The sum of all the real values of x satisfying the equation ${(x^2-7x+11)}^{x^2-6x-7}=1$is
 

$\begin{array}{l}\because {\left(x^2-7x+11\right)}^{x^2-6x-7}\\ \Rightarrow \left(\because f(x{)}^{g\left(x\right)}=1\right)\end{array}$
Case (i) $-f\left(x\right)=1$
We get $x^2-7x+11=1$
$\begin{array}{l}{x}^2-7x+10=0\\ x=2,5\end{array}$
Case (ii) $g\left(x\right)=0$
$$\begin{gathered} x^2-6x-7=0 \\ x=-1,7 \end{gathered}$$
Case (iii)  $f\left(x\right)=-1,g\left(x\right)=even$
$\begin{array}{l}{x}^2-7x+11=-1\\ x^2-7x+12=0\\ x=3,4\end{array}$
$\because x=4$ not possible 
 $\begin{array}{l}\left(g\right(4)=odd)\\ \therefore x=-1,2,3,5,7\end{array}$
Sum = 16.