08/13/2007, 02:58 PM
PS. Just a side-note, I understand that (\( \mu \)) and (slog) are different functions, but if you swap them in the above equations, then you get the formula for iterated exponentials, which I found interesting.
\( {}^{x + slog_b(a)}{b} = \exp_b^{[x]}(a) \)

