# Solution to the equation of a polynomial raised to the power of a polynomial.

The problem at hand is, find the solutions of $$xx$$ in the following equation:

$$(x2−7x+11)x2−7x+6=1 (x^2−7x+11)^{x^2−7x+6}=1$$

My friend who gave me this questions, told me that you can find $$66$$ solutions without needing to graph the equation.

My approach was this: Use factoring and the fact that $$z0=1z^0=1$$ for $$z≠0z≠0$$ and $$1z=11^z=1$$ for any $$zz$$.

Factorising the exponent, we have:

$$x2−7x+6=(x−1)(x−6) x^{2}-7x+6 = (x-1)(x-6)$$

Therefore, by making the exponent = 0, we have possible solutions as $$x∈{1,6}x \in \{1,6\}$$

Making the base of the exponent = $$11$$, we get $$x2−7x+10=0 x^2-7x+10 = 0$$
$$(x−2)(x−5) (x-2)(x-5)$$

Hence we can say $$x∈{2,5}x \in \{2, 5\}$$.

However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?

Denote $$a=x2−7x+11.a=x^2-7x+11.$$ The equation becomes $$aa−5=1,a^{a-5}=1,$$ or equivalently* $$aa=a5,a^a=a^5,$$ which has in $$R\mathbb{R}$$ the solutions $$a∈{5,1,−1}.a\in \{ {5,1,-1}\}.$$ Solving the corresponding quadratic equations we get the solutions $$x∈{1,6,2,5,3,4}.x\in \{1,6,2,5,3,4\}.$$
*Note added: $$a=0a=0$$ is excluded in both equations.