Solve for x 36^(-4x-6)=(1/6)^(2x^2-12)

$36^{- 4 x - 6} = {(\frac{1}{6})}^{2 x^{2} - 12}$

Step 1

Apply the product rule to $\frac{1}{6}$ .

$36^{- 4 x - 6} = \frac{1^{2 x^{2} - 12}}{6^{2 x^{2} - 12}}$

Step 2

One to any power is one.

$36^{- 4 x - 6} = \frac{1}{6^{2 x^{2} - 12}}$

Step 3

Move $6^{2 x^{2} - 12}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$36^{- 4 x - 6} = 6^{- (2 x^{2} - 12)}$

Step 4

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

$6^{2 (- 4 x - 6)} = 6^{- (2 x^{2} - 12)}$

Step 5

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

$2 (- 4 x - 6) = - (2 x^{2} - 12)$

Step 6

Solve for $x$ .

Simplify $2 (- 4 x - 6)$ .

Rewrite.

$0 + 0 + 2 (- 4 x - 6) = - (2 x^{2} - 12)$

Simplify by adding zeros.

$2 (- 4 x - 6) = - (2 x^{2} - 12)$

Apply the distributive property.

$2 (- 4 x) + 2 \cdot - 6 = - (2 x^{2} - 12)$

Multiply.

Multiply $- 4$ by $2$ .

$- 8 x + 2 \cdot - 6 = - (2 x^{2} - 12)$

Multiply $2$ by $- 6$ .

$- 8 x - 12 = - (2 x^{2} - 12)$

Simplify $- (2 x^{2} - 12)$ .

Apply the distributive property.

$- 8 x - 12 = - (2 x^{2}) - - 12$

Multiply.

Multiply $2$ by $- 1$ .

$- 8 x - 12 = - 2 x^{2} - - 12$

Multiply $- 1$ by $- 12$ .

$- 8 x - 12 = - 2 x^{2} + 12$

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

$- 8 x - 12 + 2 x^{2} = 12$

Subtract $12$ from both sides of the equation.

$- 8 x - 12 + 2 x^{2} - 12 = 0$

Subtract $12$ from $- 12$ .

$- 8 x + 2 x^{2} - 24 = 0$

Factor the left side of the equation.

Factor $2$ out of $- 8 x + 2 x^{2} - 24$ .

Factor $2$ out of $- 8 x$ .

$2 (- 4 x) + 2 x^{2} - 24 = 0$

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

$2 (- 4 x) + 2 (x^{2}) - 24 = 0$

Factor $2$ out of $- 24$ .

$2 (- 4 x) + 2 x^{2} + 2 \cdot - 12 = 0$

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

$2 (- 4 x + x^{2}) + 2 \cdot - 12 = 0$

Factor $2$ out of $2 (- 4 x + x^{2}) + 2 \cdot - 12$ .

$2 (- 4 x + x^{2} - 12) = 0$

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

$2 (- 4 u + u^{2} - 12) = 0$

Factor $- 4 u + u^{2} - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$2 ((u - 6) (u + 2)) = 0$

Factor.

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

$2 ((x - 6) (x + 2)) = 0$

Remove unnecessary parentheses.

$2 (x - 6) (x + 2) = 0$

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

$x - 6 = 0$

$x + 2 = 0$

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $2 (x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Solve for x 36^(-4x-6)=(1/6)^(2x^2-12)

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