Let a > b, both a and b are real and positive. Consider:If a = 4, b = 2 then a^b = b^aBut:If a = 3, b = 2 then a^b > b^aIf a = 5, b = 2 then a^b < b^aFind the locus of points where a^b = b^a and a > b > 0.

a = b*(log(a)/log(b)) ?

As already seen through the usual exp the EQ writes as b*ln(a) = a*ln(b).For b=1 we get ln(a)=0 and everything is very simple. Now look at ln(a)/ln(b)=a/b.In the positives one can get a as a multiple of b with multiplicant greater than 1,write a = b * (1 + x) with 0 < x. As b is positive it is an exp, b = exp(y), b=/=0.Inserting and some formal manipulation yields y = ln(1+x)/x. In 0 this function is1 (as a limit) and decreases strictly monotonous to 0 as x goes to infinity (!).So for any 0 < y=ln(b) < 1 there is exactly one x which solves y = ln(1+x)/x (andcan even given through a special function, we do not need that).Now substitute back to get the a for given 1=exp(0) < b < exp(1).Example: 0 < y:=ln(b)=ln(2) < exp(1). The eq ln(2) = ln(1+x)/x solves for x=1, andgives a = b * (1 + x) = 4.

Good call Avt -- you did this pretty close to the way I did.Like you, I started with:a/b = ln(a) / ln(b)We need a >b, so let a = C b, C > 1 (you chose C = 1 + x, x > 0)A little algebra gets us to a = C ^ (C/(C - 1)) b = C ^ (1/(C - 1))Let k = 1/(C - 1), k > 0a = (1 + 1/k) ^ (k+1)b = (1 + 1/k) ^ (k)which is the definition of e as k-> inf (that is, (e,e) is the limiting solution corresponding to lim k->inf).The only integer solution (a = 4, b = 2) obtains with k = 1.As you pointed out, for 1 < b < e there exists 1 solution for a; for b <= 1 there are no solutions for a.

Mathematica solves the problem:Solve[x^y==y^x, y]y = -x*ProductLog[-Log[x]/x]/Log[x]

good call Chukchi,but by looking at the graph one can see that there is at least one solution that mathematica did not deliver;y=x is a solution, and it is not depicted for values grater than the Neper's number.I also think (for obvious reasons) that there is another solution not being represented for values smaller than the Neper constant.Alex

QuoteOriginally posted by: alexandreCgood call Chukchi,I also think (for obvious reasons) that there is another solution not being represented for values smaller than the Neper constant.Alexthe other solution being a replacment y <--> x in the Product Log solution.On the graph bellow this is obtained by mirroring the decaying solution using a mirror placed in the line y=x(am I being clear?...)Alex

QuoteOriginally posted by: alexandreCOn the graph bellow this is obtained by mirroring the decaying solution using a mirror placed in the line y=x(am I being clear?...)Alexif I am allowed, and for any physicist that might read this, note the similarity of this mirroring with the "method of images", used to compute correlators in open string theory, (the end point of the boundary (that, with time, defines a line) becomes the mirror...)I was wondering, are there more solutions to the problem?Alex

hehehe my solutions of a^b = b^a are(9/4, 27/8)(625/256, 3125/1024)(7776/3125, 46656/15625)...but anybody can figure out the pattern here

QuoteOriginally posted by: krhehehe my solutions of a^b = b^a are(9/4, 27/8)(625/256, 3125/1024)(7776/3125, 46656/15625)...but anybody can figure out the pattern hereehehe nice one - but really, that does not bring us anything new, given that your particular solutions are special cases of the Chukchi's ProductLog solution... (verify it using mathematica, for example!)Alex

yes, but they are /Q... indeed I challenge you to prove that all rational solutions are of this formnow there might be some solutions over Q(sqrt(p))... haven't really looked into that

well that is very true.you've given us rational solutions to the problem.- that lead us to the question, are there more rational solutions other than the ones you are suggesting?Alex[rational numbers, more important than what one usually thinks,for example, they are crucial for the definition of some types of Conformal Field Theories (the Minimal Models)]

write c = (r/s)^q, where q is maximal, r and s have no common factors, (a,b) = ( c^(1 / (1-c) ), c^(c / (1-c) ) )consider the situation where (r^q - s^q) divides q.s^q; s < r and you will see that q is very limited as a function of r - in particular, exp(3/2) < 5 so q <= 4, and in nearly all cases q <= 2. But q=2 is actually impossible... etc

in order to make the problem more interesting you may trya^(mb) = b^(na)for instance with (m,n) = (1,2), we have the solution (3^2, 3^3) i.e. both sides are 3^54and with certain combinations of m and n, a new family of solutions can appear

yup! good one!