Question

The solution set of the equation ${\log }_{x}2{\log }_{2x}2={\log }_{4x}2$ is equal to

• A. $\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$
• B. $\left\{\frac{1}{2},2\right\}$
• C. $\left\{\frac{1}{4},2^2\right\}$
• D. None of these

Answer 1

Answer: (A)

$\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$

${\log }_{x}2{\log }_{2x}2={\log }_{4x}2$
where, $x>0,\ne 1,1/2,1/4$
$\Rightarrow {\log }_{2}x{\log }_{2}2x={\log }_{2}4x$
$\Rightarrow {\log }_{2}x({\log }_{2}2+{\log }_{2}x)=2{\log }_{2}2+{\log }_{2}x,[\because \log (ab)=\log a+\log b]$
$\Rightarrow {\log }_{2}x+{\left({\log }_{2}x\right)}^2=2+{\log }_{2}x,[\because {\log }_{a}a=1]$
$\Rightarrow ({\log }_{2}x{)}^2=2$
$\Rightarrow {\log }_{2}x=\pm \sqrt{2}$
$\Rightarrow x=2^{\sqrt{2}},2^{-\sqrt{2}}$

Hence the correct answer is option A