Write the expression so that the variable is only presented once and the exponents are positive. There is this problem 
$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}$
and i know that the answer is 
$\frac{2x^{\frac{5}{6}}}{3y}$
but i cant find the logic of how is solved, Where do the coefficients come from?
I understand that fractions are the result of multiplication of expressions, but I do not understand how all come together. 
 A: You should know that $\frac {a^{k}}{a^{m}} = a^{k-m}$.
So.....
$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}=\frac {10}{15}\cdot \frac {x^{\frac 13}}{x^{-\frac 12}}\cdot \frac {y^{-\frac 14}}{y^{\frac 34}}=$
$\frac 23\cdot x^{\frac 13 -(-\frac 12)}\cdot y^{-\frac 14-\frac 34}=$
$\frac 23\cdot x^{\frac 56}\cdot y^{-1}=$
$\frac 23 \cdot x^{\frac 56}\cdot \frac 1y=$
$\frac {2x^{\frac 56}}{3y}$
A: You need to know some basic Exponential identities :
$$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}$$
$$=\frac{10}{15} \times \frac{x^{1/3}}{x^{-1/2}} \times \frac{y^{-1/4}}{y^{3/4}}$$
Using $\frac{x^a}{x^b} = x^{a-b}$
$$=\frac{5 \times 2}{5 \times 3} \times x^{1/3+1/2} \times y^{-1/4-3/4}$$
$$=\frac{2}{3} \times x^{5/6} \times y^{-1}$$
$$=\frac{5 \times 2}{5 \times 3} \times x^{5/6} \times y^{-1}$$
$$=\frac{2x^{\frac{5}{6}}}{3y}$$
A: Note that $x^{r}x^{q} = x^{r+q}$, so if $r = -q$, then $x^{-q}x^{q} = x^{-q+q} = x^{0} = 1$. Thus, you can "remove" the negative exponent factors by multiplying the numerator & denominator by the factor using the corresponding positive exponent, as shown below
$$\begin{equation}\begin{aligned}
\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}} & = \frac{2(5)x^{\frac{1}{3}}y^{-\frac{1}{4}}}{3(5)x^{-\frac{1}{2}}y^{\frac{3}{4}}} \left(\frac{x^{\frac{1}{2}}y^{\frac{1}{4}}}{x^{\frac{1}{2}}y^{\frac{1}{4}}}\right) \\
& = \frac{2x^{\frac{1}{3} + \frac{1}{2}}y^{\frac{-1}{4} + \frac{1}{4}}}{3x^{\frac{-1}{2} + \frac{1}{2}}y^{\frac{3}{4} + \frac{1}{4}}} \\
& = \frac{2x^{\frac{1}{3} + \frac{1}{2}}}{3y^{\frac{3}{4} + \frac{1}{4}}} \\
& = \frac{2x^{\frac{5}{6}}}{3y}\end{aligned}\end{equation}\tag{1}\label{eq1}$$
