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Fraction help

  1. Apr 6, 2013 #1

    939

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    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. jcsd
  3. Apr 6, 2013 #2

    SammyS

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

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

    Invert the denominator and multiply -- as usual when dividing fractions.
     
    Last edited: Apr 6, 2013
  4. Apr 7, 2013 #3

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    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
  5. Apr 7, 2013 #4

    HallsofIvy

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    Yes, and in that case you would have
    [tex]\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)[/tex]
     
  6. Apr 7, 2013 #5
    Not quite:
    [tex]\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)[/tex]
    :smile:
     
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