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. 
 A: 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.
A: 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 $x>0$,
$$0< x\leq2^{\frac{5-\sqrt{17}}2},\ \quad 2\leq x\leq2^{\frac{5+\sqrt{17}}2}$$
A: 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
\[ 2< x \le (\sqrt{2})^{5+\sqrt{17}} \]
\[ x \le (\sqrt{2})^{5-\sqrt{17}} \]
A: $$\log_2 x - \log_2 2 + 2\log_2 x - \log_2 2 + \log_2 x \le 1$$
$$4\log_2 x - 2 \le 1$$
$$4\log_2 x \le 3$$
$$\log_2 x \le 3/4 $$
$$x \le 2 \cdot 3/4$$
