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*}

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    $\begingroup$ 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. $\endgroup$ Feb 3, 2015 at 8:25
  • $\begingroup$ Please see meta.math.stackexchange.com/questions/5020/… for information about how to format mathematics on this site. $\endgroup$ Feb 3, 2015 at 13:47

1 Answer 1


As far as I can see this is what your term looks like:

$\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{5^{\frac{1}{3}}m^{\frac{1}{6}}n^{\frac{1}{3}}} \cdot \frac{2m^{\frac{1}{6}}5^{\frac{1}{3}}}{n}$

$=\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{n^{\frac{1}{3}}}\cdot \frac{2}{n}$

$=\frac{2^{\frac{1}{3}}m n^{\frac{4}{3}}}{n^{\frac{1}{3}}} \cdot \frac{2^{\frac{3}{3}}}{n^{\frac{3}{3}}}$

$=\frac{2^{\frac{4}{3}}m n^{\frac{4}{3}}}{n^{\frac{4}{3}}}$

$=\sqrt[3]{2^4}\cdot m$


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