# Fractional Exponents and Fractions

When dealing with fractional exponents like in the question below, how do you combine them so the two "n's" in the first fraction become one? ((how do i combine $4/3$ with $1/3$)) The aim is to end up with only one 'n' in that fraction. Same goes for the 'm'. \begin{align*} & \frac{2^{\frac{1}{3}}mn^{\frac{4}{3}}}{5^{\frac{1}{3}}m^{\frac{1}{6}}n^{\frac{1}{3}}} \div \frac{n}{2m^{\frac{1}{6}}5^{\frac{1}{3}}}\\ & = \frac{2^{\frac{1}{3}}mn^{\frac{4}{3}}}{5^{\frac{1}{3}}m^{\frac{1}{6}}n^{\frac{1}{3}}} \cdot \frac{2m^{\frac{1}{6}}5^{\frac{1}{3}}}{n}\\ & = \end{align*}

• Exponent rules are the same for integers and fractions. So $n^{\frac{1}{3}}\times n^{\frac{4}{3}}=n^{\frac{5}{3}}$. And when dividing, you subtract exponents similarly. Feb 3, 2015 at 8:25
• Please see meta.math.stackexchange.com/questions/5020/… for information about how to format mathematics on this site. Feb 3, 2015 at 13:47

$\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{5^{\frac{1}{3}}m^{\frac{1}{6}}n^{\frac{1}{3}}} \cdot \frac{2m^{\frac{1}{6}}5^{\frac{1}{3}}}{n}$
$=\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{n^{\frac{1}{3}}}\cdot \frac{2}{n}$
$=\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{n^{\frac{1}{3}}} \cdot \frac{2^{\frac{3}{3}}}{n^{\frac{3}{3}}}$
$=\frac{2^{\frac{4}{3}}m n^{\frac{4}{3}}}{n^{\frac{4}{3}}}$
$=\sqrt[3]{2^4}\cdot m$