8-34 Let $p, q$ satisfy $\log _{9} p=\log _{12} q=\log _{16}(p+q)$. Then $\frac{q}{p}$ equals
(A) $\frac{4}{3}$.
(B) $\frac{1}{2}(1+\sqrt{3})$.
(C) $\frac{8}{5}$.
(D) $\frac{1}{2}(1+\sqrt{5})$.
(E) $\frac{16}{9}$.
(39th American High School Mathematics Examination, 1988)
[Solution] From the given and the change of base formula, we have
$$
\frac{\lg p}{\lg 9}=\frac{\lg q}{\lg 12}=\frac{\lg (p+q)}{\lg 16}.
$$

By the property of equal ratios, we have
$$
\frac{\lg p+\lg q+\lg (p+q)}{\lg 9+\lg 12+\lg 16}=\frac{\lg q}{\lg 12},
$$

which is $\square$
or
$$
\begin{array}{l}
\frac{\lg p q(p+q)}{\lg 12^{3}}=\frac{\lg q}{\lg 12}, \\
\lg p q(p+q)=3 \lg q
\end{array}
$$
$\therefore \lg p q(p+q)-\lg q^{3}=0$, i.e., $\quad \lg \frac{p q(p+q)}{q^{3}}=0$.
Thus, $\frac{p q(p+q)}{q^{3}}=1$, then $q^{2}-p q-p^{2}=0$,
$\therefore \left(\frac{q}{p}\right)^{2}-\left(\frac{q}{p}\right)-1=0$.
Solving this, we get
$\frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$ (discard the negative value).

Therefore, the answer is $(D)$.