10. The equation $\lg (a x+1)=\lg (x-1)+\lg (2-x)$ has a unique real solution for $x$, then the range of the real number $a$ is $\qquad$
Consider $\left(-1,-\frac{1}{2}\right] \cup\{3-2 \sqrt{3}\}$.
To make the original equation meaningful, it must satisfy:
\[
\left\{\begin{array}{l}
a x+1>0 \\
x-1>0 \\
2-x>0
\end{array} \quad\left\{\begin{array}{l}
a x>-1 \\
x>1 \\
x-1 \\
1<x<2
\end{array}\right.\right.\right.
\]
From $\lg (a x+1)=\lg (x-1)+\lg (2-x)$, we know
\[
\lg (a x+1)=\lg [(x-1)(2-x)], \quad a x+1=(x-1)(2-x)
\]

Draw the rough graphs of the functions $y=a x+1$ and $y=(x-1)(2-x)=-x^{2}+3 x-2,1<x<2$.
When the line $y=a x+1$ passes through the point $(1,0)$, $a=-1$; when the line $y=a x+1$ passes through the point $(2,0)$, $a=-\frac{1}{2}$;
When the line $y=a x+1$ is tangent to $y=(x-1)(2-x)$, let the point of tangency be $P\left(x_{0}, y_{0}\right)$. By $y'=-2 x+3$, we know that at this point $a=-2 x_{0}+3$, and we have
\[
\left\{\begin{array}{l}
y_{0}=(-2 x_{0}+3) x_{0}+1 \\
y_{0}=-x_{0}^{2}+3 x_{0}-2
\end{array} \text { Solving gives } \left\{\begin{array}{l}
x_{0}=\sqrt{3} \\
y_{0}=3 \sqrt{3}-5
\end{array}\right.\right.
\]

Therefore, $a \in\left(-1,-\frac{1}{2}\right] \cup\{3-2 \sqrt{3}\}$, so the answer is $\left(-1,-\frac{1}{2}\right] \cup\{3-2 \sqrt{3}\}$.