Solving the equation $(x^2-7x+11)^{(x^2-13x+42)} = 1$

Question

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?

Answer 1

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 $\{2,3,4,5,6.7\}$

Answer 2

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.

Answer 3

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.

Answer 4

Anything of the form $a^b=1$ can be solved by taking 3 cases.

1. a=1
2. b=0;a≠0
3. 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}.