$\displaystyle \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x} $

$\displaystyle \frac{log_2{512}}{log_2{x}} = log_2{x} $

$\displaystyle log_2{512} = log_2{x^2} $

$\displaystyle 512 = x^2 $

$\displaystyle \sqrt{512} = x $

so x = 22.6 , is this correct?

appreciate any input.

thanks