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Re: If a and b are distinct integers and a^b = b^a [#permalink]

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27 Apr 2013, 04:05

1

This post was BOOKMARKED

imhimanshu wrote:

Hi Zarrolou,

Would you mind giving a shot on this one with an algebraic/graphical way..

Regards, H

A graph here is not possible, we would need a third dimension An algebric way is also very difficult to obtain, and is way beyond the GMAT-Math

Remember: \(a^b=b^a\)

is true for the following integers (1,1) (2,2) and all the values such that a=b (of course...) AND (4,2) (2,4) \(2^4=4^2\) same thing for values <0: (-1,-1) ... AND (-4,-2) ( -2,-4) If you want you can take away this simple tip, which is in my opinion much easier than solve the equation _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Try this one - 700 Level, Number Properties [#permalink]

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28 Apr 2013, 03:32

VeritasPrepKarishma wrote:

Anyone else would like to take a shot at proving it mathematically in a different way? Try it.

Hi my friends, this is my solution: (please forgive me if my bad English confuses you) For this solution to be short, le'ts assume we already have : \(0<a<b\) and we found that \((2,4)\) are one pair that satisfies the equation now let's consider \(3=<a<b\) we can prove that the pair (a,b) that satisfies the equation does not exist if \(3=<a<b\) even when a and b are not intengers. let's see : \(a^b=b^a\) so \(ln(a^b)=ln(b^a)\) so \(b.ln(a)=a.ln(b)\) it is the same as this equation \(\frac{lna}{a}=\frac{lnb}{b}\) let's consider function : \(f(x)=\frac{lnx}{x}\) (with \(x>=3\)) we have \(\frac{df(x)}{dx}=\frac{1-lnx}{x^2}\) because \(x>=3>e\) we have \(1-lnx<0\) and then we have \(\frac{df(x)}{dx}<0\) with \(x>=3\) now if \(3=<a<b\) we always have \(\frac{lna}{a}>\frac{lnb}{b}\) or \(a^b>b^a\)(e.g \(5^6>6^5\) and \(6^7>7^6\) and so on) so the conclusion is if a and b are distinct integers with \(0<a<b\) we have one pair (2,4) with \(a>b>0\) we have another pair (4,2) but if \(3<a<b\) or \(3<=b<a\) the pair that satisfiies the equation does not exist. _________________

Life is not easy I knew that and now I don't even expect life to be easy

Re: If a and b are distinct integers and a^b = b^a, how many [#permalink]

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14 Jul 2014, 09:46

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If a and b are distinct integers and a^b = b^a, how many [#permalink]

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11 Aug 2014, 15:11

VeritasPrepKarishma wrote:

The answer is indeed D (4 solutions). Good work everyone!

Now for the explanation. I tend to get a little verbose... Bear with me.

Given \(a^b = b^a\) and a and b are distinct integers. First thing that comes to mind is that if we didn't need distinct integers then the answer would have simply been infinite since \(1^1 = 1^1, 2^2 = 2^2, 3^3 = 3^3\) and so on... Next, integers include positive and negative numbers. If a result is true for positive a and b, it will also be true for negative a and b and vice versa. The reason for this is that both a and b will be either even or both will be odd because \((Even)^{Odd}\)cannot be equal to \((Odd)^{Even}\) Also, it is not possible that a is positive while b is negative or vice versa because then one side of the equation will have negative power and the other side will have positive power.

So basically, I need to consider positive integers (I can mirror it on to the negative integers subsequently). Also, I will consider only numbers where a < b because the equation is symmetrical in a and b. So if I get a solution of two distinct such integers (e.g. 2 and 4), it will give me two solutions since a can take 2 or 4 which implies that b will take 4 or 2.

Let me take a look at 0. It cannot be 'a' since it will lead to \(0^b = b^0\), not possible. Next, a cannot be 1 either since it will lead to \(1^b = b^1\), not possible. Let us consider a = 2. \(2^3 < 3^2\); \(2^4 = 4^2\)(Got my first solution); \(2^5 > 5^2\); \(2^6 > 6^2\) and the difference keeps on widening. This is where pattern recognition comes in the picture. The gap will keep widening. Now I will consider a = 3. \(3^4 > 4^3\) (first term itself is greater); \(3^5 > 5^3\) and the gap keeps widening. I can try a couple more values but the pattern should be clear by now. \(4^5 > 5^4, 5^6 > 6^5\) and so on... and as the values keep increasing, the difference in the two terms will keep increasing...

Note: Generally, out of \(a^b\) and \(b^a\), the term where the base is smaller will be the bigger term (I am considering only positive integers here.). In very few cases will it be smaller or equal (only in case of a = 1, \(2^3\) and \(2^4\)).

So I have four solutions (2, 4), (4, 2), (-2, -4) and (-4, -2). This question is pattern recognition based.

Now, we know that if the question did not have the word 'distinct', the answer would have been different, but what if the question did not have the word 'integer'? Would it make a difference? - Something to think about...

(A lot verbose, actually!)

If the question did not have the word 'integer'? Yes, the answer will different. For example: 2^k = k^2 K which is integer could be 2 and 4, and the other k will be a number that is negative. I am sure that is a negative number, since I draw the graph of 2^k and k^2, the two lines will intersect somewhere in the negative area of X axis. _________________

......................................................................... +1 Kudos please, if you like my post

Re: If a and b are distinct integers and a^b = b^a, how many [#permalink]

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19 Aug 2015, 04:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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