There are 3 ways in which something will equal 1 when working with exponents.
1. The base is 1 (eg. \(1^{2000} = 1\))
2. -1 is the base and the exponent is even (eg. \(-1^{26} = 1\))
3. The base is to the power of 0. (eg. \(100^0 = 1\))

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

1.
To have a base of 1, \(x^2 - 5x \) must equal -6.

\(x^2 - 5x = -6\)

\(x^2 - 5x + 6 = 0\)

\((x-3)(x-2) = 0\)

x = 3 or x = 2. In this case, both values of x will yield a base of 1.


2.
To have a base of -1, we need \(x^2 - 5x\) in the base to equal -8.

\(x^2 - 5x = -8\)

\(x^2 - 5x + 8 = 0\) However, this quadratic has no real solutions.

3.
For the exponent \(x+1\) to equal 0, \(x=-1\)

Values for x which satisfy the equation:

x = -1, x = 2 or x = 3

Answer C