Question

If $6({\log }_{x}2-{\log }_{4}x)+7=0$, then the possible values of $x$ is/are

• A. $3$
• B. $\frac{2}{3}$
• C. $8$
• D. $2^{-2/3}$

Answer 1

Answer: (C), (D)

The correct options are
C $8$
D $2^{-2/3}$

$6({\log }_{x}2-{\log }_{4}x)+7=0$
$\Rightarrow 6({\log }_{x}2-\frac{1}{2}{\log }_{2}x)+7=0$
Put $y={\log }_{2}x$
$\Rightarrow 6(\frac{1}{y}-\frac{y}{2})+7=0$
$\Rightarrow 3y^2-7y-6=0$
$\Rightarrow (y-3)(3y+2)=0$
$\Rightarrow y=3,-2/3$

$\therefore x=8,2^{-2/3}$