How do you simplify ((r^2t^-3)/(r^-3t^5))^-8 and write it using only positive exponents?

Apr 16, 2017

See below.

Explanation:

Let's simplify inside the expression first.

$\setminus \frac{{r}^{2} {t}^{-} 3}{{r}^{-} 3 {t}^{5}} = {r}^{5} / {t}^{8}$

This is to the $- 8$th power, so we flip the fraction, and then multiply the exponents by $8$.

=${t}^{64} / {r}^{40}$

Apr 16, 2017

See the entire solution process below:

Explanation:

First, we will use these two rules of exponents to simplify the terms within the parenthesis:

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

${\left(\frac{{r}^{\textcolor{red}{2}} {t}^{\textcolor{red}{- 3}}}{{r}^{\textcolor{b l u e}{- 3}} {t}^{\textcolor{b l u e}{5}}}\right)}^{-} 8 = {\left({t}^{\textcolor{red}{- 3} - \textcolor{b l u e}{5}} / {r}^{\textcolor{b l u e}{- 3} - \textcolor{red}{2}}\right)}^{-} 8 = {\left({t}^{-} \frac{8}{r} ^ - 5\right)}^{-} 8$

Now, we will use this rule of exponents to eliminate the outer exponent and eliminate the negative exponents:

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

${\left({t}^{\textcolor{red}{- 8}} / {r}^{\textcolor{red}{- 5}}\right)}^{\textcolor{b l u e}{- 8}} = {t}^{\textcolor{red}{- 8} \times \textcolor{b l u e}{- 8}} / {r}^{\textcolor{red}{- 5} \times \textcolor{b l u e}{- 8}} = {t}^{64} / {r}^{40}$