The other answers are excellent, but I thought I’d add one other approach that can be faster in some circumstances – using broadcasting and masking to achieve the same result:
import numpy as np
mask = (z['b'] != 0)
z_valid = z[mask]
z['c'] = 0
z.loc[mask, 'c'] = z_valid['a'] / np.log(z_valid['b'])
Especially with very large dataframes, this approach will generally be faster than solutions based on apply().