14. Given the equation about $x$: $\lg (x-1)+\lg (3-x)=\lg (x-a)$ has only one solution, try to determine the range of the real number $a$.
14. The original equation is $\left\{\begin{array}{l}(x-1)(3-x)=x-a \\ 10\end{array}\right.$, here $f(x)=x^{2}-3 x+3-a$. Therefore, when $a=\frac{3}{4}$ or $1 \leqslant a<3$, the original equation has a unique solution.