how do you solve $(x^2-5x+5)^{x^2-36} =1$

Question

Can someone please show me how they would work it out as I have never come across this before.

$$(x^2-5x+5)^{x^2-36} =1$$

Answer 1

It's kind of a trick question; there's no general way to solve that kind of equations (save for numerically), if the right-hand side had been anything else than $1$.

However, you should know that $a^b=1$ only if $a=1$ or $b=0$ (or $a=-1$ and $b$ even), so you can break it into three ordinary quadratic equations that you can solve separately.

Answer 2

We have that $a^b=1$ if and only if $a=1$ and arbitrary $b$ or $b=0$ and arbitrary $a$ (one might have to talk about the case $a=b=0$). Having this you should be able to get two seperate equations from $$(x^2-5x+5)^{x^2-36}=1$$ which can be easily solvedfor $x$.

Edit: I should drink a cup of coffee first after getting up...one also has to consider $a=-1$ and $b=2k$ for $k\in\mathbb N$, e.g. $(-1)^2=(-1)^4=(-1)^6=\dots =1$.