jpr200012 wrote:

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

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

(2) \(y > 2\)

If x and y are nonzero integers, is \(x^y < y^x\)?(1) \(x = y^2\) --> if \(x=y=1\), then \(x^y=1=y^x\), so the answer would be NO

BUT if \(y=3\) and \(x=9\), then \(x^y=9^3<y^x=3^9\), so the answer would be YES. Not sufficient.

(2) \(y>2\). No info about \(x\), not sufficient.

(1)+(2) From (1) \(x = y^2\), thus the question becomes: is \((y^2)^y<y^{(y^2)}\)? --> is \(y^{2y}<y^{(y^2)}\)? Now, since from (2) \(y=integer>2\), then \(2y\) will always be less than \(y^2\), therefore \(y^{2y}\) will be less than \(y^{(y^2)}\). Sufficient.

Answer: C.

jpr200012 wrote:

This is a tricky question. I think it relies on you misapplying the rule: \((x^a)^b = x^{ab}\). Is this only valid if a and b are constants?

Example:

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

\(x^y < y^x\) therefore, \((y^2)^y < y^{y^2}\). How do you simplify this? The guide shows to \(y^{2y} < y^{y^2}\). The left hand side makes sense to me.

Why would \(y^{y^2}\) not simplify to \(y^{2y}\) also? Plugging in numbers, it makes sense. I just want to understand the concept.

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:

\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Hope it's clear.

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