Functions Warmup 1 (Domain)

A couple of days have gone by so I will post the solution here.
For $f$ to be well defined, the following restrictions on x have to exist:

1) The denominator cannot be zero. Therefore: $\ln x \neq 0 \Leftrightarrow \ln x \neq ln 1 \Leftrightarrow x \neq 1$
2) The argument of the logarithm must be strictly positive. Therefore: $x>0$
3) The expression under the radical must be nonnegative. Therefore: $4 - |3-x|\geqslant 0\Leftrightarrow |3-x|\leqslant 4\Leftrightarrow -4\leqslant 3-x \leqslant 4\Leftrightarrow -7\leqslant -x \leqslant 1\Leftrightarrow -1\leqslant x \leqslant 7$

By combining the three restrictions above we get $\boxed{x \in (0,1)\cup (1,7]}$ which is our final answer.