Solve for x 12^(x^2+5x-4)=12^(2x+6)

$12^{x^{2} + 5 x - 4} = 12^{2 x + 6}$

Step 1

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

$12^{x^{2} + 5 x - 4} = 12^{2 x + 6}$

Step 2

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

$x^{2} + 5 x - 4 = 2 x + 6$

Step 3

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$x^{2} + 5 x - 4 - 2 x = 6$

Subtract $2 x$ from $5 x$ .

$x^{2} + 3 x - 4 = 6$

Subtract $6$ from both sides of the equation.

$x^{2} + 3 x - 4 - 6 = 0$

Subtract $6$ from $- 4$ .

$x^{2} + 3 x - 10 = 0$

Factor $x^{2} + 3 x - 10$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 5 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 2) (x + 5) = 0$ true.

$x = 2, - 5$

Solve for x 12^(x^2+5x-4)=12^(2x+6)

Related Questions

Support LCMGCF.com by disabling your adblocker.