The set of all x for which $1+\log x<x$, is

$(1, \infty)$ $(0,1)$ $(0, \infty)$ none of these

$(1, \infty)$

$1+\log x<x$ so $\log x < x-1$ $⇒x<e^{x-1}$ $⇒x∈(1, \infty)$