Question

\[(x^{2}-5x+7)^{x^2-1}=1\]

Answer 1

I found x=1 and x=-1, but there are supposed to be two more answers, x=2 and x=3.
How do you find them?

Answer 2

You got two cases either : x^2-5x+7=1 or x^2-1=0

Answer 3

x1=2;
x2=3;
x3=-1;
x4=1;

Answer 4

You're right, the "base=1" case completely slipped my mind.

Answer 5

take logs of both sides
(x^1-1) ln(x^2- 5x + 7) = 0

x^2 -1 = 0 gives you the +1 and -1 roots

ln(x^2 - 5x + 7) = 0
so
x^2 - 5x + 7 = 1
will give you the other 2 roots

Answer 6

good one @cwrw238

Answer 7

Thank you both.