NIMCET 2016 — Mathematics PYQ

Concept:
Logarithm properties:

Calculation:

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

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

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

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

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

Let ${\log }_{2}x$ be a,

$a(1+a)={\log }_2{2}^2+a(\because {\log }_{2}2=1)$

$a+a^2=2+a$

$a=\pm \sqrt{2}$

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

$\Rightarrow x=2^{\sqrt{2}},2^{-\sqrt{2}}$