Question

Solve the following inequalities.
${\log }_{x}2{\log }_{2x}2{\log }_{2}4x>1.$

Answer 1

${\log }_{x}2.{\log }_{2x}2{\log }_{2}4x>1$

$\Rightarrow \frac{\log 2}{\log x}\times \frac{\log 2}{\log 2x}\times \frac{\log 4x}{\log 2}>1$

$\Rightarrow \log 2(\log 4+\log x)>\log x(\log 2+\log x)$

$\Rightarrow 2(\log 2{)}^2+\log 2\log x>\log x\log 2+(\log x{)}^2$

$\Rightarrow 2(\log 2{)}^2>(\log x{)}^2$

$\Rightarrow (\log x{)}^2<2(\log 2{)}^2$

$\Rightarrow (\log x{)}^2-2(\log 2{)}^2<0$

$\Rightarrow (\log x-\sqrt{2}\log 2)(\log x+\sqrt{2}\log 2)<0$

Case I:

$\log x-\sqrt{2}\log 2>0$ and $\log x+\sqrt{2}\log 2<0$

$x>2^{\sqrt{2}}$ and $x<2^{-\sqrt{2}}$

which is not possible

Case II:

$\log x-\sqrt{2}\log 2<0$ and $\log x+\sqrt{2}\log 2>0$

$x>2^{\sqrt{2}}$ and $x>2^{-\sqrt{2}}$

$\Rightarrow 2^{-\sqrt{2}}<x<2^{\sqrt{2}}$ is the solution.