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

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

Step 1

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

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

Step 2

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

$2^{x + 2} \geq 32^{- 1}$

Step 3

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

$2^{x + 2} \geq 2^{- 5}$

Step 4

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

$x + 2 \geq - 5$

Step 5

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

Subtract $2$ from both sides of the inequality.

$x \geq - 5 - 2$

Subtract $2$ from $- 5$ .

$x \geq - 7$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x \geq - 7$

Interval Notation:

$[- 7, \infty)$

Step 7