Algebraic method to solve X^x = c

[Angry Citizen]

For about a year, I've had this problem in my head. It stemmed from a very benign hypothetical sci-fi scenario, which I won't bother repeating, as it is unimportant next to the conundrum that resulted.

x^x=c is interpreted as a single variable located in both the base and the exponent, set equal to a constant (whether it be something as incredibly easy as 4, wherein x=2, or as mind-numbingly difficult as 3 [which was the problem -- x^x=3]). I don't even know what branch of mathematics to use to solve this deceptively simple problem. Anyone want to take a crack at it, or at least point me in the right direction?

Yes, I'm aware one can plug-'n-pray, or just stick it in a graphing calculator. What I'm asking is whether there's an algebraic method to solve it, or if it's an unsolved problem.

 

[Hurkyl]

It really depends on what you mean by "solve".

For many practical purposes, simply knowing that x^x=3 is solution enough, and for most of the rest, a rough approximation (with error bounds for bonus points!) is all you need.

Lambert apparently worked with similar problems enough that he decided to give their solution a name -- we now call it the "Lambert W" function.

 

[Angry Citizen]

Well, again, by 'solve', I mean discerning the value of the variable if given a constant. I ended up simply approximating using the tried and true plug-'n-pray method, but the sloppiness of having to do that is so uncharacteristic of mathematics.

Was I correct, then, that it is impossible to solve according to the criteria I mentioned?

 

[Hurkyl]

Well, again, by 'solve', I mean discerning the value of the variable if given a constant.

But as I said, the phrase "the unique value satisfying x^x=c"^* is an adequate expression for the value of x for many purposes.

I ended up simply approximating using the tried and true plug-'n-pray method, but the sloppiness of having to do that is so uncharacteristic of mathematics.

Approximation is not sloppiness (unless, of course, you do it sloppily. ) -- manipulating them it one of the most basic methods of calculus and real analysis.

To put it differently, a "clever" argument that 1.83^1.83 is approximately 3 is not worth any more than a few quick calculator strokes that prove the same thing.

"Cleverness" doesn't become useful until you can start proving interesting things -- e.g. how to estimate the rate of growth of the solution to x^x=c as c increases.

(And I edited my previous post, you may have missed the change)

*: Of course, I'm assuming c>1 and we want real solutions[/size]

 

[Tinyboss]

I don't even know what branch of mathematics to use to solve this deceptively simple problem. Anyone want to take a crack at it, or at least point me in the right direction?

It's easy to show that x^x is continuous and monotonically increasing for x>1 (this bound is not strict). Since 1^1=1, x^x specifies a bijection of [1,inf) with itself, and so an inverse function exists (related to Lambert's W function, as Hurkyl noted). There's no expression of this function in terms of elementary functions, though. We just say, "it's the inverse of this function on that domain, which exists because blah blah blah", and give it a name.

These arguments belong to the branch of mathematics called analysis, and not to algebra. There's no algebraic way of getting these numbers.

 

[Angry Citizen]

It's easy to show that x^x is continuous and monotonically increasing for x>1 (this bound is not strict). Since 1^1=1, x^x specifies a bijection of [1,inf) with itself, and so an inverse function exists (related to Lambert's W function, as Hurkyl noted). There's no expression of this function in terms of elementary functions, though. We just say, "it's the inverse of this function on that domain, which exists because blah blah blah", and give it a name.

These arguments belong to the branch of mathematics called analysis, and not to algebra. There's no algebraic way of getting these numbers.

Interesting, thanks. Hopefully that will become clearer to me in a couple years.

 

[uart]

Hi AC. The LambertW function is defined as the solution to x e^x = c. That is W(c)=x, where x satisfies x e^x=c.

In terms of this function you can express the solution to your problem. That is, find x such that x^x=c, and the solution is x = e^(W(ln(c)).