For how many pair(s) of positive integers m and n, such that

Question

For how many pair(s) of positive integers  m  and  n , such that  n > m > 1 , is  m^n  not greater than  n^m ?

A. 0
B. 1
C. 2
D. 3
E. 4

Archit3110 · Answer

I could get two possiblites (2,4) & ( 2,3) where n>m>1 and  m^n  not greater than  n^m 
i.e 16=16 and 8<9
IMO C

Lampard42 · Answer

Bunuel   chetan2u . Is substitution the only way?. Is there any other efficient method. Also please share similar questions if available.

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ScottTargetTestPrep · Answer

Let m = 2, so n ≥ 3; we see that:

2^3 = 8 is not greater than 3^2 = 9, and 2^4 = 16 is not greater than 4^2 = 16. However, when n ≥ 5, we will have 2^n greater than n^2.

Now, let m = 3, so n ≥ 4; we see that:

3^4 = 81 is greater than 4^3 = 64. In fact, for any value n ≥ 4, we will have 3^n greater than n^3.

Since m^n is greater than n^m is already true for m = 3 (and n > m), it should be true for any m > 3 (and n > m).

Therefore, we have only two pairs of (m, n) such that m^n is not greater than n^m, namely, (2, 3) and (2, 4).

Answer: C

thinkvision · Answer

nehasomani33  - this one also?

Oppenheimer1945 · Answer

You’ve to check for   m^{(1/m)}<n^{(1/n)}

Take  f(x)=x^{(1/x)}

The function increases and attains max at x=e (2.71).

As 1<m<n
Hence you’ve to check for x=2,3,4. For x beyond 2.71, function decreases hence the inequality won’t hold true.

So we have only 2 solutions.

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bumpbot · Answer

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For how many pair(s) of positive integers m and n, such that

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For how many pair(s) of positive integers m and n, such that

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