log2(x-6)+log2(x-4)-log2x=2
recall that $\displaystyle \log_a x + \log_a y = \log_a xy$ and $\displaystyle \log_a x - \log_a y = \log_a \left(\frac xy \right)$
thus,
$\displaystyle \log_2 (x - 6) + \log_2 (x - 4) - \log_2 x = \log_2 \left[ \frac {(x - 6)(x - 4)}{2x} \right] = 2$
can you continue?