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

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Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 08 Oct 2019, 20:55
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

48% (02:22) correct 52% (01:44) wrong based on 25 sessions

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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 08 Oct 2019, 21:38
Is \(A^B>B^A\)?

Only if it is known A > B or A < B, a concrete YES/NO answer can be found.

(1) \(A^A > A^B\)

Since base are same, A > B. YES
SUFFICIENT.

(2) \(B^A>B^B\)

Since base are same, A > B. YES
SUFFICIENT.

Answer D.
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 08 Oct 2019, 22:59
1
Is A^B > B^A?

Statement 1: A^A > A^B
Insufficient because if A=2, and B=1, satisfying the condition in statement 1, we have A^B (2) > B^A (1)
however, if A=-2, and B=-3, satisfying the condition in statement 1 as well, we have A^B (-1/8) < B^A (1/9).

Statement 2: B^A > B^B.
Insufficient because if A=3, B=2, satisfying the condition in statement 2, we A^B (9) > B^A (8)
however, if A=-2, and B=-3, satisfying the condition in statement 2, we have A^B (-1/8) < B^A (1/9)

Statement 1 + Statement 2:
Both statements are still insufficient.
This is because if A=3, and B=2, satisfying both statements, A^B (9) > B^A (8)
however, if A=-2, and B=-3, also satisfying both statements, A^B (-1/8) < B^A (1/9).

Hence the answer is E.
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Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post Updated on: 10 Oct 2019, 06:18
#1
A^A>B^A
base is common so A>B
for all integer values +/-ve we get A^B>B^A we get yes/no ; insufficient
#2
B^A>B^B
again base is common so A>B
hence for all integer values +/-ve we get A^B>B^A we get yes/no ; insufficient
IMO E

Is A^B>B^A

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

Originally posted by Archit3110 on 09 Oct 2019, 03:34.
Last edited by Archit3110 on 10 Oct 2019, 06:18, edited 1 time in total.
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 09 Oct 2019, 04:07
1
1
Is \(A^B > B^A\)?
All those who forget that the values of A and B can be anything that is negative, positive, fraction or integers will go wrong by marking D.

(1) \(A^A > A^B\)
A=3 and B=2...Yes
A=4 and B=3...No
A=-2 and B=-3..No

(2) \(B^A > B^B\)
A=3 and B=2...Yes
A=-2 and B=-3..No


Combined..
A=3 and B=2...Yes
A=-2 and B=-3..No

E
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 09 Oct 2019, 05:03
1
Quote:
Is A^B>B^A?

(1) \(A^A>A^B\)
(2) \(B^A>B^B\)


case: a… b… a^a… b^b… a^b… b^a… condition y/n:
I.…: 1… -1… 1… -1… 1… -1… 1>-1? yes
II.…: -2…-3…1/4…-1/27…-1/8…1/9…-1/8>1/9? no
III.…: 2… -3…4……-1/27…1/8…1/9…1/8>1/8? yes

(1) \(A^A>A^B\): all cases are valid, insufic.
(2) \(B^A>B^B\): case II. and III. are valid, insufic.

(1&2): case II. and III. are valid, insufic.

Answer (E)
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 09 Oct 2019, 06:09
1
(1) Shows that A > B
Using the above we can say...
If B is negative and A is positive and even B^A could be greater than A^B
If A and B are both positive integers A could be greater than B

Therefore 1 is NS

(2) Also Shows that A > B

Therefore 2 is NS

(1+2) Statements 1 and 2 could both be true
If B is negative and A is positive and even B^A could be greater than A^B
If A and B are both positive integers A could be greater than B

Therefore Answer must be (E) as 1 and 2 combined are insufficient
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B  [#permalink]

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New post 09 Oct 2019, 16:56
1
Is \(A^{B}>B^{A}\) ?

Statement1: \(A^{A}>A^{B}\)

if the base of exponent is greater than 1(A>1), then
--> A>B.
-------
if A=3 and B=2, then 9 > 8 (YES)
if A=5 and B=2, then 25 >32} (NO)
Insufficient


Statement2: \(B^{A}>B^{B}\)

if the base of exponent is greater than 1(B>1), then
--> A>B.
-------
if A=3 and B=2, then 9 > 8 (YES)
if A=5 and B=2, then 25 >32} (NO)
Insufficient

Taken together 1&2, Clearly insufficient.

The answer is E.
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Re: Is A^B > B^A? (1) A^A > A^B (2) B^A > B^B   [#permalink] 09 Oct 2019, 16:56
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