Question

Sometimes to solve an equation, we may use the identity $a^{{\log }_{a}b}=b,b>0,a>0,a\ne 1$

Solution set of the equation: $\frac{1}{4}{x}^{{\log }_2\sqrt{x}}={\left(2^{1/4}\right)}^{{\left({\log }_{2}x\right)}^2}$ is,

• A. $\varphi$
• B. $\left\{4^{-\sqrt{2}}\right\}$
• C. $\left\{4^{\sqrt{2}}\right\}$
• D. none of these

Answer 1

Answer: (B), (C)

$\left\{4^{\sqrt{2}}\right\}$
$\left\{4^{-\sqrt{2}}\right\}$

$\frac{1}{4}{x}^{{\log }_2\sqrt{x}}={\left(2^{1/4}\right)}^{{\left({\log }_{2}x\right)}^2}$

Above equation is valid when $x>0,x\ne 1$

Taking logarithm on both sides, we get

${\log }_2\left(\frac{1}{4}\right)+\left({\log }_2\sqrt{x}\right)\left({\log }_{2}x\right)=\frac{1}{4}{\left({\log }_{2}x\right)}^2$

$\Rightarrow {\left({\log }_{2}x\right)}^2=8$

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

$\Rightarrow x=4^{-\sqrt{2}}$ or $4^{\sqrt{2}}$
Ans: B, C