# Solving the logarithimic inequality log_2(x)/(2) + (log_2x^2)/(log_2 (2/x)) <= 1 several times but keeping getting wrong answers.

Solving the logarithimic inequality ${\mathrm{log}}_{2}\frac{x}{2}+\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}\frac{2}{x}}\le 1$
I tried solving the logarithmic inequality:
${\mathrm{log}}_{2}\frac{x}{2}+\frac{{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}\frac{2}{x}}\le 1$
several times but keeping getting wrong answers.
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Selden1f
Let ${\mathrm{log}}_{2}x=A$, then ${\mathrm{log}}_{2}{x}^{2}=2{\mathrm{log}}_{2}x=2A$ and ${\mathrm{log}}_{2}\frac{2}{x}={\mathrm{log}}_{2}2-{\mathrm{log}}_{2}x=1-A$. So the given inequality becomes:
$\left(A-1\right)+\frac{2A}{1-A}\le 1.$
Consequently we get
$\frac{4A-{A}^{2}-1}{1-A}\le 1.$
Furthermore you get
$\frac{5A-{A}^{2}-2}{1-A}\le 0.$
Hopefully you can solve from here.

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kominis3q
Put $u={\mathrm{log}}_{2}\left(\frac{x}{2}\right)={\mathrm{log}}_{2}x-1$
Note that ${\mathrm{log}}_{2}\left({x}^{2}\right)=2\left(u+1\right)$, and ${\mathrm{log}}_{2}\left(\frac{2}{x}\right)=-u$
Hence inequality becomes
$\begin{array}{r}u-\frac{2\left(u+1\right)}{u}\le 1\\ \frac{{u}^{2}-3u-2}{u}\le 0\\ \frac{\left(u-\alpha \right)\left(u-\beta \right)}{u}\le 0\end{array}$
where

As $x>0$,

We have step-by-step solutions for your answer!