8.29 If $x, y>0, \log _{y} x+\log _{x} y=\frac{10}{3}$ and $x y=144$, then $\frac{x+y}{2}$ equals
(A) $12 \sqrt{2}$.
(B) $13 \sqrt{3}$.
(C) 24 .
(D) 30 .
(E) 36 .
(41st American High School Mathematics Examination, 1990)
[Solution] Since $\log _{y} x=\frac{1}{\log _{x} y}$, the given equation can be transformed into
$$
\log _{x}^{2} y-\frac{10}{3} \log _{x} y+1=0 .
$$

Solving this, we get $\log _{x} y=3$ or $\frac{1}{3}$, which means $y=x^{3}$ or $x^{\frac{1}{3}}$.
Also, $x y=144$, so $x^{4}=144$ or $x^{\frac{4}{3}}=144$,
When $x^{4}=144$, $x^{2}=12$, then $x=2 \sqrt{3}(\because x>0)$,
and
$\therefore$
When
$$
\begin{array}{c}
y=x^{3}=x^{2} \cdot x=24 \sqrt{3}, \\
\frac{1}{2}(x+y)=13 \sqrt{3} ;
\end{array}
$$

While
$$
\begin{array}{c}
x^{\frac{4}{3}}=144 \text {, } x=\sqrt{12^{3}}=24 \sqrt{3}, \\
y=x^{\frac{1}{3}}=\left(\sqrt{12^{3}}\right)^{\frac{1}{3}}=\sqrt{12}=2 \sqrt{3}, \\
\frac{1}{2}(x+y)=13 \sqrt{3} .
\end{array}
$$

Therefore, the answer is $(B)$.