Question

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

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

Answer 1

The correct option is A $\{2^{-\sqrt{2}},2^{\sqrt{2}}\}$

We have
${\log }_{x}2\cdot {\log }_{2x}2={\log }_{4x}2$
$x>0,x\ne 1,x\ne 0.5,x\ne 0.25$

$\Rightarrow \frac{1}{{\log }_{2}x}\times \frac{1}{{\log }_{2}2x}=\frac{1}{{\log }_{2}4x},x\ne 1,x>0$
$\Rightarrow \frac{1}{{\log }_{2}x}\times \frac{1}{(1+{\log }_{2}x)}=\frac{1}{(2+{\log }_{2}x)}$
$\Rightarrow ({\log }_{2}x+2)=\left({\log }_{2}x\right)(1+{\log }_{2}x)$
$\Rightarrow ({\log }_{2}x{)}^2=2\Rightarrow {\log }_{2}x=\pm \sqrt{2}\therefore x=2^{-\sqrt{2}},2^{\sqrt{2}}$