# How do you use the laws of exponents to simplify the expression (-14a^2b^3)/ (5ab^10)?

Mar 30, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$- \frac{14}{5} \left({a}^{2} / a\right) \left({b}^{3} / {b}^{10}\right)$

Next, use these rules for exponents to simplify the $a$ terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${a}^{\textcolor{red}{1}} = a$

$- \frac{14}{5} \left({a}^{2} / a\right) \left({b}^{3} / {b}^{10}\right) \implies$

$- \frac{14}{5} \left({a}^{\textcolor{red}{2}} / {a}^{\textcolor{b l u e}{1}}\right) \left({b}^{3} / {b}^{10}\right) \implies$

$- \frac{14}{5} \left({a}^{\textcolor{red}{2} - \textcolor{b l u e}{1}}\right) \left({b}^{3} / {b}^{10}\right) \implies$

$- \frac{14}{5} \left({a}^{\textcolor{red}{1}}\right) \left({b}^{3} / {b}^{10}\right) \implies$

$- \frac{14}{5} \left(a\right) \left({b}^{3} / {b}^{10}\right) \implies$

$- \frac{14 a}{5} \left({b}^{3} / {b}^{10}\right)$

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

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

$- \frac{14 a}{5} \left({b}^{\textcolor{red}{3}} / {b}^{\textcolor{b l u e}{10}}\right) \implies$

$- \frac{14 a}{5} \left(\frac{1}{b} ^ \left(\textcolor{b l u e}{10} - \textcolor{red}{3}\right)\right) \implies$

$- \frac{14 a}{5} \left(\frac{1}{b} ^ 7\right) \implies$

$- \frac{14 a}{5 {b}^{7}}$

Mar 30, 2018

$\frac{- 14 a}{5 {b}^{7}}$
$\frac{- 14 {a}^{2} {b}^{3}}{5 a {b}^{10}}$
$\therefore = \frac{- 14 {a}^{2 - 1}}{5 {b}^{10 - 3}}$
$\therefore = \frac{- 14 a}{5 {b}^{7}}$