The function f(x)=x/(log x) increases on the interval

Clearly, f(x) is defined for x > 0

Now,

$f(x)=\frac{x}{\log x} \Rightarrow f'(x)=\frac{\log x-1}{(\log x)^2}$

∴  $f'(x)>0 \Rightarrow \log x-1>0 \Rightarrow \log x>1 \Rightarrow x>e \Rightarrow x \in(e, \infty)$