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?

  • 1
    $\begingroup$ $x=3,4{}{}{}{}$ $\endgroup$ – Kenta S Jan 11 at 15:28
  • 3
    $\begingroup$ 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. $\endgroup$ – lulu Jan 11 at 15:29
  • $\begingroup$ $4$ surely works $\endgroup$ – Math Lover Jan 11 at 15:29
  • $\begingroup$ $3$ does too as @KentaS mentioned. $\endgroup$ – 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,




i.e. x={2,5}

For b=0;




For a=-1;




But for this, we have to check if b is even or not.

Therefore for x=3,




Also for x=4,




Thus we see that b is even in both cases.

So the correct values for x are, x={2,3,4,5,6,7}.


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