If the product of all solutions of the equation $\frac{\left(2017\right)x}{2018}={\left(2017\right)}^{{\log }_x\left(2018\right)}$ can be expressed in the lowest form as $\frac{m}{n}$ then the value of $(m-n)$ is

Let ${\log }_{2018}x=y$

Taking log to base 2018 on both sides

${\log }_{2018}2017+y-{\log }_{2018}2018={\log }_{x}2018.{\log }_{2018}2017$

Let ${\log }_{2018}2017=\alpha$

$\Rightarrow \alpha +y-1=\left(\frac{1}{y}\right).\alpha \Rightarrow \alpha y+y^2-y=\alpha \Rightarrow y^2+(\alpha -1)y-\alpha =0$

Sum of roots ${\log }_{2018}{x}_1+{\log }_{2018}{x}_2=1-\alpha$

$\Rightarrow {\log }_{2018}{x}_1{x}_2={\log }_{2018}2018-{\log }_{2018}2017$

$\Rightarrow x_1{x}_2=\frac{2018}{2017}\Rightarrow m-n=2018-2017=1$