 Solution

Choose $a\in K^{\times }$ with $0<|a{|}_1<1$. For any $x\ne 0$, the powers $x^n$ tend to zero precisely when $|x{|}_1<1$. Equality of the two topologies therefore impliesIn particular $0<|a{|}_2<1$. For positive integers $n$ and arbitrary integers $m$, apply this equivalence to $x^n{a}^{-m}$. It says thatThe 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 the equivalent absolute values satisfyThe formula also holds at zero. Nontriviality is essential to choosing $a$; no Archimedean or non-Archimedean restriction was used in the proof.