(** And we can't find better approximations in general. - The upper bound is exact for powers of 2. - Concerning the lower bound, for any c>1, take a=b=2^c+1, then [log2_up (a*b) = c+c +1] while [(log2_up a) = (log2_up b) = c+1] *)
(** At least, we get back the usual equation when we multiply by 2 (or 2^k) *)

Lemma log2_up_succ_double : forall a, 0<a ->
log2_up (2*a+1) == 2 + log2 a.