Solve for x (1/125)^(-4x-10)=(1/5)^(-2x^2-16)

${(\frac{1}{125})}^{- 4 x - 10} = {(\frac{1}{5})}^{- 2 x^{2} - 16}$

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

${(\frac{1}{5})}^{3 (- 4 x - 10)} = {(\frac{1}{5})}^{- 2 x^{2} - 16}$

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

$3 (- 4 x - 10) = - 2 x^{2} - 16$

Solve for $x$ .

Simplify $3 (- 4 x - 10)$ .

Rewrite.

$0 + 0 + 3 (- 4 x - 10) = - 2 x^{2} - 16$

Simplify by adding zeros.

$3 (- 4 x - 10) = - 2 x^{2} - 16$

Apply the distributive property.

$3 (- 4 x) + 3 \cdot - 10 = - 2 x^{2} - 16$

Multiply.

Multiply $- 4$ by $3$ .

$- 12 x + 3 \cdot - 10 = - 2 x^{2} - 16$

Multiply $3$ by $- 10$ .

$- 12 x - 30 = - 2 x^{2} - 16$

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

$- 12 x - 30 + 2 x^{2} = - 16$

Add $16$ to both sides of the equation.

$- 12 x - 30 + 2 x^{2} + 16 = 0$

Add $- 30$ and $16$ .

$- 12 x + 2 x^{2} - 14 = 0$

Factor the left side of the equation.

Factor $2$ out of $- 12 x + 2 x^{2} - 14$ .

Factor $2$ out of $- 12 x$ .

$2 (- 6 x) + 2 x^{2} - 14 = 0$

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

$2 (- 6 x) + 2 (x^{2}) - 14 = 0$

Factor $2$ out of $- 14$ .

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

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

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

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

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

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

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

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

$- 7, 1$

Write the factored form using these integers.

$2 ((u - 7) (u + 1)) = 0$

Factor.

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

$2 ((x - 7) (x + 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 7) (x + 1) = 0$

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

$x - 7 = 0$

$x + 1 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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.

$x = - 1$

The final solution is all the values that make $2 (x - 7) (x + 1) = 0$ true.

$x = 7, - 1$

Solve for x (1/125)^(-4x-10)=(1/5)^(-2x^2-16)

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