# How do you simplify and write (2b)^-5 with positive exponents?

Jan 28, 2017

See the entire simplification process below:

#### Explanation:

First, use this rule for exponents to eliminate the negative exponent:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

${\left(2 b\right)}^{-} 5 = \frac{1}{2 b} ^ \left(- - 5\right) = \frac{1}{2 b} ^ 5$

Now, use these rules for exponents to complete the simplification:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

$\frac{1}{2 b} ^ 5 = \frac{1}{{2}^{1} {b}^{1}} ^ 5 = \frac{1}{{2}^{1 \times 5} {b}^{1 \times 5}} = \frac{1}{{2}^{5} {b}^{5}} = \frac{1}{32 {b}^{5}}$

Jan 28, 2017

$\frac{1}{32 {b}^{5}}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{laws of exponents}}$

color(red)(bar(ul(|color(white)(2/2)color(black)((ab)^m=a^mb^m ; a^-m=1/a^m)color(white)(2/2)|)))

$\Rightarrow {\left(2 b\right)}^{-} 5 = \frac{1}{2 b} ^ 5$

$= \frac{1}{{2}^{5} {b}^{5}} = \frac{1}{32 {b}^{5}}$