If $K$ is a $p$-adic field, and $E_q$ and $E_{q'}$ are the corresponding Tate curves for $|q|,|q'|<1$, why does $E_q$ and $E_{q'}$ being isogenous imply that there are integers $A$ and $B$ such that $q^A=q'^B$?
So we're clear on notation, this is Exercise 5.10 (b) in Silverman's Advanced Topics in Elliptic Curves, and $E_q$ is given by $y^2+xy=x^4-5s_3(q)-\frac{5s_3(q)x+7s_5(q)}{12}$, where $s_k(q):=\sum_{n\geq 1}{\frac{n^kq^n}{1-q^n}}$.
I've tried looking this up, but I couldn't find an easy proof. From the complex picture, I'd guess that the induced map $K^\times/q^{\mathbb{Z}}\rightarrow K^{\times}/q'^{\mathbb{Z}}$ is given by $a$ mapping to $a^r$ for some $r$. If this were true, then the result would follow.
I think Silverman gives us that $K^\times/q^{\mathbb{Z}}\rightarrow K^{\times}/q'^{\mathbb{Z}}$ is a continuous surjective morphism of abelian groups. Then, I think we can reduce it to the problem of showing $\mathbb{Z}_p/\log(q)\rightarrow \mathbb{Z}_p/\log(q')$ has to be multiplication by $r$ for some $r$, but I don't know how to go from here.
Serre's $l$-adic representation theory book proves this in the appendix by going through Tate modules. Is this necessary?