= Solution
Choose $a\in K^\times$ with $0<|a|_1<1$. For any $x\ne0$, the powers $x^n$ tend to zero precisely when $|x|_1<1$. Equality of the two topologies therefore implies
$$
|x|_1<1\quad\Longleftrightarrow\quad |x|_2<1.
$$
In particular $0<|a|_2<1$. For positive integers $n$ and arbitrary integers $m$, apply this equivalence to $x^n a^{-m}$. It says that
$$
\frac{\log|x|_1}{\log|a|_1}>\frac mn\quad\Longleftrightarrow\quad \frac{\log|x|_2}{\log|a|_2}>\frac mn.
$$
The inequality reverses on division by either negative logarithm. The rational cuts of the two real numbers are equal, so density of $\mathbb Q$ makes the numbers equal. Hence \b[the <equivalent absolute values> satisfy]
$$
\boxed{|x|_2=|x|_1^c,\qquad c=\frac{\log|a|_2}{\log|a|_1}>0.}
$$
The formula also holds at zero. Nontriviality is essential to choosing $a$; no Archimedean or non-Archimedean restriction was used in the proof.
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