# Y/l =(k/l)^α (e)^1-α

1. Sep 29, 2008

### albert2008

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

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

Gog Bless

2. Sep 30, 2008

### Defennder

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.

3. Sep 30, 2008

### HallsofIvy

Staff Emeritus
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:
$$\frac{a}{b}\frac{c}{d}= \frac{ac}{bd}= a\frac{c}{bd}$$

Do you believe in Gog?

4. Sep 30, 2008

### albert2008

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.

5. Sep 30, 2008

### albert2008

Last edited: Sep 30, 2008
6. Sep 30, 2008

### albert2008

Thank you so much for helping me out.
God bless!!!