Solve for x (1/100)^(-2x+36)=(1/10)^(2x^2+2)

${(\frac{1}{100})}^{- 2 x + 36} = {(\frac{1}{10})}^{2 x^{2} + 2}$

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

${(\frac{1}{10})}^{2 (- 2 x + 36)} = {(\frac{1}{10})}^{2 x^{2} + 2}$

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

$2 (- 2 x + 36) = 2 x^{2} + 2$

Solve for $x$ .

Simplify $2 (- 2 x + 36)$ .

Rewrite.

$0 + 0 + 2 (- 2 x + 36) = 2 x^{2} + 2$

Simplify by adding zeros.

$2 (- 2 x + 36) = 2 x^{2} + 2$

Apply the distributive property.

$2 (- 2 x) + 2 \cdot 36 = 2 x^{2} + 2$

Multiply.

Multiply $- 2$ by $2$ .

$- 4 x + 2 \cdot 36 = 2 x^{2} + 2$

Multiply $2$ by $36$ .

$- 4 x + 72 = 2 x^{2} + 2$

Subtract $2 x^{2}$ from both sides of the equation.

$- 4 x + 72 - 2 x^{2} = 2$

Subtract $2$ from both sides of the equation.

$- 4 x + 72 - 2 x^{2} - 2 = 0$

Subtract $2$ from $72$ .

$- 4 x - 2 x^{2} + 70 = 0$

Factor the left side of the equation.

Factor $- 2$ out of $- 4 x - 2 x^{2} + 70$ .

Reorder $- 4 x$ and $- 2 x^{2}$ .

$- 2 x^{2} - 4 x + 70 = 0$

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

$- 2 x^{2} - 4 x + 70 = 0$

Factor $- 2$ out of $- 4 x$ .

$- 2 x^{2} - 2 (2 x) + 70 = 0$

Factor $- 2$ out of $70$ .

$- 2 x^{2} - 2 (2 x) - 2 \cdot - 35 = 0$

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

$- 2 (x^{2} + 2 x) - 2 \cdot - 35 = 0$

Factor $- 2$ out of $- 2 (x^{2} + 2 x) - 2 (- 35)$ .

$- 2 (x^{2} + 2 x - 35) = 0$

Factor.

Factor $x^{2} + 2 x - 35$ 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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$- 2 ((x - 5) (x + 7)) = 0$

Remove unnecessary parentheses.

$- 2 (x - 5) (x + 7) = 0$

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

$x - 5 = 0$

$x + 7 = 0$

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 $x + 7$ 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 $- 2 (x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

Solve for x (1/100)^(-2x+36)=(1/10)^(2x^2+2)

Related Questions

Support LCMGCF.com by disabling your adblocker.