<s>[INST] 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. 
 [/INST] Put $u=\log_2\left(\dfrac x2 \right)=\log_2x-1$
Note that $\log_2(x^2)=2(u+1)$, and $\log_2\left(\dfrac 2x\right)=-u$.
Hence inequality becomes
$$ \begin{align}
u-\dfrac {2(u+1)}u \leq1\\
\dfrac{u^2-3u-2}u \leq0\\
\dfrac{(u-\alpha)(u-\beta)}u \leq 0\\
\end{align}$$
where $\alpha=\dfrac {3+\sqrt{17}}2,\ \beta=\dfrac {3-\sqrt{17}}2\ $
$$
\begin{align}
u\leq\beta &,\ \quad 0\leq u\leq\alpha,\\\
\log_2\left(\frac x2\right)\leq \dfrac {3-\sqrt{17}}2 &,\quad \ 0\leq\log_2\left(\frac x2\right)\leq\dfrac{3+\sqrt{17}}2\\
\log_2x\leq \frac{5-\sqrt{17}}2&,\quad \ 1\leq \log_2 x\leq \frac{5+\sqrt{17}}2\\
\end{align}$$
As $x>0$,
$$0< x\leq2^{\frac{5-\sqrt{17}}2},\ \quad 2\leq x\leq2^{\frac{5+\sqrt{17}}2}$$
</s>