Solve for x 10^(2x^2-6)=(1/100)^(-4x-2)

$10^{2 x^{2} - 6} = {(\frac{1}{100})}^{- 4 x - 2}$

Apply the product rule to $\frac{1}{100}$ .

$10^{2 x^{2} - 6} = \frac{1^{- 4 x - 2}}{100^{- 4 x - 2}}$

One to any power is one.

$10^{2 x^{2} - 6} = \frac{1}{100^{- 4 x - 2}}$

Move $100^{- 4 x - 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$10^{2 x^{2} - 6} = 100^{- (- 4 x - 2)}$

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

$10^{2 x^{2} - 6} = 10^{2 (- (- 4 x - 2))}$

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

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

Solve for $x$ .

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

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $- 4$ by $- 1$ .

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

Multiply $- 1$ by $- 2$ .

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

Apply the distributive property.

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

Multiply.

Multiply $4$ by $2$ .

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

Multiply $2$ by $2$ .

$2 x^{2} - 6 = 8 x + 4$

Subtract $8 x$ from both sides of the equation.

$2 x^{2} - 6 - 8 x = 4$

Subtract $4$ from both sides of the equation.

$2 x^{2} - 6 - 8 x - 4 = 0$

Subtract $4$ from $- 6$ .

$2 x^{2} - 8 x - 10 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 8 x - 10$ .

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

$2 (x^{2}) - 8 x - 10 = 0$

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

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

Factor $2$ out of $- 10$ .

$2 x^{2} + 2 (- 4 x) + 2 \cdot - 5 = 0$

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

$2 (x^{2} - 4 x) + 2 \cdot - 5 = 0$

Factor $2$ out of $2 (x^{2} - 4 x) + 2 \cdot - 5$ .

$2 (x^{2} - 4 x - 5) = 0$

Factor.

Factor $x^{2} - 4 x - 5$ 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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Write the factored form using these integers.

$2 ((x - 5) (x + 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 5) (x + 1) = 0$

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

$x - 5 = 0$

$x + 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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 $2 (x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Solve for x 10^(2x^2-6)=(1/100)^(-4x-2)

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