Confusing Problem

[knight247]

Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B

OA is E. Detailed explanations would be appreciated. Thanks

[force5]

can you please check this problem again. Is it copies correct?

[HeintzC2]

Statements 1 and 2 tell us the exact same thing, so we can automatically narrow our choices to D and E.

In statement 1:
Plugging 2 in for A we get 2^2>2^B, so B<2 or simply B<A.
However, plugging in -2 for A we get (-2)^(-2)>(-2)^B, in this case B = -1 and B = -3 both satisfy the equation.
We cannot discern a relation between A and B from statement 1 alone.

Statement 2 yields similar results. Both statements are insufficient alone, and provide no new information together.

Answer is E

Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B

OA is E. Detailed explanations would be appreciated. Thanks

[GmatKiss]

IMO:E

If u inctroduce values for A and B and check (1) and (2), we end up with E!

[navami]

Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B


1)

a>b


2)

a>b

so both the statement yeilds the same information a>b

now consider a = 4 b = 2
A^B>B^A ???
A^B = 16
B^A = 16

so A^B>B^A = FALSE


then a = 3 b = 2

A^B>B^A ???
A^B = 9
B^A = 8

so A^B>B^A = TRUE

so we can not conclude even after combining these two.

hence E

[prateek_guy2004]

Well yes this was little bit confusing... i spended around 4 mins.....Is better spend little time rather than confusing yourself.....Finally found answer should be E.