Question 1115862
Let's work from the outside in,
{{{f(x)=log((z))}}} so then
{{{z=log((y))}}} and
{{{z>=0}}} by substitution,
{{{log((y))>0}}} so then,
{{{y=log((x))}}} and
{{{y>=1}}} by substitution,
{{{log((x))>=1}}} so finally,
{{{x>=10}}}