Is there a way to reduce $N_1\cdot k\cdot\ln(V_1) + N_2\cdot k \cdot\ln(V_2)$ to $(N_1+N_2)\cdot k\cdot \ln(V)$ where $V = V_1 + V_2$ ? I have arrived at $k\cdot[\ln(V_1)^{N_1} + \ln(V_2)^{N_2}]$ but not sure what is the next step.
Reducing $N_1\cdot k\cdot\ln(V_1) + N_2\cdot k \cdot\ln(V_2)$ to $(N_1+N_2)\cdot k\cdot \ln(V)$
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algebra-precalculus
logarithms
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0I'm not sure this questions should have been migrated so fast. I guess OP is aiming at additivity of entropy(?). Maybe the question can be edited to reflect that and be migrated back. Although I'm not sure where you came up with this expression. It resembles, but is not identical to the entropy of an ideal gas. See https://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation – 2017-02-13
2 Answers
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No. For this to be true you'd need $V=V_1^{N_1/(N_1+N_2)}V_2^{N_2/(N_1+N_2)}$.
This is because $N_1k\ln V_1+N_2k\ln V_2=\ln\big(V_1^{N_1k}V_2^{N_2k}\big)=(N_1+N_2)k\ln\big(V_1^{N_1/(N_1+N_2)}V_2^{N_2/(N_1+N_2)}\big)$.
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Even more simply: let $N_1 = N_2 = k = 1$: now your expression reduces to $\ln V_1 + \ln V_2 = 2\ln\left(V_1 + V_2\right)$, which is false.