# 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?

• $x=3,4{}{}{}{}$ – Kenta S Jan 11 at 15:28
• You need the base to $\pm 1$ or you need the exponenet to be $0$. If the base is $-1$ you need the exponent to be even. – lulu Jan 11 at 15:29
• $4$ surely works – Math Lover Jan 11 at 15:29
• $3$ does too as @KentaS mentioned. – Math Lover Jan 11 at 15:31

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

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.

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.

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}.