The number of integral values of K for which the equation $e^x=\frac{K}{x-3}$  has exactly two solutions is ______________

    $$\begin{gathered} e^x=\frac{k}{x-3};x\ne 3 \\ {e}^x(x-3)=k \end{gathered}$$
 Let  $f\left(x\right)=e^x(x-3)$
 $$\begin{gathered} f\text{'}\left(x\right)=e^x(x-3)+e^x.1 \\ =e^x(x-2)=0;x=2 \\ f\left(2\right)=-e^2 \end{gathered}$$
 
       $-e^2<k<0$
$k\in \{-7,-6,-5,-4,-3,-2,-1\}$  Seven values