Cho các số thực dương x, y thỏa mãn 4logx=9logy=9log(xy/4+1) .

Đặt ${\log }_{4}x={\log }_{9}y={\log }_6\left(\frac{xy}{4}+1\right)=t\Rightarrow x=4^t,y=9^t,xy={4.6}^t-4$

$\Rightarrow 6^{2t}-{4.6}^t+4=0\Rightarrow 6^t=2\Rightarrow t={\log }_{6}2$.

Khi đó ${\log }_{4}x+{\log }_{9}y={\log }_6\left(\frac{xy}{4}+1\right)=t={\log }_{6}2\iff x=4^{{\log }_{6}2},y=9^{{\log }_{6}2}$.

Do đó $P={\left(4^{{\log }_{6}2}\right)}^{{\log }_{4}6}+{\left(9^{{\log }_{6}2}\right)}^{{\log }_{9}6}={\left(4^{{\log }_{4}6}\right)}^{{\log }_{6}2}+{\left(9^{{\log }_{9}6}\right)}^{{\log }_{6}2}=6^{{\log }_{6}2}+6^{{\log }_{6}2}=4$.