Question 23. Find the minimum value of $2 \lg (x-3)-\lg (x-4)$.

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Since the argument of a logarithm is positive,
\[
\therefore x-3>0, x-4>0 \text{. }
\]

Therefore, in the range $x>4$,
\[
\begin{array}{l}
2 \lg (x-3)-\lg (x-4)=\lg \frac{(x-3)^{2}}{x-4} . \\
\text{Let } y=\frac{(x-3)^{2}}{x-4}, \text{ then} \\
y=\frac{x^{2}-6 x+9}{x-4}=-\frac{(x-2)(x-4)+1}{x-4} \\
\quad=x-2+\frac{1}{x-4}=x-4+\frac{1}{x-4}+2 . \\
\because x>4, \\
\therefore x-4+\frac{1}{x-4} \geqslant 2 \sqrt{(x-4) \cdot \frac{1}{x-4}}=2 . \\
\therefore y \geqslant 2+2=4 .
\end{array}
\]

The equality holds if and only if $x-4=\frac{1}{x-4}$, i.e., $x=5$. Therefore, when $x=5$, $y$ attains its minimum value of 4.
\[
\therefore \text{The required minimum value is } \lg 4=2 \lg 2.
\]