Column 4 If the equation $\lg (x-1)+\lg (3-x)=\lg (a-x)$ has only one real root with respect to $x$, find the range of real number $a$.

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The above text has been translated into English, preserving the original text's line breaks and format.
The original equation is equivalent to
$$
\left\{\begin{array}{l}
x-1>0, \\
3-x>0, \\
a-x>0, \\
(x-1)(3-x)=a-x
\end{array}\right.
$$

has exactly one real root $\Leftrightarrow\left\{\begin{array}{l}1<x<3, \\ (x-1)(3-x)=a-x\end{array}\right.$ has only one real root.
That is, for what range of values of $a$ does the function $a=-x^{2}+5 x-3$ have a unique value of the independent variable $x$ in the interval $(1,3)$. By observing the graph, we know that $a=\frac{13}{4}$ or $1<a \leqslant 3$.