Two obvious desirable features of this definition are:
- When you put two systems next to each other, considering them as one system, the total number of possible microstates $\Omega_t$ is equal to the product of $\Omega$s of the two systems, $\Omega_t=\Omega_1\times \Omega_2$. But for this system the entropy is the sum of the entropies, indicating the necessity of an exponential definition.
- The $\ln$ function has the property that the entropy of a system with one microstate $(\Omega=1)$ is zero, which is desirable.
This relation can be obtained form the assumption of equal a priori probabilities, i.e. that the equilibrium corresponds to the macrostate with the maximum number of microstates:
Consider two isolated systems, separately in equilibrium, each with macrostates $E_i^{(0)}, V_i, N_i$ (energy, volume, number of particles). Each of them has a total number of $\Omega_i(N_i,V_i,E_i^{(0)})$ possible microstates.
Now we bring them into thermal contact so that they can exchange energy. After this point, we'll have $E_t=E_1'+E_2'=\text{constant}$. $N$ and $V$ for each system remain unchanged. The total number of possible microstates for each system would be $\Omega_i(N_i,V_i,E_i')$ and for the composite system: $$\Omega=\Omega_1(N_1,V_1,E_1')\times \Omega_2(N_2,V_2,E_2')=\Omega_1(E_1')\Omega_2(E_2')=$$ $$\Omega(E_t,E_1)$$
With the assumption of equilibrium occurs at the point of having maximum $\Omega$, we find the value of $E_1^*$ (and hence $E_2^*$) that maximizes $\Omega(E_t,E_1)$:
$$d\Omega=0\to\left (\frac{\partial\Omega_1(E_1)}{\partial E_1}\right )_{E_1=E_1^*} \Omega_2(E_2^*) +\Omega_1(E_1^*)\left (\frac{\partial\Omega_2(E_2)}{\partial E_2}\right )_{E_2=E_2^*}\frac{\partial E_2}{\partial E_1}=0\tag{1}$$
$$\frac{\partial E_2}{\partial E_1}=-1\to$$
$$\beta_1=\left (\frac{\partial \ln \Omega_1(E_1)}{\partial E_1}\right )_{E_1=E_1^*}=\left (\frac{\partial \ln \Omega_2(E_2)}{\partial E_2}\right )_{E_2=E_2^*}=\beta_2\tag{2}$$
Naturally we expect these quantities $\beta_1$ and $\beta_2$ to be related to temperatures of the systems. From thermodynamics we know that
$$\left(\frac{\partial S}{\partial E}\right)_{N,V}=\frac{1}{T}\tag{3}$$
Comparing $(2)$ and $(3)$, we can conclude that:
$$\frac{\partial S}{\partial (\ln \Omega)}=k$$
or
$$\Delta S=k\ln \Omega$$
where $k$ is a constant.