(** 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_mul_pow2 : forall a b, 0<a -> 0<=b -> log2 (a*2^b) == b + log2 a.