= Solution
Any extending absolute value is also non-Archimedean. Indeed its values on integers are the given base-field values, bounded by one, so the <integer criterion for a non-Archimedean absolute value> applies. For the criterion's sufficiency, the binomial theorem gives $|x+y|^n\le(n+1)\max(|x|,|y|)^n$; taking $n$th roots yields the ultrametric inequality.
Fix a $K$-basis of $L$ and its coordinate maximum norm. Every extending absolute value is a vector-space norm satisfying $\|av\|=|a|\|v\|$. The <finite-dimensional non-Archimedean norm equivalence over a complete field> makes it equivalent to that coordinate norm. Here is the relevant proof: induct on dimension. The span $W$ of all but the last basis vector $e$ is complete by the induction hypothesis, hence closed in the given norm. Therefore $\delta=\inf_{w\in W}\|e-w\|>0$. For $v=w+ae$, one has $|a|\le\|v\|/\delta$ and $\|w\|\le\max(\|v\|,|a|\|e\|)$. The induction bound on $W$ bounds all its coordinates by a constant times $\|v\|$. The reverse bound follows from the triangle inequality. This proof applies to every complete base field here, without a local compactness assumption.
Thus two extending absolute values satisfy $c|x|_1\le|x|_2\le C|x|_1$ for fixed positive constants. Apply this to $x^n$ and take $n$th roots:
$$
c^{1/n}|x|_1\le|x|_2\le C^{1/n}|x|_1.
$$
Letting $n\to\infty$ proves equality. The <bounded comparison forces equality of multiplicative absolute values> argument gives the required <unique extension of an absolute value to a finite extension>, without proving existence.
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