Question

The equation $2{\log }_2\left({\log }_{2}x\right)+{\log }_{1/2}{\log }_2\left(2\sqrt{2x}\right)=1$ has

• A. Product of all its solution = 4
• B. A rational solution which is not an integer
• C. Has a natural solution
• D. Has no prime solutions.

Answer 1

The correct option is B A rational solution which is not an integer

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

$⟹{\log }_2\left(\right({\log }_{2}x{)}^2)-{\log }_2{\log }_2(2\sqrt{2x})=1$

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

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

$⟹({\log }_{2}x{)}^2=2{\log }_2(2\sqrt{2x})$

$⟹({\log }_{2}x{)}^2={\log }_2(\left(2\sqrt{2x}{)}^2\right)$

$⟹({\log }_{2}x{)}^2={\log }_2(8x)$

$⟹({\log }_{2}x{)}^2=3+{\log }_2(x)$

Let ${\log }_{2}x=y$

$⟹y^2-y-3=0$

$⟹y=\frac{1\pm \sqrt{1+12}}{2}$

$⟹y=\frac{1\pm \sqrt{13}}{2}$

Hence, solution is rational but not an integer and their product is $-3$.