What is the method of solving these types of exponential equations

Question

I've seem to come to a sort of a road block while doing my homework, the following questions are very similar to this one so any insight in how to solve these types of equations would be of much help. $(x-2)^{x^2-x}=(x-2)^{12}$

Answer 1

You want solutions to $(x-2)^{x^2-x}=(x-2)^{12} $.

If $x = 2$ this is true, so that is a solution.

If $x \ne 2$, you can, as suggested by Peter Szilas, divide by $(x-2)^{12}$ to get $1 =(x-2)^{x^2-x-12} $.

For this to be true we must have either $x-2=1$ or $-1$, so that $x = 3$ or $1$ (this works since both exponents are even), or $x^2-x-12 = 0$, so that, since $x^2-x-12 = (x-4)(x+3)$, $x = 4$ or $x = -3$.

Therefore the solutions are $x = 1, 2, 3, 4, -3$.