$$8n^2 = 64n\log_{2}n$$
Been a while since I have used logarithms. I am actually comparing the running time of two algorithms and can obviously solve graphically but for the life of me can't remember the necessary steps to solve by hand. Thanks so much in advance.
Since $n>0$ or the logarithm doesn't exist, the equation is the same as $$ n=8\log_2n $$ If we consider the function $f(x)=x-8\log_2x$, we have $$ \lim_{x\to0^+}f(x)=\infty=\lim_{x\to\infty}f(x) $$ Moreover $$ f'(x)=1-\frac{8}{x\log 2} $$ (natural logarithm), which vanishes for $x=\frac{8}{\log2}=8\log_2e$.
Since $$ f(8\log_2e)=8\log_2e-8\log_2(8\log_2e)=8(\log_2e-3-\log_2\log_2e)<0 $$ there are two values where $f(x)=0$. One value is $\approx1.1$, the other one is $\approx43.56$.