Logging off

$\large 81^{\frac{1}{\log_5 9}} = 81^{\log_9 5} = (9^2)^{ \log_9 5}= 9^{\log_9 25}=25$

\large 3^{\frac{3}{\log_\sqrt{6} 3}}=3^{3 \log_3 \sqrt{6}}= (\sqrt{6})^{3}=6\sqrt{6}

$\large (\sqrt{7})^{\frac{2}{\log_{25} 7}} = 7^{\frac{1}{2}(2 \log_7 25)} = 25$

$\large 125^{\log_{25} 6} = (5^3)^{\log_{25} 6} = 5^ {\log_{25} 6^{3}} = 5^ {\log_{5} 6^{\frac{3}{2}}} = 6^{\frac{3}{2}}=6\sqrt{6}$

Putting in the given equation

$\large x=\frac{(25+6\sqrt6)(25-6\sqrt{6})}{409}= \frac{625-216}{409}=1$

$\implies \log_2 x = \log_2 1 = \boxed{0}$