26. Let $p, q$ satisfy $\log _{4}(p)=\log _{12}(q)$ $=\log _{18}(p+q)$. Then $\frac{q}{p}$
(A) $\frac{4}{3}$
(B) $\frac{1}{2}(1+\sqrt{3})$.
(C) $\frac{8}{5}$.
(D) $\frac{1}{2}(1+\sqrt{5})$.
(E) $\frac{16}{y}$
26
D
26. Let the given equation be $t$, then we have $p=9^{t}$, $q=12^{t}$, and $16^{t}=p+q=9^{t}+12^{t}$. Dividing the last equation by $9^{t}$, we get $\frac{16^{\mathrm{t}}}{9^{\mathrm{t}}}=\left(\frac{4^{\mathrm{t}}}{3^{\mathrm{t}}}\right)^{2}=\left(\frac{12^{\mathrm{t}}}{9^{\mathrm{t}}}\right)^{2}=\left(\frac{q}{p}\right)^{2}$. Let $x=\frac{q}{p}$, from the above equation we have $1+x=x^{2}$. $\therefore x=\frac{1}{2}(1+\sqrt{5})$ (x must be positive) hence the answer is (D).