If $x \frac{d y}{d x}=y(\log y-\log x+1)$ then the solution of the equation is :

$\log \frac{y}{x}=c x$

$\frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)$

Put $y=vx \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

∴  $v+x \frac{d v}{d x}=v(\log v+1)$

$\Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$

$\Rightarrow \log (\log v)=\log x+\log c=\log c x$

$\Rightarrow \log v=cx \Rightarrow \log \left(\frac{y}{x}\right)=c x$

Hence (4) is the correct answer.