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

Question

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

(1) u = 1
(2) v > 2

MathRevolution · Accepted Answer

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.

rishi02 · Answer

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

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=4xfD6xYlN1k

gauravsaggis1 · Answer

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.

adiagr · Answer

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

GGMU · Answer

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

GGMU · Answer

Bunuel  Can you please post the OA?

Looks tricky. IMO B should be the answer.

Can anyone solve it by picking up the numbers?

Bunuel · Answer

The OA is C. Please refer to the solutions above and ask if anything remains unclear.

GGMU · Answer

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.

Bunuel · Answer

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

Mo2men · Answer

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

LakerFan24 · Answer

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?

Aadi01 · Answer

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

TheNightKing · Answer

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.

Irising · Answer

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!

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!

giovannisumano · Answer

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}

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=xjymJ_kF2Ak

ScottTargetTestPrep · Answer

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

mohakgarg10 · Answer

This is incorrect since in statement 2 you assumed u = -1 when the question clearly says both u and v > 0.

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

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

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