Solution to the equation of a polynomial raised to the power of a polynomial. The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^{x^2−7x+6}=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation. 
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.  
Factorising the exponent, we have:
$$ x^{2}-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x \in \{1,6\} $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x \in \{2, 5\} $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?

Take natural logarithm from both sides:
$$\ln (x^2−7x+11)^{x^2−7x+6}=\ln1  \Rightarrow \\
(x^2-7x+6)\cdot \ln |x^2-7x+11|=0 \Rightarrow \\
1) \ x^2-7x+6=0 \Rightarrow x_{1,2}=1,6; \\
2) \ \ln |x^2-7x+11|=0 \Rightarrow |x^2-7x+11|=1 \Rightarrow x^2-7x+11=\pm 1 \Rightarrow \\
x_{3,4,5,6}=2,5,3,4.$$
Note: The found solutions satisfy the domain of the equation.
