> Because the log of any base is precisely the same up to constants

i.e.,

  \log_{b_1} n = \frac{1}{\log_{b_2} b_1} \log_{b_2} n

where

  \frac{1}{\log_{b_2} b_1}

is the constant fixed for a given choice of `b_1` and `b_2`, i.e., the bases.