# How to solve x^x = 13 using logarithms

1. May 24, 2006

### minase

I was trying to solve a problem using logarithm. It says x^x =13 I tried to solve it using logarithm but I couldn't. I used a graphing calculator solver and I found it is 2.6410619... Is it possible to solve using logarithm? If not is there another way to solve besides trying and checking?

2. May 24, 2006

### vsage

I'm not aware of any way to solve such a problem other than numerically (IE using some numeric approximation). Ah I see VietDao has a solution.

Last edited by a moderator: May 24, 2006
3. May 24, 2006

### VietDao29

You can solve it by Lambert W-function, i.e the inverse function of f(x) = xex.
$$x ^ x = 13$$
$$\Leftrightarrow \ln (x ^ x) = \ln (13)$$
$$\Leftrightarrow x \ln x = \ln (13)$$
$$\Leftrightarrow (\ln x ) e ^ {\ln x} = \ln (13)$$
$$\Leftrightarrow \ln x = W( \ln (13))$$
$$\Leftrightarrow x = e ^ {W( \ln (13))}$$
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Or you can get an approximation of it by Newton's Method.
First, we'll try to rearrange the equation to give the form of f(x) = 0
So
$$\x ^ x = 13$$
$$\Leftrightarrow \x ^ x - 13 = 0$$
Let f(x) = xx - 13
Then we differentiate f(x) with respect to x to give:
f'(x) = xx(ln(x) + 1)
And we choose an arbitrary x0, say x0 = 2 (we should graph it first to chose x0 precisely, we choose x0 since there's one solution near 2).
And we apply the formula:
$$x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}$$, and let n increase without bound to obtain the desired solution, i.e:
$$x = \lim_{n \rightarrow \infty} x_n$$.
The root of that equation is about 2.641061916.
:)