Question

The number of solutions for real x, which satisfy the equation $2lo{g}_{2}lo{g}_{2}x+lo{g}_{\frac{1}{2}}lo{g}_2\left(2\sqrt{2}x\right)=1$:

• A. 1
• B. 2
• C. 4
• D. none of these

Answer 1

The correct option is A 1

$2lo{g}_{2}lo{g}_{2}x+lo{g}_{\frac{1}{2}}lo{g}_2\left(2\sqrt{2x}\right)=1\Rightarrow 2lo{g}_{2}lo{g}_{2}x-lo{g}_{2}lo{g}_2\left(2\sqrt{2x}\right)=lo{g}_{2}2\Rightarrow lo{g}_2(lo{g}_{2}x{)}^2-lo{g}_2[lo{g}_{2}2\sqrt{2x}]=lo{g}_{2}2\Rightarrow lo{g}_2\frac{(lo{g}_{2}x{)}^2}{\left[lo{g}_{2}2\sqrt{2}x\right]}=lo{g}_{2}2\Rightarrow \frac{(lo{g}_{2}x{)}^2}{lo{g}_2\left(2\sqrt{2}x\right)}=2$
$\Rightarrow (lo{g}_{2}x{)}^2=2lo{g}_2(2\sqrt{2}x)=2lo{g}_2\left(2^{\frac{3}{2}}x\right)\Rightarrow (lo{g}_{2}x{)}^2=2[\frac{3}{2}+lo{g}_{2}x]=3+2lo{g}_{2}x\Rightarrow (lo{g}_{2}x{)}^2-2lo{g}_{2}x-3=0lo{g}_{2}x=-1orlo{g}_{2}x=3\Rightarrow x=\frac{1}{2}orx=8$
But for $x=\frac{1}{2},lo{g}_{2}lo{g}_2\left(\frac{1}{2}\right)$ is undefined
$\therefore$ Only possible value of x = 8.