Question

If $\frac{({\log }_{2}x{)}^2}{2}-{\log }_2{x}^{1/4}={\log }_x\sqrt{2}+\frac{5}{4},$ then the value(s) of $x$ is/are

• A. $4$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $2\sqrt{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $4$
B $\frac{1}{2}$
C $\frac{1}{\sqrt{2}}$

$\frac{({\log }_{2}x{)}^2}{2}-{\log }_2{x}^{1/4}={\log }_x\sqrt{2}+\frac{5}{4}\Rightarrow \frac{({\log }_{2}x{)}^2}{2}-\frac{1}{4}{\log }_{2}x-\frac{5}{4}=\frac{1}{2}{\log }_{x}2\Rightarrow \frac{({\log }_{2}x{)}^2}{2}-\frac{1}{4}{\log }_{2}x-\frac{5}{4}=\frac{1}{2\left({\log }_{2}x\right)}$

Assuming ${\log }_{2}x=t,$ we get
$\frac{t^2}{2}-\frac{t}{4}-\frac{5}{4}=\frac{1}{2t}\Rightarrow 2t^3-t^2-5t-2=0\Rightarrow (t+1)(2t^2-3t-2)\Rightarrow (t+1)(t-2)(2t+1)=0\Rightarrow t=-1,2,-\frac{1}{2}\Rightarrow {\log }_{2}x=-1,2,-\frac{1}{2}\Rightarrow x=2^{-1},2^2,2^{-\frac{1}{2}}\therefore x=\frac{1}{2},4,\frac{1}{\sqrt{2}}$