If  the product of all solutions of the equation  $\frac{\left(2009\right)x}{2010}={\left(2009\right)}^{{\log }_x\left(2010\right)}$  can be expressed in the lowest form as  $\frac{m}{n}$  then the value of  $(m-n)$  is 

 $\frac{\left(2009\right)x}{2010}={\left(2009\right)}^{{\log }_x\left(2010\right)}$
Taking log on both the sides on base  
 ${\log }_x\left(2009\right)+1-{\log }_x\left(2010\right)={\log }_x\left(2010\right).{\log }_x\left(2009\right)$
 ${\log }_x\left(2009\right)[1-{\log }_x\left(2010\right)]+[1-{\log }_x\left(2010\right)]=0$
 $[1+{\log }_x\left(2009\right)][1-{\log }_x\left(2010\right)]=0$
$\Rightarrow x=\frac{1}{2009}orx=2010;$ product  $=\frac{2010}{2009}$
 $\Rightarrow (m-n)=1$