I know that q= 120L^{1 1/2}K - (L^{3}/√K) has homogenous degree of 2 1/2 but I'm unsure how that has been worked out. I think its the square root thats throwing me, could someone show me how its 2 1/2?

Thanks in advance.

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- Mar 25th 2013, 10:49 AMJackpotWorking out the level of homogeneity
I know that q= 120L

^{1 1/2}K - (L^{3}/√K) has homogenous degree of 2 1/2 but I'm unsure how that has been worked out. I think its the square root thats throwing me, could someone show me how its 2 1/2?

Thanks in advance. - Mar 25th 2013, 05:43 PMhollywoodRe: Working out the level of homogeneity
In this case, homogeneous means that the sum of the exponents of L and K for the first term is the same as the sum of the exponents of L and K for the second term.

For the first term, L has exponent $\displaystyle 1\,\frac{1}{2}$ and K has exponent 1, so the sum is $\displaystyle 2\,\frac{1}{2}$. For the second term, L has exponent 3 and K has exponent $\displaystyle -\frac{1}{2}$ (since $\displaystyle K^{-\frac{1}{2}}=\frac{1}{\sqrt{K}}$), so the sum is again $\displaystyle 2\,\frac{1}{2}$.

- Hollywood - Mar 26th 2013, 05:07 AMJackpotRe: Working out the level of homogeneity
Thank you, for some reason I was adding those terms together.