$$\begin{cases} \log_{2}^{2}(-\log_{2}x) + \log_{2}\log_{2}^{2}x \leq 3 & \\-4 |x^2-1|-3\geq \frac{1}{x^2-1}& \end{cases}$$

What I've tried:

Make substitution $t=x^2-1$ and solve second inequality:

$-4|t|-3 \geq \frac{1}{t} <=> -4|t|-3-\frac{1}{t} \geq 0$

Where I got interval for $t$: $-\frac{1}{4}\leq t < 0$

And then for $x$: $x \in (-1; -\frac{\sqrt{3}}{2}] \cup [\frac{\sqrt{3}}{2}; 1)$

But I have no thoughts how to solve first inequality.

  • 2
    $\begingroup$ the first one seems to be invalid ,since if we set $t=\log_2x$ :$\\ \log_2^2(-t)+2\log _2t\le 3$ we see that $t\le 0$ in order for $(\log_2^2(-t))^2$ to be defined,but this immediatley,disqualifies $2\log_2t$ which works for $t>0$ so obviously we have a contradiction $\endgroup$ – Jonas Kgomo Apr 16 '14 at 13:07
  • 1
    $\begingroup$ For your first inequality if you write $ \log _2 \left(\log _2^2 x\right)$ as $ 2\log _2 \left(\log _2 (x)\right)$ (which I think restrict $x\in \mathbb{R}$) then you would have: $\log _2^2 \left(-\left(\log _2 x\right)\right)+2 \left(\log _2 \left(\log _2 x\right)\right)\leq 3$. After asking Mathematica to find an instance of and $x$ such that the equation is satisfied it returns nothing. So it seems to be impossible. $\endgroup$ – Jay Apr 16 '14 at 13:08

$$ \log_{2}^{2}(-\log_{2}x) + \log_{2}\log_{2}^{2}x \leq 3 $$ $$ \log_{2}^{2}(-\log_{2}x) + 2\log_{2}(\log_{2}x) \leq 3 $$

Determine domain of LHS: $$ -\log_2(x)>0 $$ $$ \log_2(x)<0 $$ $$ \rightarrow x<1 $$

Also: $$ \log_2(x)>0 $$

So domain of this inequality is: $x\in \phi$. Inequality doesn't have any solutions.


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