Bunuel
Is y^x > x^y?
(1) y > x
(2) x is a positive number
Target question: Is y^x > x^y? Statement 1: y > x There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 2 and y = 4. In this case, y^x - 4^2 = 16 and x^y = 2^4 = 16. So, the answer to the target question is
NO, y^x is NOT greater than x^yCase b: x = 1 and y = 2. In this case, y^x - 2^1 = 2 and x^y = 1^2 = 1. So, the answer to the target question is
YES, y^x IS greater than x^ySince we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x is a positive numberSince we have no information about the value of y, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Before I perform any kind of analysis, it would help to check whether the counter-examples I used for statement 1 apply when the two statements are combined.
Notice that both of the cases we used for statement one satisfy statement 2's condition that x is a positive number.
So, the same counter-examples will satisfy the two statements COMBINED.
Can other words:
Case a: x = 2 and y = 4. In this case, y^x - 4^2 = 16 and x^y = 2^4 = 16. So, the answer to the target question is
NO, y^x is NOT greater than x^yCase b: x = 1 and y = 2. In this case, y^x - 2^1 = 2 and x^y = 1^2 = 1. So, the answer to the target question is
YES, y^x IS greater than x^ySince we cannot answer the
target question with certainty, the combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent