Hi all

I can see here, that

$\displaystyle 8n^2 = 64nlog_2n $

can be arranged into $\displaystyle 8n^2 - 64nlog_2n = 0 $ (if I'm correct on that), but I'm just not sure where to go from here.

If I'm wrong, then I tried diving by 64n on each side to get

$\displaystyle n/8 = log_2n $,

then applying one of the log laws on each side to get

$\displaystyle 2^n^/^8 = 2^l^o^g^_^2^n $

Finally: $\displaystyle 2^n^/^8 = n $

Then got stuck here. =(