Inequality Algebric Solution required

[navdeep_bajwa]

Manhatten challenge problem

Can you solve this algebrically


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

(1) x^x > y

(2) x > y^y

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question

[winsayou]

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

Sol from (x^y) * (y^-x) = 1
Therefore [x^y = y^x] --> x= y only

clearly that we cannot convert these inequlities to equation. Therefore, we need to check that which choice x and y cannot be true in the quesion.
(1) x^x>y
if x=y --> y^y>y this is always true
So there is a pobabilty, but it didn't prove that (x^y) * (y^-x) = 1
(2) x>y^y
if x=y --> y>y^y this is not true (opposite from (1))
So no! probablity that so (x^y) * (y^-x) = 1

(2) sufficient but (1) is not.

I'm not sure. Please read carefully.
Tell me if I'm wrong.

[Ben Ku]

We are left with the rephrased question, Is x^y = y^x? winsayou said the only way this would work is if x = y. There are actually a few other possibilities: (2, 4), (4, 2), (-2, -4) and (-4, -2).

So we need to ask: Does x = y or (2, 4) or (4, 2) or (-2, -4) or (-4, -2).

Statement (1) states x^x > y. This statement on its own does not result in any of our four numerical solutions.

We have to ask, if x^x > y, do we know if x = y? Let's test some values:
If x = 3 and y = 3, so 9 > 3, then the answer is YES.
If x = 3 and y = 1, so 9 > 1, then the answer is NO.
Because this statement yields mixed answers, statement (1) is insufficient.

Statement (2) states x > y^y. This statement on its own does not result in any of our four numerical solution.

We have to ask, if x > y^y, do we know if x = y? Let's test some values:
If x = 9 and y = 3, then the answer is NO.
So are there ANY situations where x = y? Suppose x = y, can x > x^x? This can NEVER be the case because the number will not be larger than a power of the number. Therefore, statement (2) is sufficient, because x will NEVER be y.

The answer is B.

winsayou's response is close. A few cautions:
(1) winsayou found that the question can be ask if x = y. However, the numerical solutions of 2^4 = 4^2 and (-2)^(-4) = (-4)^(-2) was not thought of.
(2) for the statements, winsayou ASSUMED that "if x = y" to evaluate the statements. This is the wrong process; you should not assume the question to evaluate the statement. Instead, you need to use do it the other way around: use the statements to evaluate the question.

Hope that helps.