How do I simplify this k/l ratio?

[939]

Homework Statement

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)

Homework Equations

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

The Attempt at a Solution

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

 

[SammyS]

Homework Statement

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)

Homework Equations

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

The Attempt at a Solution

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

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.

 

[939]

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.

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

 

[HallsofIvy]

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)

 

[skiller]

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)

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)