$\large \log_{2}x*\log_{\frac{x}{64}}2=\log_{\frac{x}{16}}2$

$\large \log_{2}x *\frac{\log_{2}2}{\log_{2}\frac{x}{64}}=\frac{\log_{2}2}{\log_{2}\frac{x}{16}}$

$\large \log_{2}x *\frac{1}{\log_{2}x-\log_{2}64}=\frac{1}{\log_{2}x-\log_{2}16}$

$\large \log_{2}x *\frac{1}{\log_{2}x-6}=\frac{1}{\log_{2}x-4}$

$\large \log_{2}x=\frac{\log_{2}x-6}{\log_{2}x-4}$

$\large (\log_{2}x)^{2}-5\log_{2}x+6=0$

$\large (\log_{2}x-3)(\log_{2}x-2)=0$

So, $\large (\log_{2}x-3)=0$ or $\large (\log_{2}x-2)=0$

So, $\large \log_{2}x=3$ or $\large \log_{2}x=2$

So,$\large x=8$ or $\large x=4$