Solving x to the power of x

How would you go about solving something like

x^x = 10

Ive found elsewhere how to differentiate x^x; the derivative is (1 + ln x)x^x

I cant find a way of solving x^x algebrically but i can find the solution to the problem with this calculator method that i stumbled upon.

Start with any number, lets use 5.
Then find the 5th root of 10, root(5)10 = 1.584...
Then use this result as the base and do it again, root(1.584...)10 = 4.275...
Then use this result as the base and do it again, root(4.275...)10 = 1.713...

You get the picture.
The value for this sequence converges on 2.486...
(convergence, hmm. maybe we need limits or whatever they are.)

And 2.486...^2.486... = 10

So thats it solved, but wheres the maths behind it?

Arent us mathmaticians weird... looking for the solution to a problem we already have a way of solving. But thats not good enough for me.

Reply 1

nota bene

15

x^x=e^(xln(x)) ?

edit: although I am not if that would help:s

Reply 2

DFranklin

18

Crisis143

So thats it solved, but wheres the maths behind it?
You are calculating

\displaystyle x_{n+1} = 10^{1/x_n}

.

If this converges then (abusing language a little) you have

x_{n+1} = x_n

and so

x_n = 10^{1/x_n}

. Raising both sides to the power of

x_n

gives

x_n^{x_n} = 10

.

As to why it converges, that is more complicated I think. If we consider y(x) = 10^{1/x}, we see ln(y) = (ln 10 / x) and so y' = -(y ln 10) / x^2. If you do some faffing around (that I can't really be bothered to), you can show |y'| < .999 for x > 2.4 or so. Then the mean value theorem tells us that |y(x0)-y(x1)| < .999|x1-x0|, and then there's a theorem called the contraction mapping theorem that says under those circumstances the sequence defined by

x_{n+1} = y(x_n)

must converge. Possibly there's an easier way.

Edit: Actually, I don't think that explanation of convergence works, because we can end up on the "small side" of the root and y is not a contraction mapping there. I think you'd need to consider the 2nd iterate, which I'm fairly sure is a contraction mapping, although proving it would be a pain and a half...

Reply 3

jpowell

15

x^x = a

A solution is x =

\frac{log[a]}{W[log[a]]}

where W is the Lambert W function. http://en.wikipedia.org/wiki/Lambert_W_function

W is not expressible in terms of elementary functions. The wikipedia article has more information.

Reply 4

Dystopia

10

jpowell

x^x = a

A solution is x =

\frac{log[a]}{W[log[a]]}

where W is the Lambert W function. http://en.wikipedia.org/wiki/Lambert_W_function

W is not expressible in terms of elementary functions. The wikipedia article has more information.

Interesting. I've been looking for something like that.

Reply 5

Richardcz

Hi! I think you can solve it graphically. Try this:

1) first take 10th logarithm of both sides, you will get log(x) * x = 1
2) re-write it as log(x) = 1/x
3) draw graphs for both functions

There should be only one intercept point.

I think you can't get a precise number with any formulas. Problems of this type are solved by iterating guesses.

Reply 6

Lord of the Flies

19

Original post by Richardcz

Hi! I think you can solve it graphically. Try this:

1) first take 10th logarithm of both sides, you will get log(x) * x = 1
2) re-write it as log(x) = 1/x
3) draw graphs for both functions

There should be only one intercept point.

I think you can't get a precise number with any formulas. Problems of this type are solved by iterating guesses.