Hi,

Given information: x,y are none-zero integers, is x^y<y^x?

1. As x = y^2, then I'll replace all "x" with "y" in the above inequality.

x^y = (y^2)^y = y^(2*y)
y^x = y^(y^2) = y^(y*y)

Try to plug in some easy values:
If y = 1 then y^(2y) = y^(y^2) = 1, or x^y = y^x
If y = 3 then
y^(2y) = 3^6
y^(y^2) = 3^9
and 3^6<3^9, or x^y<y^x
Hence 1 is insufficient.

2. y>2, there're two unknown here, but only one condition that y>2, hence it's insufficient to confirm the inequality.

Example:
x = 1, y = 3, then 1^3<3^1
x = -2, y =4, then -2^4>4^(-2)

1&2.
x = y^2 and y>2

x^y = (y^2)^y = y^(2*y)
y^x = y^(y^2) = y^(y*y)

To compare A^B with A^C: if A>1, C>B>0, then A^C>A^B
Since:
y>2
y^2>2y>0 (as y>2)
Then y^(2*y) is always smaller than y^(y*y), or x^y<y^x

=>Suff.
Pick C.

The answer is C.

Thanks a lot. And guess what? im taking my gmat on the 2nd of Nov. So we both are more or less in the same situation! Good Luck with your prep. May you get that 700+ score!

Excellent care and attention to details, limestone.

Very nice problem, rb... thanks for sharing.

Regards,
Fabio.

x^y < y^x

I) statement X = Y ^ 2

Substitute x with y ^ 2

(y ^ 2 )^ y < y ^ (y ^ 2 )

y ^ 2y < y ^ y ^ 2

Base is same y so

2y < y ^ 2

2y can be greater than y if y = 1 or can be smaller if y is greater than 1 or less than 0

So insufficient

II ) y > 2

Pretty sure,

I did this way , But i am not sure whether my approach is correct or not.
Please guys Limestone and fabio

Share your view on my solution.

goyalsau wrote: y ^ 2y < y ^ y ^ 2

Base is same y so

2y < y ^ 2

I did this way , But i am not sure whether my approach is correct or not.
Please guys Limestone and fabio
Share your view on my solution.

No Sir, not good!

Let us understand this with care:

when you go from (say) x> y to a^x > a^y you are using the exponential function f(z) = a^z in the following sense:

x > y you would like to imply that f(x) > f(y) that is, a^x > a^y , correct?

So, the question is: when x> y implies f(x) > f(y) ?
Answer: when the function f is strictly increasing, that is, for greater values (in the domain), you get greater values as their images...

The fact is that f(z) = a^z is strictly increasing when a is any constant GREATER THAN 1, therefore you could use this property as limestone did BECAUSE you know that the parameter, I mean, the constant a (in the problem´s case y) is greater than 1, because it is greater than 2 (statement (2)).

Just to be sure you got it: try 3 > 2 then (1/2)^3 > (1/2) ^2 (MISTAKE!) to see how the parameter be greater than 1 is important.

For the sake of completeness, when the parameter a is between 0 and 1, the function is strictly decreasing, that is, when x > y then certainly a^x < a^y ... therefore the case with (1/2) above is not a single "exception", it is a good representant of all cases where the parameter is between 0 and 1.

Crystal Clear?

Regards,
Fabio.