Question

If ${\log }_{2x}2+{\log }_{4}2x=-\frac{3}{2}$, then $x$ is equal to

• A. $x=2^{-2}$
• B. $x=2^{-3}$
• C. $x=2^{-1}$
• D. none of these

Answer 1

Answer: (A), (B)

$x=2^{-2}$
$x=2^{-3}$

${\log }_{2x}2+{\log }_{4}2x=-\frac{3}{2}$
${\log }_{2x}2+\frac{1}{{\log }_{2x}4}=-\frac{3}{2}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
${\log }_{2x}2+\frac{1}{2{\log }_{2x}2}=-\frac{3}{2}[\because \log a^m=m\log a]$
$\Rightarrow {2\left({\log }_{2x}2\right)}^2+3{\log }_{2x}2+1=0$
Put ${\log }_{2x}2=t$
$\Rightarrow 2t^2+3t+1=0$
$\Rightarrow t=-1,-\frac{1}{2}$
$\Rightarrow {\log }_{2x}2=-1,{\log }_{2x}2=-\frac{1}{2}$
$\Rightarrow (2x{)}^{-1}=2,(2x{)}^{-\frac{1}{2}}=2$
$\Rightarrow x=2^{-2},2^{-3}$