Number of real roots of equation  $3^{{\log }_3(x^2-4x+3)}=(x-3)$ is

 $\because 3^{{\log }_3(x^2-4x+3)}=(x-3)\mathrm{.....}\left(i\right)$
Eq.(i) is defined, if  $x^2-4x+3>0$
$\Rightarrow (x-1)(x-3)>0$
$\Rightarrow x<1orx>\mathrm{3.....}\left(ii\right)$
Eq.(i) reduces to  $x^2-4x+3=x-3\Rightarrow x^2-5x+6=0$
$\therefore x=2,\mathrm{3.....}\left(iii\right)$
From Eqs.(ii) and (iii), use get  $x\in \varphi$
Number of real roots = 0