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

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

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

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

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

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

Solve for $x$ .

Divide each term in $2 (x^{2} + 4 x) = - 6$ by $2$ and simplify.

Divide each term in $2 (x^{2} + 4 x) = - 6$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x^{2} + 4 x$ by $1$ .

$x^{2} + 4 x = \frac{- 6}{2}$

Simplify the right side.

Divide $- 6$ by $2$ .

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

Add $3$ to both sides of the equation.

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

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

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

$x + 1 = 0$

$x + 3 = 0$

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.

$x = - 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

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

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