Solve for x 2^(x+2)>1/32

$2^{x + 2} > \frac{1}{32}$

Raise $32$ to the power of $1$ .

$2^{x + 2} > \frac{1}{32^{1}}$

Move $32^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{x + 2} > 32^{- 1}$

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

$2^{x + 2} > 2^{- 5}$

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

$x + 2 > - 5$

Move all terms not containing $x$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$x > - 5 - 2$

Subtract $2$ from $- 5$ .

$x > - 7$

The result can be shown in multiple forms.

Inequality Form:

$x > - 7$

Interval Notation:

$(- 7, \infty)$