# Fraction help

1. Apr 6, 2013

### 939

1. The problem statement, all variables and given/known data

Simplify this:
y = ((k^1/2)/(l^1/4))/((l^1/5)/((k^1/8))

Ideally I would like to make rather than (k/l)/(l/k) simply (k/l)

2. Relevant equations

y = ((k^1/2)/(l^1/4))/((l^1/5)/((k^1/8))

3. The attempt at a solution

y = (k^3/8)/(l^1/20)

2. Apr 6, 2013

### SammyS

Staff Emeritus
Wow, that's hard to read. Let's render it in LaTeX.

$\displaystyle y=\frac{(k^{1/2})/(l^{1/4})}{(l^{1/5})/((k^{1/8})}$
$\displaystyle =\frac{\displaystyle \frac{k^{1/2}}{l^{1/4}}}{\displaystyle \frac{l^{1/5}}{k^{1/8}}}$​
Several ways to do this.

Invert the denominator and multiply -- as usual when dividing fractions.

Last edited: Apr 6, 2013
3. Apr 7, 2013

### 939

Thank =))... But isn't it possible to merely move the powers from the denominator to the numerator with a negative in front of them?

Last edited: Apr 7, 2013
4. Apr 7, 2013

### HallsofIvy

Staff Emeritus
Yes, and in that case you would have
$$\frac{k^{1/2}l^{-1/4}}{l^{1/5}k^{-1/8}}= \left(k^{1/2}l^{-1/4}\right)\left(k^{-1/5}l^{1/8}\right)$$

5. Apr 7, 2013

### skiller

Not quite:
$$\frac{k^{1/2}l^{-1/4}}{l^{1/5}k^{-1/8}}= \left(k^{1/2}l^{-1/4}\right)\left(k^{1/8}l^{-1/5}\right)$$