8.40 Let $\frac{\lg a}{p}=\frac{\lg b}{q}=\frac{\lg c}{r}=\lg x, x \neq 1$, if $\frac{b^{2}}{a c}=x^{y}$, then $y$ equals
(A) $\frac{q^{2}}{p+r}$.
(B) $\frac{p+r}{2 q}$.
(C) $2 q-p-r$.
(D) $2 q-p r$.
(E) $q^{2}-p r$.
(18th American High School Mathematics Examination, 1967)
［Solution］Let $a=x^{p}, b=x^{q}, c=x^{r}$, therefore
$$
\begin{aligned}
& \frac{b^{2}}{a c}=\frac{x^{2 q}}{x^{p} \cdot x^{r}}=x^{2 q-p-r}=x^{y}, \\
\therefore \quad & y=2 q-p-r .
\end{aligned}
$$

Hence, the correct choice is $(C)$.