# Solve the equation

1. Oct 21, 2015

### Zhang Jiawen

a<b, a^b=b^a, how to solve such kind of equation?

2. Oct 21, 2015

### andrewkirk

I doubt there's a general technique. Why not try a few small integers and see what happens though?

3. Oct 21, 2015

### andrewkirk

I take it back. There is a technique. Start by taking logs.

4. Oct 21, 2015

### Yatin

What exactly do you mean by 'solve'? Even by taking log, i think 'a' will always exist in terms of 'b' and vice versa.

5. Oct 22, 2015

### aikismos

Hmmm. I don't see a general technique right away... obviously if a=b, then the solution is 1, but if a<b...

To check out the boundaries of the trichotomy of reals:

$(-2)^{-1} = (-1)^{-2} \rightarrow -\frac{1}{2} = 1$
$(-1)^{0} = (0)^{-1} \rightarrow 1 = 0$
$(0)^{1} = (1)^{0} \rightarrow 0 = 1$
$(1)^{2} = (2)^{1} \rightarrow 1 = 2$

To check out large gaps:
$(2)^{2000} = (2000)^{2} \rightarrow big = small$
$(-2)^{-2000} = (-2000)^{-2} \rightarrow small = big$

Take the b-th root of both sides:
$a = b^{\frac{a}{b}}$

Take the log base a of both sides and then exponentiate on the value a:
$b = a \cdot log_a b$
$a^b = a^a \cdot b$

Set the last with the first on $a^b$ to get:
$a^a = b^{b-1}$

Yeah, I'm thinking that this doesn't violate what I thought was a rule that it's not possible to solve one equation with two unknowns.

6. Oct 22, 2015

### andrewkirk

It can't be solved in the sense of identifying a unique solution (a,b). But it can be solved in the sense of finding a range for a for which a solution exists, and the corresponding b can then be found for any a in that range. I think finding the value of b would require numerical techniques as I don't think there's an analytic solution.

Having said that, there's a neat pair of small positive integers that is one solution.