We use that \frac{1}{\Psi}\frac{\partial \Psi}{\partial \alpha_i} = \frac{\partial \ln{\Psi}}{\partial \alpha_i} , and find
\begin{align*} \ln{\Psi({\mathbf{X}})} &= -\ln{Z} - \sum_m^M \frac{(X_m - a_m)^2}{2\sigma^2} + \sum_n^N \ln({1 + e^{b_n + \sum_i^M \frac{X_i w_{in}}{\sigma^2}})}. \end{align*}