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. 

Here are the steps
\[
\log_2 \frac{x}{2}+\frac{\log_2 x^{2}}{\log_2 \frac{2}{x}} \le 1
\]
\[
\log_2 x -\log_2 2+\frac{2\log_2 x}{\log_2 2-\log_2 x} \le 1
\]
\[
\log_2 x -1+\frac{2\log_2 x}{1-\log_2 x} \le 1
\]
Let $\alpha= \log_2 x$, then
\[
\alpha -1+\frac{2\alpha}{1-\alpha} \le 1
\]
\[
\alpha -2+\frac{2\alpha}{1-\alpha} \le 0
\]
\[
\frac{(1-\alpha)(\alpha -2)+2\alpha}{1-\alpha} \le 0
\]
\[
\frac{5\alpha -\alpha^{2}-2}{1-\alpha} \le 0
\]
After solving for $\alpha$, we have the solutions
\[ 1<\alpha \le \frac{1}{2}(5+\sqrt{17}) \]
\[ \alpha \le \frac{1}{2}(5-\sqrt{17}) \]
Which is
\[ 1<\log_2 x \le \frac{1}{2}(5+\sqrt{17}) \]
\[ \log_2 x \le \frac{1}{2}(5-\sqrt{17}) \]
Thus
\[ 2< x \le (\sqrt{2})^{5+\sqrt{17}} \]
\[ x \le (\sqrt{2})^{5-\sqrt{17}} \]
