# Working out the level of homogeneity

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• March 25th 2013, 11:49 AM
Jackpot
Working out the level of homogeneity
I know that q= 120L1 1/2 K - (L3/√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.
• March 25th 2013, 06:43 PM
hollywood
Re: 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 $1\,\frac{1}{2}$ and K has exponent 1, so the sum is $2\,\frac{1}{2}$. For the second term, L has exponent 3 and K has exponent $-\frac{1}{2}$ (since $K^{-\frac{1}{2}}=\frac{1}{\sqrt{K}}$), so the sum is again $2\,\frac{1}{2}$.

- Hollywood
• March 26th 2013, 06:07 AM
Jackpot
Re: Working out the level of homogeneity
Thank you, for some reason I was adding those terms together.