Question

Solve $6({\log }_{x}2-{\log }_{4}x)+7=0$

Answer 1

$6\left({\log }_{x}2-{\log }_{4}x\right)+7=0$

$=6\left({\log }_{x}2-\frac{1}{2}{\log }_{2}x\right)+7=0$

$Lety={\log }_{2}x$

$6\left(\frac{1}{y}-\frac{1}{2}y\right)+7=0$

$6\left(\frac{2-y^2}{2y}\right)+7=0$

$6-3y^2+7y=0$

$3y^2-7y-6=0$

$3y^2+2y-9y-6=0$

$y\left(3y+2\right)-3\left(3y+2\right)=0$

$\left(3y+2\right)\left(y-3\right)=0$

$y=\frac{-2}{3},3$

$\log \frac{x}{2}=\frac{-2}{3}or\log \frac{x}{2}=3$

$if{\log }_{2}x=\frac{-2}{3}$

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

$x=\frac{1}{2^{2/3}}$

$If{\log }_{2}x=3$

$x=2^3$

$x=8$

$so$

$x=\frac{1}{2^{2/3}}andx=8$