Question

The number of real solutions of the equation: $2{\log }_2{\log }_{2}x+{\log }_{1/2}{\log }_2\left(2\sqrt{2}x\right)=1$

• A. $1$
• B. $2$
• C. $3$
• D. more than $3$

Answer 1

The correct option is A $1$

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

Put, $x=2^y\Rightarrow {\log }_{2}x=y$

Then, $2{\log }_2\left(y\right)+{\log }_{1/2}\left({\log }_{2}2\sqrt{2\times 2^y}\right)=1$

Or, $2{\log }_2\left(y\right)+{\log }_{1/2}\left({\log }_2{2}^{\frac{y+3}{3}}\right)=1$

$\Rightarrow$ ${\log }_2{y}^2-{\log }_2\left(\frac{y+3}{2}\right)=1$

$\Rightarrow$ ${\log }_2\left(\frac{y^2}{\frac{y+3}{2}}\right)=1$

$\Rightarrow$ $\frac{y^2}{\frac{y+3}{2}}=2$

$\Rightarrow$ $y^2=y+3$

$\Rightarrow$ $y^2–y-3=0$

Solving for y, we get,

$y=\frac{1\pm \sqrt{13}}{2}$ giving us 1 real value.

And, $x=2^{\frac{1\pm \sqrt{13}}{2}}$

This gives us 1 real solution.