# How do you simplify (ab)^-1/(cd^-2) and write it using only positive exponents?

Feb 15, 2017

See the entire simplification process below:

#### Explanation:

First, let's simplify the numerator using these two rules for exponents:

$a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left(a b\right)}^{-} \frac{1}{c {d}^{-} 2} = {\left({a}^{\textcolor{red}{1}} {b}^{\textcolor{red}{1}}\right)}^{\textcolor{b l u e}{- 1}} / \left(c {d}^{-} 2\right) = \frac{{a}^{\textcolor{red}{1} \times \textcolor{b l u e}{- 1}} {b}^{\textcolor{red}{1} \times \textcolor{b l u e}{- 1}}}{c {d}^{-} 2} = \frac{{a}^{-} 1 {b}^{-} 1}{c {d}^{-} 2}$

Now, we can deal with the terms with negative exponents using these rules for exponents:

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

$\frac{{a}^{\textcolor{red}{- 1}} {b}^{\textcolor{red}{- 1}}}{c {d}^{\textcolor{red}{- 2}}} = {d}^{\textcolor{red}{- - 2}} / \left({a}^{\textcolor{red}{- - 1}} {b}^{\textcolor{red}{- - 1}} c\right) = {d}^{\textcolor{red}{- - 2}} / \left({a}^{\textcolor{red}{- - 1}} {b}^{\textcolor{red}{- - 1}} c\right) = {d}^{\textcolor{red}{2}} / \left({a}^{\textcolor{red}{1}} {b}^{\textcolor{red}{1}} c\right)$

Now, we can finalize the simplification by applying this rule for exponents:

${a}^{\textcolor{red}{1}} = a$

${d}^{\textcolor{red}{2}} / \left({a}^{\textcolor{red}{1}} {b}^{\textcolor{red}{1}} c\right) = {d}^{2} / \left(a b c\right)$