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)) \neq 0 $, so that $ 2 - \log(x-2) \neq 1 $. Equivalently, $ \log(x-2) \neq 1 $, so $ x \neq 12 $. Thus, the answer is $ x \in \boxed{(2,12) \cup (12,102)} $