Question

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

• A. $x=8$ or $x=2^{-\frac{2}{3}}$
• B. $x=7$ or $x=2^{-\frac{2}{3}}$
• C. $x=6$ or $x=2^{-\frac{2}{3}}$
• D. none of these

Answer 1

The correct option is A $x=8$ or $x=2^{-\frac{2}{3}}$

$6({\log }_{x}2-{\log }_{4}x)+7=0$
or $6({\log }_{x}2-\frac{1}{2}{\log }_{2}x)+7=0$
or $6(\frac{1}{y}-\frac{y}{2})+7=0$ (where $y={\log }_{2}x$)
or $6\left(\frac{2-y^2}{2y}\right)+7=0$
or $3\left(\frac{2-y^2}{y}\right)+7=0$
or $6-3y^2+7y=0$
or $3y^2-7y-6=0$
or $y^2+2y-9y-6=0$
or $(y-3)(3y+2)=0$
or $y=3$ or $y=-\frac{2}{3}$
$\Rightarrow$ ${\log }_{2}x=3$ or $-\frac{2}{3}$
or $x=8$ or $x=2^{-\frac{2}{3}}$