# How do you simplify ((3x^5y^4)/(x^0y^-3))^2 and write it using only positive exponents?

Feb 16, 2017

See the entire simplification process below:

#### Explanation:

First, I would use this rule of exponents to simplify the denominator of the fraction: ${a}^{\textcolor{red}{0}} = 1$

${\left(\frac{3 {x}^{5} {y}^{4}}{{x}^{\textcolor{red}{0}} {y}^{-} 3}\right)}^{2} = {\left(\frac{3 {x}^{5} {y}^{4}}{1 {y}^{-} 3}\right)}^{2} = {\left(\frac{3 {x}^{5} {y}^{4}}{y} ^ - 3\right)}^{2}$

Next, I would use this rule of exponents to simplify the terms within parenthesis: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

${\left(\frac{3 {x}^{5} {y}^{\textcolor{red}{4}}}{y} ^ \textcolor{b l u e}{- 3}\right)}^{2} = {\left(3 {x}^{5} {y}^{\textcolor{red}{4} - \textcolor{b l u e}{- 3}}\right)}^{2} = {\left(3 {x}^{5} {y}^{7}\right)}^{2}$

Now, I would use these two rules of exponents to complete the simplification: $a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left(3 {x}^{5} {y}^{7}\right)}^{2} = {\left({3}^{\textcolor{red}{1}} {x}^{\textcolor{red}{5}} {y}^{\textcolor{red}{7}}\right)}^{\textcolor{b l u e}{2}} = {3}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} {x}^{\textcolor{red}{5} \times \textcolor{b l u e}{2}} {y}^{\textcolor{red}{7} \times \textcolor{b l u e}{2}} = {3}^{2} {x}^{10} {y}^{14} = 9 {x}^{10} {y}^{14}$