Logarithm warms up

First note that $\log_{2}{1}=0$ .Now let $\log_{2}{x}$ be $a$ .The equation becomes: $a^{2}-6a+8=0 \\ \implies a^2-2a-4a+8=0\\ \implies (a-2)(a-4)=0 \\ \implies a=2,4 \\ \implies \log_{2}{x}=2,4\\ \therefore x=4,16$

$(\log_2 x)^2 - 6 \cdot \log_2 x + 8 = \log_2 1 \\ \log_2 x=m\Rightarrow m^2-6m+8=0 \\ \underbrace{\underbrace{(m-2)}_{x=4}\underbrace{(m-4)}_{x=16}}_{\text{Sum of values of }x=\boxed{20}}=0$