determine all the values of x, such that (x^2-15)^x^2-4 = 1.

Answer 1

We know (any base)^0 = 1
so x^2 - 4 = 0
x = ± 2

Answer 2

Also, 1 to the power of any finite number is also 1.
So can you find the remaining solutions?

Answer 3

If x = 4 or -4, x^2 -15 = 1, so no matter what x^2-4 is, the function
(x^2-15)^x^2-4 = 1 equals 1 when x = + or - 4.

Therefore -4 and 4 are also solutions.

Thanks to MathMate for pointing that out.