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Re: Which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 10:38

2

This post received KUDOS

Answer should be C. Here is how:

If x and y are both integers, which is larger, x^x or y^y ?

Statement A: x=y+1

So x and y are consecutive integers. Remember they can be positive, negative or 0 (from the question stem).

Suppose y=1 and x=2 , then x^x is larger , but suppose y=-2 and x =-1 then y^y is larger.

Hence Insufficient.

Statement B: x^y>x and x is positive.

Knowing that x>0, x^y>x is only possible if y>1 . Please note even when y=0 , x>0 and x is an integer. So now we know that y is positive and x is positive but we do not know which is larger. Hence Insufficient

Combined:

From Statement 2 we know that x and y are both >0 and from Statement 1 we know that x is bigger. So YES. x^x is bigger than y^y

Sufficient. Hence Answer C . Seriously Kudos Hungry _________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Re: Which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 11:08

Expert's post

omerrauf wrote:

Answer should be C. Here is how:

If x and y are both integers, which is larger, x^x or y^y ?

Statement A: x=y+1

So x and y are consecutive integers. Remember they can be positive, negative or 0 (from the question stem).

Suppose y=1 and x=2 , then x^x is larger , but suppose y=-2 and x =-1 then y^y is larger.

Hence Insufficient.

Statement B: x^y>x and x is positive.

Let's re-arrage this a bit.

x^y>x so x^y-x>0 so x*(y-1)>0 and we know that x is +ve Now in order for x*(y-1)>0 the factor (y-1) has to be >0 so we know that y>1 but we do not know which is bigger, x or y.

Hence Insufficient

Combined:

From Statement 2 we know that x and y are both >0 and from Statement 1 we know that x is bigger. So YES. x^x is bigger than y^y

Sufficient. Hence Answer C Seriously Kudos Hungry

The red part is not correct.

If x and y are both integers, which is larger, x^x or y^y?

(1) x = y + 1 --> if y is positive integer then x^x=(y+1)^{y+1}>y^y but if y=-2 then x=-1 and x^x=-1<\frac{1}{4}=y^y

(2) x^y > x and x is positive --> since x is positive then x^{y-1}>1 --> since x and y are integers then y>1. If x=1 and y=2 then x^x<y^y but if x=3 and y=2 then x^x>y^y. Not sufficient.

(1)+(2) From (2) y>1, so it's a positive integer then from (1) x^x=(y+1)^{y+1}>y^y. Sufficient.

Re: If x and y are both integers, which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 20:05

Had already corrected that before you wrote bunuel. While writing the solution I mistakenly took x^y-x for xy-x but corrected it when i was reading my solution after I had posted. Thankyou anyways! _________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Re: Which is larger, x^x or y^y? [#permalink]
27 Dec 2013, 09:09

Bunuel wrote:

omerrauf wrote:

Answer should be C. Here is how:

If x and y are both integers, which is larger, x^x or y^y ?

Statement A: x=y+1

So x and y are consecutive integers. Remember they can be positive, negative or 0 (from the question stem).

Suppose y=1 and x=2 , then x^x is larger , but suppose y=-2 and x =-1 then y^y is larger.

Hence Insufficient.

Statement B: x^y>x and x is positive.

Let's re-arrage this a bit.

x^y>x so x^y-x>0 so x*(y-1)>0 and we know that x is +ve Now in order for x*(y-1)>0 the factor (y-1) has to be >0 so we know that y>1 but we do not know which is bigger, x or y.

Hence Insufficient

Combined:

From Statement 2 we know that x and y are both >0 and from Statement 1 we know that x is bigger. So YES. x^x is bigger than y^y

Sufficient. Hence Answer C Seriously Kudos Hungry

The red part is not correct.

If x and y are both integers, which is larger, x^x or y^y?

(1) x = y + 1 --> if y is positive integer then x^x=(y+1)^{y+1}>y^y but if y=-2 then x=-1 and x^x=-1<\frac{1}{4}=y^y

(2) x^y > x and x is positive --> since x is positive then x^{y-1}>1 --> since x and y are integers then y>1. If x=1 and y=2 then x^x<y^y but if x=3 and y=2 then x^x>y^y. Not sufficient.

(1)+(2) From (2) y>1, so it's a positive integer then from (1) x^x=(y+1)^{y+1}>y^y. Sufficient.

Answer: C.

P.S. Not a GMAT style question.

Just curious, why not a GMAT style question?

Thanks

Cheers! J

gmatclubot

Re: Which is larger, x^x or y^y?
[#permalink]
27 Dec 2013, 09:09