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If x and y are integers, does (x)^y * (y)^-x = 1?

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If x and y are integers, does (x)^y * (y)^-x = 1? [#permalink] New post 06 Jun 2012, 22:44
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Responding to a pm:

Question: If x and y are integers, does (x)^y * (y)^-x = 1?

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

Let's re-arrange the question first:
Is (x)^y * (y)^{-x} = 1?
Is (x)^y = (y)^x?

Check this post for a detailed discussion on this: try-this-one-700-level-number-properties-103461.html#p805817

So, (x)^y = (y)^x when x = y or x and y take values 2,4 or -2,-4

Look at the statements now:

(1) (x)^x > y
We know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case. (x)^y is not equal to (y)^x.
But does it hold for any values which will make (x)^y = (y)^x?
Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.
Not sufficient.

(2) x > (y)^y
Again, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.
But does it hold for any values which will make (x)^y = (y)^x?
Let's see. If x = y, x cannot be greater than y^y. Check for a few values to figure out the pattern.
If x = 4 and y = 2, x is not greater than y^y.
Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.
Therefore, if x > (y)^y, (x)^y = (y)^x cannot hold for any values of x and y. Hence answer to the question stays NO.
Sufficient.

Answer (B).
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Re: Tricky question on Exponents [#permalink] New post 06 Jun 2012, 22:58
Thank you Karishma for such a detailed reply.

Thanks.
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Re: Tricky question on Exponents [#permalink] New post 06 Jun 2012, 23:28
VeritasPrepKarishma wrote:
Responding to a pm:

Question: If x and y are integers, does (x)^y * (y)^-x = 1?

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

Let's re-arrange the question first:
Is (x)^y * (y)^{-x} = 1?
Is (x)^y = (y)^x?

Check this post for a detailed discussion on this: try-this-one-700-level-number-properties-103461.html#p805817

So, (x)^y = (y)^x when x = y or x and y take values 2,4 or -2,-4

Look at the statements now:

(1) (x)^x > y
We know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case. (x)^y is not equal to (y)^x.
But does it hold for any values which will make (x)^y = (y)^x?
Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.
Not sufficient.

(2) x > (y)^y
Again, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.
But does it hold for any values which will make (x)^y = (y)^x?
Let's see. If x = y, x cannot be greater than y^y. Check for a few values to figure out the pattern.
If x = 4 and y = 2, x is not greater than y^y.
Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.
Therefore, if x > (y)^y, (x)^y = (y)^x cannot hold for any values of x and y. Hence answer to the question stays NO.
Sufficient.

Answer (B).

good Q.. thanks for the explanation...
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Re: Tricky question on Exponents   [#permalink] 06 Jun 2012, 23:28
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