Solve for x 8^(-2x^2-10)=(1/64)^(-10x-16)

$8^{- 2 x^{2} - 10} = {(\frac{1}{64})}^{- 10 x - 16}$

Step 1

Apply the product rule to $\frac{1}{64}$ .

$8^{- 2 x^{2} - 10} = \frac{1^{- 10 x - 16}}{64^{- 10 x - 16}}$

Step 2

One to any power is one.

$8^{- 2 x^{2} - 10} = \frac{1}{64^{- 10 x - 16}}$

Step 3

Move $64^{- 10 x - 16}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$8^{- 2 x^{2} - 10} = 64^{- (- 10 x - 16)}$

Step 4

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

$8^{- 2 x^{2} - 10} = 8^{2 (- (- 10 x - 16))}$

Step 5

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

$- 2 x^{2} - 10 = 2 (- (- 10 x - 16))$

Step 6

Solve for $x$ .

Simplify $2 (- (- 10 x - 16))$ .

Rewrite.

$- 2 x^{2} - 10 = 0 + 0 + 2 (- (- 10 x - 16))$

Simplify by adding zeros.

$- 2 x^{2} - 10 = 2 (- (- 10 x - 16))$

Apply the distributive property.

$- 2 x^{2} - 10 = 2 (- (- 10 x) - - 16)$

Multiply.

Multiply $- 10$ by $- 1$ .

$- 2 x^{2} - 10 = 2 (10 x - - 16)$

Multiply $- 1$ by $- 16$ .

$- 2 x^{2} - 10 = 2 (10 x + 16)$

Apply the distributive property.

$- 2 x^{2} - 10 = 2 (10 x) + 2 \cdot 16$

Multiply.

Multiply $10$ by $2$ .

$- 2 x^{2} - 10 = 20 x + 2 \cdot 16$

Multiply $2$ by $16$ .

$- 2 x^{2} - 10 = 20 x + 32$

Subtract $20 x$ from both sides of the equation.

$- 2 x^{2} - 10 - 20 x = 32$

Subtract $32$ from both sides of the equation.

$- 2 x^{2} - 10 - 20 x - 32 = 0$

Subtract $32$ from $- 10$ .

$- 2 x^{2} - 20 x - 42 = 0$

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} - 20 x - 42$ .

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

$- 2 x^{2} - 20 x - 42 = 0$

Factor $- 2$ out of $- 20 x$ .

$- 2 x^{2} - 2 (10 x) - 42 = 0$

Factor $- 2$ out of $- 42$ .

$- 2 x^{2} - 2 (10 x) - 2 \cdot 21 = 0$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (10 x)$ .

$- 2 (x^{2} + 10 x) - 2 \cdot 21 = 0$

Factor $- 2$ out of $- 2 (x^{2} + 10 x) - 2 (21)$ .

$- 2 (x^{2} + 10 x + 21) = 0$

Factor.

Factor $x^{2} + 10 x + 21$ 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 $21$ and whose sum is $10$ .

$3, 7$

Write the factored form using these integers.

$- 2 ((x + 3) (x + 7)) = 0$

Remove unnecessary parentheses.

$- 2 (x + 3) (x + 7) = 0$

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

$x + 3 = 0$

$x + 7 = 0$

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.

$x = - 3$

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

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

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $- 2 (x + 3) (x + 7) = 0$ true.

$x = - 3, - 7$

Solve for x 8^(-2x^2-10)=(1/64)^(-10x-16)

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