(** And we can't find better approximations in general. - The lower bound is exact for powers of 2. - Concerning the upper bound, for any c>1, take a=b=2^c-1, then log2 (a*b) = c+c -1 while (log2 a) = (log2 b) = c-1 *)
(** At least, we get back the usual equation when we multiply by 2 (or 2^k) *)

Lemma log2_eq_succ_iff_pow2 : forall a, 0<a ->
(log2 (S a) == S (log2 a) <-> exists b, S a == 2^b).