The proof of (5) only goes through for $~$n\in\mathbb{N}$~$.
You can prove a version of (8) from (5), namely, $~$f(b)=1\Rightarrow f(b^q)=q$~$ for $~$q\in\mathbb{Q}$~$, but this doesn't pin down $~$f$~$ completely, unless you include a continuity condition.