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If a and b are distinct integers and a^b = b^a, how many

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Re: If a and b are distinct integers and a^b = b^a [#permalink] New post 27 Apr 2013, 02:39
Expert's post
imhimanshu wrote:
Question: If a and b are distinct integers and a^b = b^a, how many solutions does the ordered pair (a, b) have?

(A) None
(B) 1
(C) 2
(D) 4
(E) Infinite

Hi Experts,
I'm stumped by this question, and have seen quite a few questions that usually test understanding of exponential questions. From the first impression of this question, I found that this question will test positive,negative, 0, even, odd every possibility. Can someone please post a detailed solution of this.

Thanks


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Re: If a and b are distinct integers and a^b = b^a [#permalink] New post 27 Apr 2013, 03:05
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
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Re: If a and b are distinct integers and a^b = b^a [#permalink] New post 27 Apr 2013, 22:53
Expert's post
imhimanshu wrote:

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



Check this out: http://www.veritasprep.com/blog/2013/01 ... cognition/

It uses a graphical and pattern approach to discuss this question.
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Re: Try this one - 700 Level, Number Properties [#permalink] New post 28 Apr 2013, 02:32
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.
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Re: Try this one - 700 Level, Number Properties   [#permalink] 28 Apr 2013, 02:32
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