Solve for x (x+2)^(x+4)=(x+2)^(x^2+3x+1)

${(x + 2)}^{x + 4} = {(x + 2)}^{x^{2} + 3 x + 1}$

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

${(x + 2)}^{x + 4} = {(x + 2)}^{x^{2} + 3 x + 1}$

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

$x + 4 = x^{2} + 3 x + 1$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 3 x + 1 = x + 4$

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

Subtract $x$ from both sides of the equation.

$x^{2} + 3 x + 1 - x = 4$

Subtract $x$ from $3 x$ .

$x^{2} + 2 x + 1 = 4$

Subtract $4$ from both sides of the equation.

$x^{2} + 2 x + 1 - 4 = 0$

Subtract $4$ from $1$ .

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

Factor $x^{2} + 2 x - 3$ 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 $2$ .

$- 1, 3$

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 $0$ .

$x - 1 = 0$

Add $1$ to 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 (x+2)^(x+4)=(x+2)^(x^2+3x+1)

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