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k^3 = 3^k

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by shankar.ashwin » Sun Nov 13, 2011 11:21 am
a^b = b^a would be true only when a=b IMO. So k can only be 3 to satisfy the condition. B IMO
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by vaibhavgupta » Sun Nov 13, 2011 11:27 am
venmic wrote:For how many integers k is k^3 = 3^k ?
(A) None
(B) One
(C) Two
(D) Three
(E) More than three

Thanks for your help


+1 for B
OA
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by neelgandham » Sun Nov 13, 2011 11:55 am
shankar.ashwin wrote:a^b = b^a would be true only when a=b IMO. So k can only be 3 to satisfy the condition. B IMO
Though the answer is correct, I think the statement mentioned by you is incorrect.
Let us take an example as shown below

Let the value of a be 2 and that of b(=k) be 4
2^4 = 4^2 = 16.

I am not sure how useful it will be but here is something I came up with,

If a^b = b^a
b log a = a log b (Applying log on both sides)
(log a)/a = (log b)/b
Let b = a^n, where n is an Integer, then
(log a)/a = n(log a)/b
1/a = n/b
b = an

So, we have two equations now
b = an and b = a^n
From the above a should satisfy the equation a^n = an

a = 3 in this case
3^n = 3n

Method 1: Draw the graphs of 3n ad 3^n.One can see that they intersect at only one integer point where n = 1. So, the value of b(= k) is 3^n = 3^1 = 3

Method 2:
Solving for n , 3^n = 3n
If n = 0, 3^n = 3 and 3n = 0
If n = 1, 3^n = 3 and 3n = 3
If n = 2, 3^n = 9 and 3n = 6
if n = 3, 3^n = 27 and 3n = 9 ..You observe that the difference between the values of 3^n and 3n is increasing drastically, so there would be no number n > 1 which will satisfy the equation.
If n = negative, 3^n = positive and 3n = negative (So, no negative number will satisfy the equation)

The value of n = 1.
b(=k) = an and b(=k) = a^n, where a = 3 so b(=k) = 3^1 or 3*1 = 3
Answer : B, only one integer 3

p.s: Just to make you happy :
a^n = an; 2^n = 2n, satisfies for both two values of n
n = 1; 2^n = 2 and 2n = 2
[/b]n = 2; 2^n = 4 and 2n = 4 ;)

p.p.s: The above explanation is put up in a hurry, so please ignore the shortcuts, sms lingo(if any). Correct me if I am wrong.
Last edited by neelgandham on Sun Nov 13, 2011 12:03 pm, edited 1 time in total.
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by shankar.ashwin » Sun Nov 13, 2011 12:03 pm
Dude!! You're insane :) I wouldn't even be thinking using log in the GMAT :)

I came up with a^b = b^a would be true only when a=b, but it works only when 'a' and 'b' are prime numbers I guess, in this case they are.

neelgandham wrote:
shankar.ashwin wrote:a^b = b^a would be true only when a=b IMO. So k can only be 3 to satisfy the condition. B IMO
Though the answer is correct, I think the statement mentioned by you is incorrect.
2^4 = 4^2 = 16. In this case the value of a = 2 and b = 4

I am not sure how useful it will be but here is something I came up with,

If a^b = b^a
b log a = a log b (Applying log on both sides)
(log a)/a = (log b)/b
Let b = a^n, then
(log a)/a = n(log a)/b
1/a = n/b
b = an

So, we have two equations now
b = an and b = a^n
From the above a should satisfy the equation a^n = an

a = 3 in this case
3^n = 3n

Method 1: Draw the graphs of 3n ad 3^n and they intersect at only one integer point where n = 1. So, the value of b is 3

Method 2:

Solving for n , 3^n = 3n
If n = 0, 3^n = 3 and 3n = 0
If n = 1, 3^n = 3 and 3n = 3
If n = 2, 3^n = 9 and 3n = 6
if n = 3, 3^n = 27 and 3n = 9 ..You observe that the difference between the values of 3^n and 3n is increasing drastically, so there would be no number n > 1 which will satisfy the equation.
If n = negative, 3^n = positive and 3n = negative (So, no negative number will satisfy the equation)

The value of n = 1.
b = an and b = a^n, where a = 3 so b = 3^1 or 3*1 = 3
Answer : B, only one integer 3

p.s: Just to make you happy :
a^n = an; 2^n = 2n, satisfies for both two values of n
n = 1; 2^n = 2 and 2n = 2
[/b]n = 2; 2^n = 4 and 2n = 4 ;)

p.p.s: The above explanation is put up in a hurry, so please ignore the shortcuts, sms lingo(if any). Correct me if I am wrong.
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by neelgandham » Sun Nov 13, 2011 12:20 pm
shankar.ashwin wrote:Dude!! You're insane :) I wouldn't even be thinking using log in the GMAT :)
I came up with a^b = b^a would be true only when a=b, but it works only when 'a' and 'b' are prime numbers I guess, in this case they are.
I know that it is a waste of time ;)(took me around 3.5 minutes to solve it) but couldn't help. I think what you said is correct 'a^b = b^a would be true only when a=b, but it works only when 'a' and 'b' are prime numbers I guess, in this case they are.'

Good logical reasoning skills, I say !
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