Question

If $3^{{\log }_3({\log }_{2}x{)}^2}={\log }_{2}x-({\log }_{2}x{)}^2+1$, then $x$ is/are

• A. $2^{1/2}$
• B. $2$
• C. $2^{-1/2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B), (C)

$2^{-1/2}$

$3^{{\log }_3({\log }_{2}x{)}^2}={\log }_{2}x-({\log }_{2}x{)}^2+1$
$\Rightarrow ({\log }_{2}x{)}^2={\log }_{2}x-({\log }_{2}x{)}^2+1(\because a^{{\log }_{a}b}=b)$
$\Rightarrow 2({\log }_{2}x{)}^2-{\log }_{2}x-1=0$
$\Rightarrow (2{\log }_{2}x+1)({\log }_{2}x-1)=0$
$\Rightarrow {\log }_{2}x=-\frac{1}{2},{\log }_{2}x=1\Rightarrow x=2^{-1/2},2$