# How do you simplify (sqrt49(a^2)(b^-8))?

Oct 12, 2015

$\frac{7 {a}^{2}}{b} ^ 8$

#### Explanation:

Your starting expression looks like this

$\sqrt{49} \cdot {a}^{2} \cdot {b}^{- 8}$

The first thing to notice here is that $49$ is a perfect square, which means that you can write

$49 = 7 \cdot 7 = {7}^{2}$

$\sqrt{49} = \sqrt{{7}^{2}} = 7$

The expression becomes

$7 \cdot {a}^{2} \cdot {b}^{- 8}$

Next, rewrite the negative exponent by using

${x}^{- n} = \frac{1}{x} ^ n \text{ }$, for $x \ne 0$

${b}^{- 8} = \frac{1}{b} ^ 8$
$\sqrt{49} \cdot {a}^{2} \cdot {b}^{- 8} = \textcolor{g r e e n}{7 \cdot {a}^{2} \cdot \frac{1}{b} ^ 8}$