Solve for x (1/9)^(-2x^2+6x)=(1/81)^(-2x-36)

${(\frac{1}{9})}^{- 2 x^{2} + 6 x} = {(\frac{1}{81})}^{- 2 x - 36}$

Step 1

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

${(\frac{1}{9})}^{- 2 x^{2} + 6 x} = {(\frac{1}{9})}^{2 (- 2 x - 36)}$

Step 2

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

$- 2 x^{2} + 6 x = 2 (- 2 x - 36)$

Step 3

Solve for $x$ .

Simplify $2 (- 2 x - 36)$ .

Apply the distributive property.

$- 2 x^{2} + 6 x = 2 (- 2 x) + 2 \cdot - 36$

Multiply.

Multiply $- 2$ by $2$ .

$- 2 x^{2} + 6 x = - 4 x + 2 \cdot - 36$

Multiply $2$ by $- 36$ .

$- 2 x^{2} + 6 x = - 4 x - 72$

Move all terms containing $x$ to the left side of the equation.

Add $4 x$ to both sides of the equation.

$- 2 x^{2} + 6 x + 4 x = - 72$

Add $6 x$ and $4 x$ .

$- 2 x^{2} + 10 x = - 72$

Add $72$ to both sides of the equation.

$- 2 x^{2} + 10 x + 72 = 0$

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} + 10 x + 72$ .

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

$- 2 x^{2} + 10 x + 72 = 0$

Factor $- 2$ out of $10 x$ .

$- 2 x^{2} - 2 (- 5 x) + 72 = 0$

Factor $- 2$ out of $72$ .

$- 2 x^{2} - 2 (- 5 x) - 2 \cdot - 36 = 0$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (- 5 x)$ .

$- 2 (x^{2} - 5 x) - 2 \cdot - 36 = 0$

Factor $- 2$ out of $- 2 (x^{2} - 5 x) - 2 (- 36)$ .

$- 2 (x^{2} - 5 x - 36) = 0$

Factor.

Factor $x^{2} - 5 x - 36$ 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 $- 36$ and whose sum is $- 5$ .

$- 9, 4$

Write the factored form using these integers.

$- 2 ((x - 9) (x + 4)) = 0$

Remove unnecessary parentheses.

$- 2 (x - 9) (x + 4) = 0$

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

$x - 9 = 0$

$x + 4 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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 $- 2 (x - 9) (x + 4) = 0$ true.

$x = 9, - 4$

Solve for x (1/9)^(-2x^2+6x)=(1/81)^(-2x-36)

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