题目
若 $4^{\log _6 x}+x=9^{\log _6 x}, \quad 6^{\log _4 y}+y=9^{\log _4 y}$. 求 $\dfrac{x}{y}$.
解析
设 $m=\log _6 x ， n=\log _4 y $, 则 $ x=6^m, y=4^n$,则
$$
\begin{equation}
4^{\log _6 x}+x=9^{\log _9 x} \implies 4^m+6^m=9^m \implies\left(\frac{4}{9}\right)^m+\left(\frac{6}{9}\right)^m=1\qquad (1)
\end{equation}
$$
$$
\begin{equation}
6^{\log _4 y}+y=9^{\log _4 y} \implies 6^n+4^n=9^n \implies\left(\frac{4}{9}\right)^n+\left(\frac{6}{9}\right)^n=1
\end{equation}
$$
令$f(x)=\left(\dfrac{4}{9}\right)^x+\left(\dfrac{6}{9}\right)^x, f(x)$ 单调递减，由$f(m)=f(n)=1 $ 得$ m=n$.
由(1) $$\left(\frac{2}{3}\right)^{2 m}+\left(\frac{2}{3}\right)^m-1=0 \implies\left(\frac{2}{3}\right)^m=\frac{-1+\sqrt{5}}{2} $$
所以
$$\frac{x}{y}=\frac{6^m}{4^n}=\left(\frac{6}{4}\right)^m=\left(\frac{3}{2}\right)^m=\frac{2}{\sqrt{5}-1}=\frac{1+\sqrt{5}}{2} .$$