# How do you simplify (g^6h^2m)/(hg^7)?

Apr 17, 2017

$\frac{h m}{g}$

#### Explanation:

$\frac{{g}^{6} {h}^{2} m}{h} {g}^{7} = {g}^{6} / {g}^{7} \cdot {h}^{2} / h \cdot m = \frac{1}{g} \cdot h \cdot m = \frac{h m}{g}$

Apr 17, 2017

See the entire solution process below:

#### Explanation:

First, use this rule of exponents to modify the $h$ term in the denominator:

$a = {a}^{\textcolor{b l u e}{1}}$

$\frac{{g}^{6} {h}^{2} m}{h {g}^{7}} = \frac{{g}^{6} {h}^{2} m}{{h}^{\textcolor{b l u e}{1}} {g}^{7}}$

Next, use these rules of exponents to simplify the $g$ and $h$ terms in the numerator and denominator:

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

$\frac{{g}^{6} {h}^{2} m}{{h}^{\textcolor{red}{1}} {g}^{7}} = \frac{{h}^{\textcolor{red}{2} - \textcolor{b l u e}{1}} m}{g} ^ \left(\textcolor{b l u e}{7} - \textcolor{red}{6}\right) = \frac{{h}^{1} m}{g} ^ 1$

Now, we can use the reverse of the first rule to finalize the simplification:

${a}^{\textcolor{red}{1}} = a$

$\frac{{h}^{1} m}{{g}^{1}} = \frac{h m}{g}$