# How do you simplify ((2a^-2b^4c^2)/(-4a^-2b^-5c^-7))^-1?

May 17, 2017

See a solution process below:

#### Explanation:

First, rewrite this expression as:

${\left(\left(\frac{2}{-} 4\right) \left({a}^{-} \frac{2}{a} ^ - 2\right) \left({b}^{4} / {b}^{-} 5\right) \left({c}^{2} / {c}^{-} 7\right)\right)}^{-} 1 \implies$

${\left(- \frac{1}{2} \cdot 1 \left({b}^{4} / {b}^{-} 5\right) \left({c}^{2} / {c}^{-} 7\right)\right)}^{-} 1 \implies$

${\left(- \frac{1}{2} \left({b}^{4} / {b}^{-} 5\right) \left({c}^{2} / {c}^{-} 7\right)\right)}^{-} 1$

Next, use this rule of exponents to simplify the terms within the parenthesis:

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

${\left(- \frac{1}{2} \left({b}^{4} / {b}^{-} 5\right) \left({c}^{2} / {c}^{-} 7\right)\right)}^{-} 1 \implies {\left(- \frac{1}{2} \left({b}^{\textcolor{red}{4}} / {b}^{\textcolor{b l u e}{- 5}}\right) \left({c}^{\textcolor{red}{2}} / {c}^{\textcolor{b l u e}{- 7}}\right)\right)}^{-} 1 \implies$

${\left(- \frac{1}{2} \left(\frac{1}{b} ^ \left(\textcolor{b l u e}{- 5} - \textcolor{red}{4}\right)\right) \left(\frac{1}{c} ^ \left(\textcolor{b l u e}{- 7} - \textcolor{red}{2}\right)\right)\right)}^{-} 1 \implies {\left(- \frac{1}{2 {b}^{-} 9 {c}^{-} 9}\right)}^{-} 1$

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

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

${\left(- \frac{1}{2 {b}^{-} 9 {c}^{-} 9}\right)}^{-} 1 \implies {\left(- \frac{1}{{2}^{\textcolor{red}{1}} {b}^{\textcolor{red}{- 9}} {c}^{\textcolor{red}{- 9}}}\right)}^{\textcolor{b l u e}{- 1}} \implies - \frac{1}{{2}^{\textcolor{red}{1} \times \textcolor{b l u e}{- 1}} {b}^{\textcolor{red}{- 9} \times \textcolor{b l u e}{- 1}} {c}^{\textcolor{red}{- 9} \times \textcolor{b l u e}{- 1}}} \implies$

$- \frac{1}{{2}^{\textcolor{red}{- 1}} {b}^{9} {c}^{9}} \implies - {2}^{\textcolor{red}{- - 1}} / \left({b}^{9} {c}^{9}\right) \implies$

$- \frac{2}{{b}^{9} {c}^{9}}$