How to prove $\exp(\log x)=x$ in $\mathbb{Q}_p$
$\mathbb{Q}_p$ is the $p$-adic completion of $\mathbb{Q}$.
We define:
$$\text{exp}(x) =\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
$$\text{log}(x)=\sum_{n=1}^{\infty}\frac{(x-1)^n(-1)^{(n-1)}}{n}$$
Then how to prove that $\exp(\log(x))=\log(\exp(x))=x$ in the convergence domain?
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$\begingroup$The relations are easy to see; care has to be taken for the region of convergence of the functions. For example, the $p$-adic exponential function $\exp_p$ converges in the disk $$ D_p=\{x\in \mathbb{Q}_p\mid |x|_p <p^{-1/(p-1)}\} $$ The we have the following result:
Proposition: The maps $\exp_p\colon D_p\rightarrow 1+D_p$ and $\log_p:1+D_p\rightarrow D_p$ are inverse to each other, i.e., we have $$ \log_p(\exp_p(x))=x,\; \exp_p(\log_p(1+x))=1+x. $$ Proof: By considering the corresponding formal power series, we know that the relations given here hold. So the only thing we have to make sure is that all involved power series converge. This is easily verified. We have $$ |\exp_p(x)-1|_p\le |x|_p,\; |\log_p(1+x)|_p\le |x|_p. $$
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