Is x^y>y^x?

[ghong14]

Hi this is a data sufficiency question that I came across. I believe it was an example question from Thursday with Ron. I have it in my notes and don't know exactly where it came from.

Is x^y>y^x?

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

My question is since we know that x=y^2 then the question becomes is (y^2)^y>(y)^(y^2). Then why can't we simplify the preceding expression to y^2y>y^2y. I think it is permissible to simplify (y^2)^y to y^(2y) correct? why can't we just simplify y^(y^2) to y^(2y)?

Thanks.

[RonPurewal]

why can't we just simplify y^(y^2) to y^(2y)?

Thanks.

Well, the most important answer is "you can't because you can't". (Why doesn't (a - b) - c = a - (b - c) work? Because it doesn't work.)

In this case, though, you can actually see why it doesn't work. The purple parts are the same, so, effectively, you'd be saying that y^2 and 2y are the same thing.
Try plugging in y = 3, for instance; you'd have 3^9 = 3^6. Clearly not happening.

In general, always investigate by plugging in specific values when you're not sure whether something will work.

[ghong14]

In this case what exponential rule are we applying to allow us to simplify (y^2)^y to y^(2y). Why doesn't this rule apply to y^(y^2) to y^(2y)?

Thanks

[RonPurewal]

In this case what exponential rule are we applying to allow us to simplify (y^2)^y to y^(2y). Why doesn't this rule apply to y^(y^2) to y^(2y)?

Thanks

When a power is raised to another power, you can multiply the powers.
This is the familiar rule from high school. E.g., (x^3)^5 = x^15.

If you forget that rule, or (like me) just aren't so hot at remembering rules in the first place, just think about how something like that would multiply out:
(x^3)^5 = (x^3)(x^3)(x^3)(x^3)(x^3)
= (xxx)(xxx)(xxx)(xxx)(xxx)
= x^15
and the same thing will happen every time. So, it works for variables, too.

If you write stuff out like that, it's also clear why the same thing doesn't extend to something like x^(3^5). That's a totally different animal -- you'd just have to evaluate 3^5 and then write out that number of x's.

[julchik.m]

Hi, is there any quick way to establish that the second statement is insufficient in this case? Thanks!

[RonPurewal]

Hi, is there any quick way to establish that the second statement is insufficient in this case? Thanks!

I'd just plug numbers.
Let y = 5. Then x can still be anything.
If x = 1, then ... is 1^5 > 5^1? No.
If x = 2, then ... is 2^5 > 5^2? Yes.
Done.

[CHYNARAT644]

So what is OA for this question? is it C?

[RonPurewal]

So what is OA for this question? is it C?

Yes.