Solve for x 25^(2x)=5^(x^2-12)

$25^{2 x} = 5^{x^{2} - 12}$

Step 1

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

$5^{2 (2 x)} = 5^{x^{2} - 12}$

Step 2

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

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

Step 3

Solve for $x$ .

Multiply $2$ by $2$ .

$4 x = x^{2} - 12$

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

$4 x - x^{2} = - 12$

Add $12$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $4 x - x^{2} + 12$ .

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

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

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $4 x$ .

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

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - (- 4 x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 4 x)$ .

$- (x^{2} - 4 x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2} - 4 x) - 1 (- 12)$ .

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

Factor.

Factor $x^{2} - 4 x - 12$ 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.

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

Remove unnecessary parentheses.

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

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

$x = 6, - 2$

Solve for x 25^(2x)=5^(x^2-12)

Related Questions

Support LCMGCF.com by disabling your adblocker.