Question

Number of integral solutions of the equation
${\log }_{0.5x}\left(x^2\right)-14{\log }_{16x}\left(x^3\right)+40{\log }_{4x}\sqrt{x}=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

${\log }_{0.5x}\left(x^2\right)-14{\log }_{16x}\left(x^3\right)+40{\log }_{4x}\sqrt{x}=0$
$\Rightarrow x>0,x\ne 2,x\ne \frac{1}{16},x\ne \frac{1}{4}$
Now,
$\frac{{\log }_x{x}^2}{{\log }_x\left(\frac{x}{2}\right)}-14\cdot \frac{{\log }_x{x}^3}{{\log }_{x}16x}+40\cdot \frac{{\log }_x\sqrt{x}}{{\log }_{x}4x}=0$
$\Rightarrow \frac{2}{1-{\log }_{x}2}-14\cdot \frac{3}{4{\log }_{x}2+1}+40\cdot \frac{1/2}{2{\log }_{x}2+1}=0$

Let $y={\log }_{2}x$
$\Rightarrow \frac{1}{1-1/y}-\frac{21}{4/y+1}+\frac{10}{2/y+1}=0$
$\Rightarrow \frac{y}{y-1}-\frac{21y}{4+y}+\frac{10y}{2+y}=0$
$\Rightarrow \frac{-5y(2y^2-3y-2)}{(y-1)(4+y)(2+y)}=0$
$\Rightarrow y(2y+1)(y-2)=0$
$\Rightarrow y=-\frac{1}{2},0,2$

$\Rightarrow {\log }_{2}x=-\frac{1}{2}\text{or}{\log }_{2}x=0\text{or}{\log }_{2}x=2\Rightarrow x=\frac{1}{\sqrt{2}},1,4$

Only two integral solutions.