Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$

My approach :

Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, $1$ can be written as $\log_2\log_24$

$\implies 2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)= \log_2\log_24$

$\implies \log_2\log_2x^2 -\log_{2}\log_2(2\sqrt{2}x)= \log_2\log_24$

$\implies \log_2 \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_2\log_24$

$\implies \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_24$

Please suggest whether is it the right approach... thanks...

  • 1
    $\begingroup$ Some MathJax advice: Named math operators should appear upright, and the common ones have their own code for this purpose (e.g. \sin, \log - see entry 11 in our MathJax guide). $\endgroup$ – Zev Chonoles Oct 11 '13 at 4:55
  • $\begingroup$ why the functional equation tag? I cant see how does this relates to functional equations. $\endgroup$ – jimjim Oct 11 '13 at 5:00

I think you went off the rails in the second line. Here's what I get

$$2 \log_2{\log_2{x}} = \log_2{\log_2^2{x}}$$

so that

$$2 \log_2{\log_2{x}} + \log_{1/2}{\log_2{(2 \sqrt{2} x)}} = \log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}}$$

and the equation becomes

$$\log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}} = 1 \implies \frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}} = 2$$


$$\log_2^2{x} = 2 \log_2{2^{3/2}} + 2 \log_2{x} \implies \log_2^2{x} - 2 \log_2{x}-3=0$$

This implies that $\log_2{x}=3$ or $\log_2{x}=-1$. In the former case, we have that $x=8$; in the latter, we have $x=1/2$. However, in the latter case, we have a false solution, as $\log_2{\log_2{(1/2)}} = \log_2{(-1)}$ which is outside the realm of the reals. Thus, the only solution is at $x=8$.

  • $\begingroup$ what is $x$ in the later case, why it is invalid? $\endgroup$ – lab bhattacharjee Oct 11 '13 at 5:08
  • $\begingroup$ @labbhattacharjee: see above; my initial analysis was right. $\endgroup$ – Ron Gordon Oct 11 '13 at 5:16

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