Solve for x (1/81)^(3x+36)=(1/9)^(3x^2-9x)

${(\frac{1}{81})}^{3 x + 36} = {(\frac{1}{9})}^{3 x^{2} - 9 x}$

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

${(\frac{1}{9})}^{2 (3 x + 36)} = {(\frac{1}{9})}^{3 x^{2} - 9 x}$

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

$2 (3 x + 36) = 3 x^{2} - 9 x$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - 9 x = 2 (3 x + 36)$

Simplify $2 (3 x + 36)$ .

Apply the distributive property.

$3 x^{2} - 9 x = 2 (3 x) + 2 \cdot 36$

Multiply.

Multiply $3$ by $2$ .

$3 x^{2} - 9 x = 6 x + 2 \cdot 36$

Multiply $2$ by $36$ .

$3 x^{2} - 9 x = 6 x + 72$

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

Subtract $6 x$ from both sides of the equation.

$3 x^{2} - 9 x - 6 x = 72$

Subtract $6 x$ from $- 9 x$ .

$3 x^{2} - 15 x = 72$

Subtract $72$ from both sides of the equation.

$3 x^{2} - 15 x - 72 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} - 15 x - 72$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 15 x - 72 = 0$

Factor $3$ out of $- 15 x$ .

$3 (x^{2}) + 3 (- 5 x) - 72 = 0$

Factor $3$ out of $- 72$ .

$3 x^{2} + 3 (- 5 x) + 3 \cdot - 24 = 0$

Factor $3$ out of $3 x^{2} + 3 (- 5 x)$ .

$3 (x^{2} - 5 x) + 3 \cdot - 24 = 0$

Factor $3$ out of $3 (x^{2} - 5 x) + 3 \cdot - 24$ .

$3 (x^{2} - 5 x - 24) = 0$

Factor.

Factor $x^{2} - 5 x - 24$ 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 $- 24$ and whose sum is $- 5$ .

$- 8, 3$

Write the factored form using these integers.

$3 ((x - 8) (x + 3)) = 0$

Remove unnecessary parentheses.

$3 (x - 8) (x + 3) = 0$

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

$x - 8 = 0$

$x + 3 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $3 (x - 8) (x + 3) = 0$ true.

$x = 8, - 3$

Solve for x (1/81)^(3x+36)=(1/9)^(3x^2-9x)

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