Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality: 
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$ 
several times but keeping getting wrong answers.
Let $\log_2 x=A$, then $\log_2 x^2=2\log_2 x=2A$ and $\log_2\frac{2}{x}=\log_22- \log_2x=1-A$. So the given inequality becomes:
$$(A-1)+\frac{2A}{1-A} \leq 1.$$
Consequently we get
$$\frac{4A-A^2-1}{1-A} \leq 1.$$
Furthermore you get
$$\frac{5A-A^2-2}{1-A} \leq 0.$$
Hopefully you can solve from here.
$$\log_2 x - \log_2 2 + 2\log_2 x - \log_2 2 + \log_2 x \le 1$$
$$4\log_2 x - 2 \le 1$$
$$4\log_2 x \le 3$$
$$\log_2 x \le 3/4 $$
$$x \le 2 \cdot 3/4$$