Solve for x 1/2=2^((x)(x+2))

$\frac{1}{2} = 2^{(x) (x + 2)}$

Step 1

Rewrite the equation as $2^{(x) (x + 2)} = \frac{1}{2}$ .

$2^{(x) (x + 2)} = \frac{1}{2}$

Step 2

Raise $2$ to the power of $1$ .

$2^{(x) (x + 2)} = \frac{1}{2^{1}}$

Step 3

Move $2^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{(x) (x + 2)} = 2^{- 1}$

Step 4

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

$(x) (x + 2) = - 1$

Step 5

Solve for $x$ .

Simplify $x (x + 2)$ .

Apply the distributive property.

$x \cdot x + x \cdot 2 = - 1$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 2 = - 1$

Move $2$ to the left of $x$ .

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

Add $1$ to both sides of the equation.

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$