Question

The equality ${\log }_{x}2.{\log }_{2x}2={\log }_{4x}2$ holds true for

• A. only $x=2^{\sqrt{2}}$
• B. only $x=2^{-\sqrt{2}}$
• C. $x=2^{-\sqrt{2}}$ or $x=2^{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is C $x=2^{-\sqrt{2}}$ or $x=2^{\sqrt{2}}$

${\log }_{x}2.{\log }_{2x}2={\log }_{4x}2$

Converting each of them to log base 2 ,

$\frac{1}{{\log }_{2}x}\frac{1}{{\log }_{2}2x}=\frac{1}{{\log }_{2}4x}$

$\Rightarrow {\log }_{2}4x={\log }_{2}2x{\log }_{2}x$

$\Rightarrow {\log }_{2}4+{\log }_{2}x=({\log }_{2}2+{\log }_{2}x){\log }_{2}x$

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

$\Rightarrow ({\log }_{2}x{)}^2=2$

$\Rightarrow {\log }_{2}x=\pm \sqrt{2}$

$\Rightarrow x=2^{\pm \sqrt{2}}$