Question

Value of $x$ which satisfies the following equation: $\frac{1}{4}{x}^{{\log }_2\sqrt{x}}=\left(2^{\frac{1}{4}}\right)2^{\frac{({\log }_{2}x{)}^2}{4}}$

• A. $3$
• B. $10$
• C. $100$
• D. $8$

Answer 1

The correct option is D $8$

Given $\frac{1}{4}{x}^{{\log }_2\sqrt{x}}=\left(2^{\frac{1}{4}}\right)2^{\frac{({\log }_{2}x{)}^2}{4}}$
clearly $x>0$
Taking log on both sides with base 2, we get
${\log }_2\left(\frac{1}{4}\right)+\left({\log }_2\sqrt{x}\right)\left({\log }_{2}x\right)=\frac{1}{4}+\frac{1}{4}{\left({\log }_{2}x\right)}^2[\because \log a^m=m\log a]$
$\Rightarrow -2{\log }_2\left(2\right)+\frac{1}{2}\left({\log }_{2}x\right)\left({\log }_{2}x\right)=\frac{1}{4}+\frac{1}{4}{\left({\log }_{2}x\right)}^2[\because {\log }_{a}a=1]$
or ${\left({\log }_{2}x\right)}^2=9$ or ${\log }_{2}x=\pm 3$
or $x=2^3,2^{-3}=8,\frac{1}{8}$
Ans: D