Solution of $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ Find the solution of ${\left( {{x^2} - 7x + 11} \right)^{\left( {{x^2} - 13x + 42} \right)}} = 1$
My approach is as follow $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$
Case 1: $f\left( x \right) = 1;x = 2,5$
Case 2: $f\left( x \right) \ne 0;g\left( x \right) = 0;x = 6,7\& f\left( 6 \right) \ne 0;f\left( 7 \right) \ne 0$
Case 3: $f\left( x \right) =  - 1;g\left( x \right) =$ even ;$x = 3,4\& f\left( 3 \right) = 12;f\left( 4 \right) = 6$
I have one doubt in Case 3, I am analysing the case $x^x$, in such scenario $x>0$.
Also if $y=a^x$, then if $a=0,x \in R-0$ and $a>0$
$\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ is also a type of exponential function, they I presume that f(x) is not negative so why they have taken $f(x)=-1$ for case 3, please elaborate

Eric is correct, but you need to exclude the case where $|f(x)|=0$, because $0^{0}$ is indeterminate form. $f(x)$ can be negative, as long as $g(x)$ makes $\phi(x)$ defined and real, in our case as long as $g(x)$ (even).
