Hard problem : calculus : prove and solve for a and b?

Question

a^1/a = b^1/b

Answer 1

taking ln
lna/a=lnb/b so give a value to b>0 and call lnb/b=k
now you have to find a
ln a-k*a=0
Let´s take the function
y= ln x-k*x
lim y x==>0+ = -infinity
suppose first k<0
lim y x==> +infinity is + infinity
y´= 1/x-k >0 so y is increasing and the is only one solution
note that k<0 implies b<1.So for each given b <1 you´ll find one a =b
If k>0
lim y x==> +infinity is - infinity
y´=1/x-k =0 so x = 1/k and f(x) =-lnk-1
if -ln k-1>0 or ln k <-1 so k<1/e there are two solutions for each b if lnb/b<1/e
ifk=1/e which implies b=e there is one solution and no solution if
k>1/e excluding the obvious a=b
Ex. take b= 2 k= ln2/2=0.3466>0 and <1/e=0.3679 so there are two values of a one would be 2and the other 4.Normally
You have to find the other with a calculator or by approximation methods of calculus

Answer 2

since you are performing the same operation on both sides it's like the other guy said 1=1, it doesn't matter what a and b are, because they must be the same (positive) number for that to work

Answer 3

Here is my proof in real domain.
a^1/a = b^1/b, a>0, b>0
Raise to the power of ab,
a^b = b^a
Take ln,
lnb/lna = lna/lnb
(lnb)^2 - (lna)^2 = 0
(lnb + lna)(lnb - lna) = 0
lnb + lna = 0 => a = b = 1
or
lnb = lna => a = b
Combine: a = b

Answer 4

Thats actually easier than you make it out to be: Since both numerators are raised to the 1st power, they are themselves (a=a, b=b). So you really have a/a = b/b or 1=1

Unless I am reading it wrong