$8 \cdot 30$ If $\log _{b}{ }^{2} x+\log _{x}^{2} b=1, b>0, b \neq 1, x \neq 1$, then $x$ equals
(A) $\frac{1}{b^{2}}$.
(B) $\frac{1}{b}$.
(C) $b^{2}$.
(D) $b$.
(E) $\sqrt{b}$.
(15th American High School Mathematics Examination, 1964)
[Solution] Let $\log _{b}{ }^{2} x=m, \log _{x}{ }^{2} b=n$, then
$$
x=b^{2 m}, \quad b=x^{2 n} .
$$
$\therefore \quad\left(x^{2 n}\right)^{2 m}=b^{2 m}$, i.e., $x^{4 m n}=x$.
$\therefore \quad 4 m n=1$, so $n=\frac{1}{4 m}$.
Given $m+n=1$, i.e., $m+\frac{1}{4 m}=1$,
simplifying gives $4 m^{2}-4 m+1=0$, yielding $m=\frac{1}{2}$, i.e., $x=b$. Therefore, the answer is $(D)$.