# How do you simplify the expression ((x^7y^-2)/(3y^-3))^-2 using the properties?

Jun 15, 2017

See a solution process below:

#### Explanation:

First, to eliminate the outer exponent, use these rules for exponents:

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

${\left(\frac{{x}^{7} {y}^{-} 2}{3 {y}^{-} 3}\right)}^{-} 2 \implies {\left(\frac{{x}^{\textcolor{red}{7}} {y}^{\textcolor{red}{- 2}}}{{3}^{\textcolor{red}{1}} {y}^{\textcolor{red}{- 3}}}\right)}^{\textcolor{b l u e}{- 2}} \implies \frac{{x}^{\textcolor{red}{7} \cdot \textcolor{b l u e}{- 2}} {y}^{\textcolor{red}{- 2} \cdot \textcolor{b l u e}{- 2}}}{{3}^{\textcolor{red}{1} \cdot \textcolor{b l u e}{- 2}} {y}^{\textcolor{red}{- 3} \cdot \textcolor{b l u e}{- 2}}} \implies$

$\frac{{x}^{-} 14 {y}^{4}}{{3}^{-} 2 {y}^{6}}$

Next, use this rule of exponents to simplify the $y$ terms:

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

$\frac{{x}^{-} 14 {y}^{\textcolor{red}{4}}}{{3}^{-} 2 {y}^{\textcolor{b l u e}{6}}} \implies {x}^{-} \frac{14}{{3}^{-} 2 {y}^{\textcolor{b l u e}{6} - \textcolor{red}{4}}} \implies {x}^{-} \frac{14}{{3}^{-} 2 {y}^{2}}$

Then use this rule of exponents to eliminate the negative exponent for the $x$ term:

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

${x}^{\textcolor{red}{- 14}} / \left({3}^{-} 2 {y}^{2}\right) \implies \frac{1}{{3}^{-} 2 {x}^{\textcolor{red}{- - 14}} {y}^{2}} \implies \frac{1}{{3}^{-} 2 {x}^{14} {y}^{2}}$

Now, use this rule of exponents to eliminate the negative exponent for the $3$ term:

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

$\frac{1}{{3}^{\textcolor{red}{- 2}} {x}^{14} {y}^{2}} \implies {3}^{\textcolor{red}{- - 2}} / \left({x}^{14} {y}^{2}\right) \implies {3}^{2} / \left({x}^{14} {y}^{2}\right) \implies \frac{9}{{x}^{14} {y}^{2}}$