(5) was intended to assume that $~$n \in \mathbb R^{\ge 1},$~$ or possibly $~$\in \mathbb R^{\ge 0}$~$ if you want an easy way to prove (6). In that case, how does (8) not follow from (5)? (If $~$f(x^y)=yf(x)$~$ in general, then $~$f(b^n)=nf(b)$~$ and $~$f(b)=1 \implies f(b^n)=n,$~$ unless I'm missing something.)