Anurag@Gurome wrote:Thus either (x = 0 and y = 0) or (x = 4 and y = 2)
For both of the cases x^y is equal to y^x.

I have already explained in other ocasions that 0^0 is not defined in Mathematics, therefore the question stem must exclude this possibility (implicitly or explicitly) at the very beginning (if the expression could appear, of course).

The problem "as it is" considers implicitly that x and y are not simultaneously equal to zero, otherwise the question itself would not have meaning/sense.

You cannot compare 0^0 to any other "thing" , not even to "itself", because to make a comparison both things to be compared must be meaningful. (It is not a matter of opinion or style. It is simply the rigorous way Math works!)

Conclusion for sttms (1) and (2) together: (x = 4 and y = 2) is the only possible solution, therefore we answer (in the affirmative) the question asked, therefore [spoiler] (C) [/spoiler].

Regards,
Fabio.

P.S.: there ARE special situations where it is convenient to "define LOCALLY" and for very particular and limited reasons 0^0 as 1, but I repeat, there is no ("general") definition possible to 0^0.