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Y/l =(k/l)^α (e)^1-α

  1. Sep 29, 2008 #1
    Dear People would someone be able to explain in more detail this problem. I can't follow the answer.
    a=alpha (its a superscript)

    Y/L =(K/L)^α (E)^1-α

    These are steps to get to final answer

    Y/L=K^a/L^-a * E/E^a Step 1 - rewrite

    Y=K^a/L^a * EL/E^a Step2 multiply by L

    How did L^-a became L^a after multiplying by L. Is this rule of negative exponent??

    Now with some algebra, we can rearrange it; Step 3

    Y=K^a * EL/(EL)^a

    please help me understand in detail how you end up with K^a after multiplying by L. How can L cancel out L^a

    Answer

    Y=K^a (EL)^1-a

    Gog Bless
     
  2. jcsd
  3. Sep 30, 2008 #2

    Defennder

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    Homework Helper

    This doesn't follow from the given equation. It's Ka/La

    You could write it as

    Y/L = KaL-a * E/Ea

    Simply multiply the right side by L. So we now have L1-a on the right side. Put the one with the positive power, L as the numerator and L-a as La in the denominator. Remember that L-a = 1/La. Then move this term onto the E/Ea fraction. This can be done because the order of multiplication doesn't matter.
     
  4. Sep 30, 2008 #3

    HallsofIvy

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    There is no "problem" here. Just an equation. What do you want to do with it?

    It doesn't. First you may have copied step 1 incorrectly: (K/L)^a= K^a/L^a or K^aL^(-a), not K^a/L^-a. Second, the two terms, (K/L)^a and E^(1-a) are multiplied, not added- the "distributive law" does not apply and you do not multiply both terms by L.
    Multiplying both sides of Y/L= K^a/L^a * E/E^a by L gives Y= (K^a/L^a)* (EL)/E^a. Whoever is doing this chose to multiply the second term by the L, not the first, so that stays the same.

    All they did here was take the "L^a" from the denominator of the first fraction and put it in the denominator of the second. Since you multiply fractions by multiplying numerators and denominators separatly, it doesn't matter:
    [tex]\frac{a}{b}\frac{c}{d}= \frac{ac}{bd}= a\frac{c}{bd}[/tex]

    Do you believe in Gog?
     
  5. Sep 30, 2008 #4
    I want to thank all of you both taking your time and helping me out. I need to go back and review my calculus book.
     
  6. Sep 30, 2008 #5
     
    Last edited: Sep 30, 2008
  7. Sep 30, 2008 #6
    Thank you so much for helping me out.
    God bless!!!
     
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