If x and y are integers, does x^y * y^(-x) = 1?

If x and y are integers, does x^y * y^(-x) = 1?

(1) x^x > y
(2) x > y^y

SOLUTION:

Let's re-arrange the question first:

Is \((x)^y * (y)^{-x} = 1\)?
Is \((x)^y = (y)^x\)?

Check this post for a detailed discussion on this: try-this-one-700-level-number-properties-103461.html#p805817

So, \((x)^y = (y)^x\) when x = y or x and y take values 2,4 or -2,-4

Look at the statements now:

(1) \((x)^x > y\)
We know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case. \((x)^y\) is not equal to \((y)^x\).
But does it hold for any values which will make \((x)^y = (y)^x\)?
Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.
Not sufficient.

(2) \(x > (y)^y\)
Again, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.
But does it hold for any values which will make \((x)^y = (y)^x\)?
Let's see. If x = y, x cannot be greater than \(y^y\). Check for a few values to figure out the pattern.
If x = 4 and y = 2, x is not greater than \(y^y\).
Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.
Therefore, if \(x > (y)^y\), \((x)^y = (y)^x\) cannot hold for any values of x and y. Hence answer to the question stays NO.
Sufficient.

Answer (B).