Vincen wrote: ↑

Wed Nov 17, 2021 7:49 am

If \(x\) and \(y\) are nonzero integers, is \(x^y < y^x ?\)

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

Answer: C

Source: Official Guide

Target question: Is x^y < y^x ?

Given: x and y are nonzero integers

Statement 1: x = y²
Take the original target question, and replace x with y² to get: Is (y²)^y < y^(y²)?
Simplify the left side power to get: Is y^(2y) < y^(y²)?

Let's TEST some values of y.
Case a: If y = 1, then y^(2y) = 1 and y^(y²) = 1. In this case, the answer to the target question is NO, it is NOT the case that y^(2y) < y^(y²)
Case b: If y = 3, then y^(2y) = 3^6 and y^(y²) = 3^9. In this case, the answer to the target question is YES, it IS the case that y^(2y) < y^(y²)
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y > 2
Since we have no information about x, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1, we changed the target question to read Is y^(2y) < y^(y²)?
Now let's divide both sides of the inequality by y^(2y) to get: Is 1 < [y^(y²)/y^(2y)]?
Apply the Quotient Law to get: Is 1 < y^(y² - 2y)?
Notice that, if y is an INTEGER greater than 2 (i.e., statement 2), then y² - 2y is greater than 1
So, y^(y² - 2y) MUST be greater than 1
So, the answer to the target question is YES it IS the case that 1 < y^(y² - 2y)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent