Question

If $9^{{\log }_3\left({\log }_{3}x\right)}={\log }_{2}x-{\left({\log }_{2}x\right)}^2+1$, then $x$ is equal to

• A. $1$
• B. $1/2$
• C. $3$
• D. None of these

Answer 1

The correct option is D None of these

We have

${\log }_3\left({\log }_{2}x\right)\Rightarrow 3^{2{\log }_3\left({\log }_{2}x\right)}={\log }_{2}x-{\left({\log }_{2}x\right)}^2+1$

$\Rightarrow 3^{{\log }_3{\left({\log }_{2}x\right)}^2}={\log }_{2}x-{\left({\log }_{2}x\right)}^2+1$
$\Rightarrow 2{\left({\log }_{2}x\right)}^2-{\log }_{2}x-1=0$

$\Rightarrow (2{\log }_{2}x+1)({\log }_{2}x-1)=0$

$\Rightarrow {\log }_{2}x=-\frac{1}{2},{\log }_{2}x=1$

$\Rightarrow x=2^{-1/2},2$