Reducing $N_1\cdot k\cdot\ln(V_1) + N_2\cdot k \cdot\ln(V_2)$ to $(N_1+N_2)\cdot k\cdot \ln(V)$

Question

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.

Answer 1

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 $N_1k\ln V_1+N_2k\ln V_2=\ln\big(V_1^{N_1k}V_2^{N_2k}\big)=(N_1+N_2)k\ln\big(V_1^{N_1/(N_1+N_2)}V_2^{N_2/(N_1+N_2)}\big)$.

Answer 2

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.