jmc

The inner logarithm is only defined if $x-2>0$, so $x>2$. Furthermore, the outer logarithm is only defined if $2-\log (x-2)>0$, which implies that $2>\log (x-2)$, so that $100>x-2$. Thus, $x<102$. Finally, it must also be true that $\log (2-\log (x-2))\ne 0$, so that $2-\log (x-2)\ne 1$. Equivalently, $\log (x-2)\ne 1$, so $x\ne 12$. Thus, the answer is $x\in \boxed{(2,12)\cup (12,102)}$