Kapranov’s theorem takes care of $\operatorname{Log}(w(V_{K}))=\operatorname{Log}_{K}(V_{K})$. We need to prove that the values $u(a_{j})$ take care of the arguments of $W(V_{K})$. Let $x\in\operatorname{Log}_{K}(V_{K})$. By Kapranov’s theorem it means that there is a set of indices $j_{1},\dots,j_{l}$ such that $\operatorname{val}(a_{j_{1}})=\dots=\operatorname{val}(a_{j_{l}})\geq\operatorname{val}(a_{j})$ for any other index $j$. Let $z\in(K^{*})^{n+1}$ be a point such that $\operatorname{Log}_{K}(z)=x$. The lowest powers of $t$ in the Puiseux series $f(z)$ are contributed by the monomials $a_{j_{1}}z^{j_{1}},\dots,a_{j_{l}}z^{j_{l}}$. If $f(z)=0$ then the coefficients at these lowest powers are such that their sum is zero. Conversely, the higher powers of $t$ can be arranged to make $f(z)=0$ without the change of $W(z)$ as in the proof of Kapranov’s theorem. ∎