$6 \cdot 64$ If $a>1, b>1$, and $\lg (a+b)=\lg a+\lg b$, then the value of $\lg (a-1)+$ $\lg (b-1)$
(A) equals $\lg 2$.
(B) equals 1.
(C) equals 0.
(D) is not a constant independent of $a$ and $b$.
(China High School Mathematics League, 1998)
[Solution] $\because \quad \because \quad \lg (a+b)=\lg a+\lg b$,
$\therefore \quad a+b=a b$, which means $(a-1)(b-1)=1$.
Therefore, $\quad \lg (a-1)+\lg (b-1)=0$.
Hence, the answer is $(C)$.