^{2}) + 2 \cdot - 35 = 0$

Factor $2$ out of $2 (- 2 x + x^{2}) + 2 \cdot - 35$ .

$2 (- 2 x + x^{2} - 35) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 (- 2 u + u^{2} - 35) = 0$

Factor $- 2 u + u^{2} - 35$ 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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Write the factored form using these integers.

$2 ((u - 7) (u + 5)) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$2 ((x - 7) (x + 5)) = 0$

Remove unnecessary parentheses.

$2 (x - 7) (x + 5) = 0$

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

$x - 7 = 0$

$x + 5 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $2 (x - 7) (x + 5) = 0$ true.

$x = 7, - 5$

Solve for x (1/9)^(-2x-42)=(1/3)^(-2x^2-14)

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