Solve for x (1/81)^(-3x-9)=9^(-3x^2-15x)

${(\frac{1}{81})}^{- 3 x - 9} = 9^{- 3 x^{2} - 15 x}$

Step 1

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

$\frac{1^{- 3 x - 9}}{81^{- 3 x - 9}} = 9^{- 3 x^{2} - 15 x}$

Step 2

One to any power is one.

$\frac{1}{81^{- 3 x - 9}} = 9^{- 3 x^{2} - 15 x}$

Step 3

Move $81^{- 3 x - 9}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$81^{- (- 3 x - 9)} = 9^{- 3 x^{2} - 15 x}$

Step 4

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

$9^{2 (- (- 3 x - 9))} = 9^{- 3 x^{2} - 15 x}$

Step 5

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

$2 (- (- 3 x - 9)) = - 3 x^{2} - 15 x$

Step 6

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.

$- 3 x^{2} - 15 x = 2 (- (- 3 x - 9))$

Simplify $2 (- (- 3 x - 9))$ .

Apply the distributive property.

$- 3 x^{2} - 15 x = 2 (- (- 3 x) - - 9)$

Multiply.

Multiply $- 3$ by $- 1$ .

$- 3 x^{2} - 15 x = 2 (3 x - - 9)$

Multiply $- 1$ by $- 9$ .

$- 3 x^{2} - 15 x = 2 (3 x + 9)$

Apply the distributive property.

$- 3 x^{2} - 15 x = 2 (3 x) + 2 \cdot 9$

Multiply.

Multiply $3$ by $2$ .

$- 3 x^{2} - 15 x = 6 x + 2 \cdot 9$

Multiply $2$ by $9$ .

$- 3 x^{2} - 15 x = 6 x + 18$

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

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

$- 3 x^{2} - 15 x - 6 x = 18$

Subtract $6 x$ from $- 15 x$ .

$- 3 x^{2} - 21 x = 18$

Subtract $18$ from both sides of the equation.

$- 3 x^{2} - 21 x - 18 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} - 21 x - 18$ .

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

$- 3 x^{2} - 21 x - 18 = 0$

Factor $- 3$ out of $- 21 x$ .

$- 3 x^{2} - 3 (7 x) - 18 = 0$

Factor $- 3$ out of $- 18$ .

$- 3 x^{2} - 3 (7 x) - 3 \cdot 6 = 0$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (7 x)$ .

$- 3 (x^{2} + 7 x) - 3 \cdot 6 = 0$

Factor $- 3$ out of $- 3 (x^{2} + 7 x) - 3 (6)$ .

$- 3 (x^{2} + 7 x + 6) = 0$

Factor.

Factor $x^{2} + 7 x + 6$ 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 $6$ and whose sum is $7$ .

$1, 6$

Write the factored form using these integers.

$- 3 ((x + 1) (x + 6)) = 0$

Remove unnecessary parentheses.

$- 3 (x + 1) (x + 6) = 0$

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

$x + 1 = 0$

$x + 6 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- 3 (x + 1) (x + 6) = 0$ true.

$x = - 1, - 6$

Solve for x (1/81)^(-3x-9)=9^(-3x^2-15x)

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