Solving the equation $(x^2-7x+11)^{(x^2-13x+42)} = 1$ I have the following equation. $$(x^2-7x+11)^{(x^2-13x+42)} = 1$$ The question is to find the number of positive integer values possible for $x$. The answer is $6$. But I am only able to find $4 ( 2,5,6,7)$. What are the other two possible values?
 A: I will give only the two other solutions the you missed. It is trick. Look carefully for the equation. RHS equal 1. if $x^2-13x+42$ is even number. then $x^2-7x+11=-1$ is a solution. By factoring the last equation , we have $x=3$ and $x=4$. So, the set of solution is $\{2,3,4,5,6.7\}$
A: You forgot that $(-1)^{2n} = 1$ for positive $n$. (The exponent is even). You have different cases for $x^y = 1$:
$$x \not= 0, y = 0$$
$$x = 1, y = \text{any real number} $$
$$x = -1, y = \text{even number} $$
You missed the last case.
A: Partial answer : $a^b = 1 \rightarrow e^{b ln(a)} = 1 \rightarrow b ln(a) = 2 \pi i n$ for sone $n \in \mathbb{Z}$, where $ln$ denotes the principal branch of the natural logarithm.
Since $a$ is always a real number, $Im(ln(a)) = \begin{cases}
0 \text{ if $a > 0$}\\
\pi \text{ if $a < 0$}
\end{cases}$
Since you want the real part of $b ln(a)$ to be 0, and $b$ is a real number, you want the real part of $ln(a)$ to be 0 as well. This happens when $|a| = 1$ As $a$ is real, that means $a = 1$ or $a = -1$.
Try to finish it from here.
A: Anything of the form $a^b=1$ can be solved by taking 3 cases.

*

*a=1

*b=0;a≠0

*a=-1;b=2n (n belongs to Integers)

Thus for,
a=1;
x²-7x+11=1
x²-7x+10=0
i.e. x={2,5}
For b=0;
x²-13x+42=0
(x-6)(x-7)=0
x={6,7}
For a=-1;
x²-7x+11=-1
x²-7x+12=0
x={3,4}
But for this, we have to check if b is even or not.
Therefore for x=3,
3²-13*3+42
=9-39+42
=12=even.
Also for x=4,
4²-13*4+42
=16-52+42
=6=even.
Thus we see that b is even in both cases.
So the correct values for x are, x={2,3,4,5,6,7}.
