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<bwe 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>=3now 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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