Question

The number of solutions of the equation ${\log }_{\left(x/2\right)}{x}^2+40{\log }_{4x}\sqrt{x}-14{\log }_{16x}{x}^3=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

By changing the base to 2 the given equation becomes

$\frac{{\log }_2{x}^2}{{\log }_{2}x/2}+\frac{40{\log }_2\sqrt{x}}{{\log }_{2}4x}-14\frac{{\log }_2{x}^3}{{\log }_{2}16x}=0$

$\Rightarrow \frac{2{\log }_{2}x}{{\log }_{2}x-1}+20\frac{{\log }_{2}x}{2+{\log }_{2}x}-42\frac{{\log }_{2}x}{4+{\log }_{2}x}=0$

Let $t={\log }_{2}x$, then we have

$2t(4+t)(2+t)-42t(t-1)(t+2)+20t(t-1)(t+4)=0$

$\Rightarrow 2t[t^2+6t+8-21t^2-21t+42+10t^2+30t-40]=0$

$\Rightarrow t[2t^2-3t-2]=0$

$\Rightarrow t=0$, $t=2$, $t=\frac{-1}{2}$

$\Rightarrow x=1$, $x=4$, $x=\frac{1}{\sqrt{2}}$