8.23 $a>1, a$ is a constant, the product of all real roots of $x^{\lg x}=a$ is
(A) 1 .
(B) $\pm 1$.
(C) $\lg ^{2} a$.
(D) $\pm \lg a$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] From the given, we know $\lg x \cdot \lg x=\lg a$, i.e., $\lg ^{2} x=\lg a$, $\therefore \quad \lg x= \pm \sqrt{\lg a}$, thus $\lg x_{1}=\sqrt{\lg a}, \quad \lg x_{2}=-\sqrt{\lg a}$. Therefore, $\lg x_{1} x_{2}=\lg x_{1}+\lg x_{2}=\sqrt{\lg a}-\sqrt{\lg a}=0$. Hence,
$$
x_{1} x_{2}=1 \text {. }
$$

So the answer is $(A)$.