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

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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

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

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

You can consider the case when u = v. In this case v^u is NOT greater than u^v.

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?

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

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.

Solution:

Question Stem Analysis:

We need to determine which of u^v and v^u is greater, given that both u and v are positive.

Statement One Alone:

Since u = 1, u^v = 1^v = 1 (1 raised to any power is 1) and v^u = v^1 = v (any number raised to the power of 1 is itself). However, we can’t determine which one is greater. For example, if v = 0.5, then u^v > v^u since 1 > 0.5. However, if v = 2, then v^u > u^v since 2 > 1. Statement one alone is not sufficient.

Statement Two Alone:

Knowing only that v > 2 is not sufficient to determine whether u^v > v^u or v^u > u^v. For example, if u = 1, and v = 3, then u^v = 1^3 = 1 and v^u = 3^1 = 3. However, if u = 3 and v = 3, then both u^v and v^u are equal to 27. Statement two alone is not sufficient.

Statements One and Two Together:

From statement one, we see that u^v = 1 and v^u = v. From statement two, we see that v > 2, so v^u > 2, and hence it’s greater than u^v. Both statements together are sufficient.

Answer: 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.

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.

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.

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

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 \(v^{u} < u^{v}, i.e. (1/2)^1 < 1^{1/2}\)

Answer: Option C

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