6.41 Given $a=\log _{0.1} b$, then $\lg b^{a}$ equals
(A) $a^{2}$.
(B) $-2 a$.
(C) $-a^{a}$.
(D) $-a^{2}$.
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics league, 1986)
[Solution] From $a=\log _{0.1} b=\frac{\lg b}{\lg 0.1}=-\lg b$; also $\lg b^{a}=a \lg b$, we get $\lg b^{a}=a \cdot(-a)=-a^{2}$. Therefore, the answer is $(D)$.