The original condition has 2 variables, so we need 2 variables as well. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer.

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If u > 0 and v > 0, which is greater, u^v or v^u?

(1) u = 1

Since we are not told if u and v are different numbers there are two possibilites:
a) u=v= 1 in which case u^v = v^u
b) They are different numbers in which case v^u will always be greater since u^v = 1 (1 raised to anything is 1)

(2) v > 2

It does not give us any information about u. For example u could be 3 and v could be 4 and vice versa

On combining we know u^v will always be 1 . Therefore v^u is greater. SUFFICIENT

Ans = C

from statement 1) u=1 so, (1)^v= 1 and v^1. if v=1 then both are equal otherwise v^1 > 1^v. so not sufficient.
2) statement 2. not sufficient. as we know the value of v only.

1) +2) sufficient because this will eliminate the first case when v=1 from first statement.
answer C.

Statement 1
when u = 1, u raised to any power is always 1. Also v^u is always v.

If v>1 then v^u > u^v

If v[b]<1, then v^u < u^v [/b]

Clearly not sufficient.

Statement 2

v > 2

say u = 0.5

(3)^0.5 = 1.732
(0.5)^3 = 0.125

v^u > u^v


v=4; u=-1

v^u = 4^(-1) = (1/4)
u^v = (-1)^4 =1

v^u < u^v

Not sufficient.

Combining these two.
v > 2; u = 1
when u = 1, u raised to any power is always 1. Also v^u is always v.

since v is always > 2, and u is always 1. hence

v^u is always greater than u^v

adiagr According to ques u>0 so it should not be taken as -1

Bunuel Can you please post the OA?

Looks tricky. IMO B should be the answer.

Can anyone solve it by picking up the numbers?

The OA is C. Please refer to the solutions above and ask if anything remains unclear.
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Bunuel Thanks for your reply.

In statement 2:

u=1 and v=3 Then v^u is greater
u=4 and v=3 Then v^u is greater

Please tell a situation when it will not hold true.

You can consider the case when u = v. In this case v^u is NOT greater than u^v.
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If u > 0 and v > 0, which is greater, u^v or v^u?

(1) u = 1

Let v = 10........then 1^10 < 10^1

Let v= 1...........then 1^1 = 1^1

Insufficient

(2) v > 2

same example as above:
Let v = 10........then 1^10 < 10^1

Let v= 10...........then 10^10 = 10^10

Insufficient

Combine 1 & 2

V > 2 & u= 1..........So first example applies

Let v = 10........then 1^10 < 10^1

Sufficient

Answer: C

i know this says its a <600 level problem, but i could not disagree more. is it just me or does this seem to be at least 600 level?

What is this? I did not understand it at all. Please elaborate. The method seems tactful but I am not getting the same

\(u^v\) ?? \(v^u\)

Statement 1 : u=1

\(1^v\) ?? \(v^1\)

Since we don't know anything about v (except it is positive) statement 1 is insufficient.

Statement 1 : v>2. Let's say 3

\(u^3\) ?? \(3^u\)

Since we don't know anything about u (except it is positive) statement 2 is insufficient.

Combined :

\(1^3\) < \(3^1\)

Hence Sufficient.

Note: Even if you take value of v as 2.5, the relation will still hold true.
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When testing cases for statement 2, the first three cases I used all turned out to show u^v is less than v^u. At that point, my timer already exceeded 2 minutes. I could go on and test more cases and eventually to find that there are cases where u^v is equal or great than v^u. But the problem is, I don't have that much time to spend during the exam. I find myself in this kind of situation constantly when working on DS arithmetic questions.

Does anyone have a recommendation on what numbers to use when testing cases? For those of you used v=u, how did you come up with that? For those of you used u=3, v=4, how did you come up with that? Thanks!

You wrote "...b) They are different numbers in which case v^u will always be greater since u^v = 1 (1 raised to anything is 1)"

I think this is not correct, since v could be less than 1, in which case \(v^{u} < u^{v}, i.e. (1/2)^1 < 1^{1/2}\)