# How do you simplify (8x^4y^-3)(1/2x^-5y^2) and write it using only positive exponents?

Feb 26, 2017

See the entire simplification process below:

#### Explanation:

First, rewrite this expression as:

$\left(8 \times \frac{1}{2}\right) \left({x}^{4} {x}^{-} 5\right) \left({y}^{-} 3 {y}^{2}\right)$

$4 \left({x}^{4} {x}^{-} 5\right) \left({y}^{-} 3 {y}^{2}\right)$

Next, use this rule of exponents to combine the $x$ and $y$ terms: ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$4 \left({x}^{\textcolor{red}{4}} \times {x}^{\textcolor{b l u e}{- 5}}\right) \left({y}^{\textcolor{red}{- 3}} \times {y}^{\textcolor{b l u e}{2}}\right) = 4 {x}^{\textcolor{red}{4} + \textcolor{b l u e}{- 5}} {y}^{\textcolor{red}{- 3} + \textcolor{b l u e}{2}} = 4 {x}^{-} 1 {y}^{-} 1$

Now, use these rules for exponents to complete the simplification: ${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$ and ${a}^{\textcolor{red}{1}} = a$

$4 {x}^{\textcolor{red}{- 1}} {y}^{\textcolor{red}{- 1}} = \frac{4}{{x}^{\textcolor{red}{- - 1}} {y}^{\textcolor{red}{- - 1}}} = \frac{4}{{x}^{\textcolor{red}{1}} {y}^{\textcolor{red}{1}}} = \frac{4}{x y}$