What is the sum of all integer values of $x$ such that ${(x^2-17x+71)}^{(x^2-34x+240)}=1$?

Question

I've tried it like this:

taking $\log$ to the base $10$ on both the sides

$$(x^2-34x+240)\cdot\log(x^2-17x+71)=\log1=0$$ From which it follows that

$$ x^2-34x+240=0\text{ or }x^2-17x+71=1$$ And hence $$x=24,10\text{ or }x=10,7$$

So, the answer should be $7+10+24=41$

but it is wrong.

Answer 1

You had a good starting idea — but you made an oversight, since your calculuation unnecessarily forces the base to be positive.

A simpler way to execute your idea is to first solve this problem:

Find all integer solutions for $y$ and $z$ to the equation $y^z = 1$.

and this won't have the problem that logarithms introduced.

Answer 2

More generally, let's see how $a^b = 1$.

• $b = 0, a \neq 0$ : $x^2 -34x + 240 = 0 \implies x = \{10,24\}$ and none of them make $a = 0$ so this is two solutions.
• $a = 1, \forall b$ : $x^2 - 17 x + 71 = 1 \implies x = \{ 7,10\}$ which is one additional solution. (We already had $10$)
• $a = -1, b \text{ is even}$ : $x^2 - 17 x + 71 = 1 \implies x = \{8,9\}$, putting these two solutions into the exponent to check the parity leads that only $8$ works.

Overall we have $ x= \{10,24,7,8 \}$