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. 

$$\log_2 x - \log_2 2 + 2\log_2 x - \log_2 2 + \log_2 x \le 1$$
$$4\log_2 x - 2 \le 1$$
$$4\log_2 x \le 3$$
$$\log_2 x \le 3/4 $$
$$x \le 2 \cdot 3/4$$
