Cho x,y>0 thỏa mãn log 6 x=log 9 y

* Hướng dẫn giải

Chọn B.

Đặt ${\log }_{6}x={\log }_{9}y={\log }_4\left(2x+2y\right)=t\Rightarrow \left\{\begin{array}{l}x=6^t\\ y=9^t\\ 2x+2y=4^t\end{array}\right.$

$\Rightarrow {2.6}^t+{2.9}^t=4^t\iff {\left(\frac{2}{3}\right)}^{2t}-2.{\left(\frac{2}{3}\right)}^t-2=0\iff \left[\begin{array}{l}{\left(\frac{2}{3}\right)}^t=1+\sqrt{3}\left(n\right)\\ {\left(\frac{2}{3}\right)}^t=1-\sqrt{3}\left(l\right)\end{array}\right.\Rightarrow {\left(\frac{2}{3}\right)}^t=1+\sqrt{3}.$

Vậy $\frac{x}{y}=1+\sqrt{3}.$