Solve for x 3^(x^2+4x)=9^-2

$3^{x^{2} + 4 x} = 9^{- 2}$

Step 1

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

$3^{x^{2} + 4 x} = 3^{2 \cdot - 2}$

Step 2

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

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

Step 3

Solve for $x$ .

Multiply $2$ by $- 2$ .

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

Add $4$ to both sides of the equation.

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

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$