# How do you simplify the expression (8e^-4f^-2)/(18ef^-5) using the properties?

Jun 7, 2017

See a solution process below:

#### Explanation:

first, simplify the constants as:

$\frac{8 {e}^{-} 4 {f}^{-} 2}{18 e {f}^{-} 5} = \frac{\left(2 \times 4\right) {e}^{-} 4 {f}^{-} 2}{\left(2 \times 9\right) e {f}^{-} 5} =$

$\frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 4\right) {e}^{-} 4 {f}^{-} 2}{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 9\right) e {f}^{-} 5} = \frac{4 {e}^{-} 4 {f}^{-} 2}{9 e {f}^{-} 5}$

Next, use these rules of exponents to simplify the $e$ terms:

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

$\frac{4 {e}^{-} 4 {f}^{-} 2}{9 e {f}^{-} 5} = \frac{4 {e}^{\textcolor{red}{- 4}} {f}^{-} 2}{9 {e}^{\textcolor{red}{1}} {f}^{-} 5} =$

$\frac{4 {f}^{-} 2}{9 {e}^{\textcolor{red}{1} - \textcolor{red}{- 4}} {f}^{-} 5} = \frac{4 {f}^{-} 2}{9 {e}^{\textcolor{red}{1} + \textcolor{red}{4}} {f}^{-} 5} =$

$\frac{4 {f}^{-} 2}{9 {e}^{5} {f}^{-} 5}$

Now, use this rule of exponents to simplify the $f$ term:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\frac{4 {f}^{\textcolor{red}{- 2}}}{9 {e}^{5} {f}^{\textcolor{b l u e}{- 5}}} = \frac{4 {f}^{\textcolor{red}{- 2} - \textcolor{b l u e}{- 5}}}{9 {e}^{5}} =$

$\frac{4 {f}^{\textcolor{red}{- 2} + \textcolor{b l u e}{5}}}{9 {e}^{5}} = \frac{4 {f}^{3}}{9 {e}^{5}}$