Question 17.

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

There can be a solution when 

$(x^2 - 5x + 7) = 1$ or 

$x^2 - 5x + 6 = 0$

or $x = 3$ and $x = 2$

There can also be a solution when $x + 1 = 0$ or $x = -1$

Hence three possible solutions exist.