Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $8$

Answer 1

The correct option is D $8$

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

Let $u={\log }_{2}x$, then
$\Rightarrow 2{\log }_{2}u+{\log }_{1/2}\left({\log }_2{2}^{\frac{3}{2}}x\right)=1$

$\because \log ab=\log a+\log b$

$\Rightarrow 2{\log }_{2}u+{\log }_{1/2}({\log }_2{2}^{\frac{3}{2}}+{\log }_{2}x)=1$

$\Rightarrow 2{\log }_{2}u+{\log }_{1/2}({\log }_2{2}^{\frac{3}{2}}+u)=1$

$\Rightarrow {\log }_2{u}^2-{\log }_2(\frac{3}{2}+u)=1\dots (b^{{\log }_b(x{)}^n}=x^n)$

$\Rightarrow {\log }_2\left(\frac{u^2}{\frac{3}{2}+u}\right)={\log }_{2}2\dots (\log a-\log b=\log \frac{a}{b})$

$\Rightarrow u^2=2(\frac{3}{2}+u)$

$\Rightarrow u^2-2u-3=0$

$\Rightarrow u=-1,3$

$\Rightarrow x=\frac{1}{2},8$

But at $x=\frac{1}{2},$ the equation is not satisfied.

Hence, $x=8$ .