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?

Solution 1:

Denote $a=x^2-7x+11.$ The equation becomes $a^{a-5}=1,$ or equivalently* $$a^a=a^5,$$ which has in $\mathbb{R}$ the solutions $a\in \{ {5,1,-1}\}.$ Solving the corresponding quadratic equations we get the solutions $x\in \{1,6,2,5,3,4\}.$

*Note added: $a=0$ is excluded in both equations.

Solution 2:

The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.

$(-1)^{2k}=1$, $\forall k\in \mathbb{Z}$

By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):

$x^{2}-7x+11=-1$ AND $x^{2}-7x+6=-6$, you would have then $(-1)^{-6}=\frac{1}{(-1)^6}=1$

And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:

$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.

So $(1,2,3,4,5,6)$ are the six solutions.

Solution 3:

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.

Solution 4:

The possibilities are

• $p^0$: $x^2-7x+6=0\to 1,6$,

• $1^q$: $x^2-7x+10=0\to 2,5$,

• $(-1)^q$: $x^2-7x+12=0\to 3,4$, and $q$ is even.