p-adic absolute value of the p-adic logarithm over $\mathbb{Q}$

Question

How does one show that $|\log_p(r)|_p\le p^{-1}$ for $r\in\mathbb{Q}^*$?

where $\mathbb{Q}^*$ is the set of invertible rationals. And $\log_p(p)=0$ as in Iwasawa logarithm

Answer 1

If $r=p$ or $r={1\over p}$ then $|\log_{p}{(r)}|_p=|\pm1|_p=1$, otherwise let $\log_{p}(r)=p^m x$ so that $|\log_p(r)|_p=p^{-m}\leq p^{-1}$. This then implies that $m\geq1$ and furthermore $r=p^{p^m x}$, but since $r\in\mathbb{Q}$, then $x\in\mathbb{Z}$ necessarily, so that:

$$|\log_p(r)|_p=|p^m x|_p=p^{-m}\leq p^{-1}$$as desired. So I belive that only these specific types of values for $r$ will make the inequality hold.

Answer 2

If $a\in \Bbb{Z}_p$ then $$\log_p(1+pa)=\sum_{k\ge 1} \frac{(-1)^{k-1}a^kp^k}{k}$$ $k < p^k$ implies that $\frac{p^k}{k}$ is always in $p\Bbb{Z}_p$ so $\log_p(1+pa)\in p\Bbb{Z}_p$.

Then $\log_p$ is extended to the whole of $\Bbb{Q}_p^*$ with $$\log_p(p^r \zeta_{p-1}^m(1+pa)) = \log_p (1+pa) \in p \Bbb{Z}_p$$