Eh? How do I solve xe^(a/x) = b?

[WarPhalange]

Eh? How do I solve xe^(a/x) = b?

Homework Statement

xe^(-a/x) = b, where a and b are numbers that are given and I'm trying to solve for x.

The Attempt at a Solution

All I can think of are Taylor series, which won't work in this case because a is ~4000 and b = ~0.1, so I'd need to expand to a lot of terms.

I could try to do it numerically, but would rather not... I know the answer ends up being ~520.

 

[gabbagabbahey]

You have a transcendental equation here, with no algebraic solution. Are you sure this is the correct equation?

 

[WarPhalange]

Yes. I'm trying to solve the equation that's in the book.

It's for charge carrier density in a semi-conductor

n^2 = B*T^3*exp(-E/kT)

Where n, B, E, and k are constants that are given. So I divide by B, then cube root it to obtain T*exp(-a/T) = b

A similar problem is later on in the homework set, where the professor says to calculate it numerically. I mean, the problem is identical except you first have to find n with another equation, then you're back to that. But he makes no mention of calculating it numerically here.

I think, though, that he still wants us to, because like you said, no easy solution. It would take a team of Russian mathematicians in one of Russia's finest Gulag's at least a year to find the solution*. I'll just do this numerically also. In the least I needed confirmation that it's not a trivial answer. Thanks.

*I'm taking a shot in the dark here.

 

[Gib Z]

There is actually no possible way of expressing that number in terms of commonly known constants. It would only take, however, someone with a big ego typing onto Physicsforums to define, eg, The GibZ Constant, whose definition is the unique solution to the above equation, numerically approximately *blah blah*. Those are your options, but in the end, for a "useful" answer you'll need a numerical approximation, no way around it.

 

[WarPhalange]

Found an answer:

http://en.wikipedia.org/wiki/Lambert_W_function

Too bad it's useless... and still amounts to a numerical computation. Oh well. Thanks guys.