# How do you simplify ((2a)/(9c))^-2 and write it using only positive exponents?

Jul 7, 2017

See a solution process below:

#### Explanation:

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

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

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

Now, use these rules of exponents to eliminate the negative exponents:

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

$\frac{{2}^{\textcolor{red}{- 2}} {a}^{\textcolor{red}{- 2}}}{{9}^{\textcolor{red}{- 2}} {c}^{\textcolor{red}{- 2}}} \implies \frac{{9}^{\textcolor{red}{- - 2}} {c}^{\textcolor{red}{- - 2}}}{{2}^{\textcolor{red}{- - 2}} {a}^{\textcolor{red}{- - 2}}} \implies \frac{{9}^{2} {c}^{2}}{{2}^{2} {a}^{2}} \implies \frac{81 {c}^{2}}{4 {a}^{2}}$

Jul 7, 2017

$\frac{81 {c}^{2}}{4 {a}^{2}}$

#### Explanation:

${\left(\frac{2 a}{9 c}\right)}^{-} 2$

;.=1/(((2a)/(9c))^2)

;.=1/((4a^2)/(81c^2))

;.=1/1 xx (81c^2)/(4a^2)

;.= (81c^2)/(4a^2)

Jul 7, 2017

$\frac{81 {c}^{2}}{4 {a}^{2}}$

#### Explanation:

One of the laws of indices states that:

${\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{+ m}$

So, by inverting the fraction, the index changes sign.

${\left(\frac{2 a}{9 c}\right)}^{\textcolor{m a \ge n t a}{- 2}} = {\left(\frac{9 c}{2 a}\right)}^{\textcolor{m a \ge n t a}{2}}$

All the indices are now positive and it only remains to square each factor.

${\left(\frac{9 c}{2 a}\right)}^{2} = \frac{81 {c}^{2}}{4 {a}^{2}}$