The minimum value of $\frac{1}{x}\log x$ in the interval [2, $\infty$] is

let $f\left(x\right)=\frac{\log x}{x}$
$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{1-\log x}{x^2}=0\Rightarrow x=e\\ f^{\prime \prime }\left(x\right)=\frac{-x-2x(1-\log x)}{x^4}\end{array}$
Put x = e
$f^{\prime \prime }\left(e\right)=\frac{-e}{e^4}<0$
f(x) is maximum at x = e and minimum value does not exist.