I tried looking for ways to solve this equation and came across something like Lambert's W function, which, by the way, I did not understand a bit, because I've never learned it nor do I have a decent mathematical background. I also came across one more method called the Newton's method, but never used it either. Can the procedure to solve this equation be made a little bit clearer? I just plugged in the value on my calculator and it solved it for me, I guess it must have used some trial and error. Please advise.

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    $\begingroup$ @J.M. I would not say that this question is a duplicate, because the OP indicates that he does not understand the solution that is given. Perhaps this question should be closed because the OP is in over his head, but I'm not able to be the judge of that. $\endgroup$ – gebruiker Mar 17 '16 at 12:22
  • $\begingroup$ @J.M, How can a question asked on July 8th of a year be a duplicate of a question asked 14 days later on the same year? That is duplicate or this one is? $\endgroup$ – Ragavan N Oct 24 '18 at 8:41

Please excuse me if I go back too far in not assuming things which you already understand very well.

You first have to clarify what $x^x$ really means. In general, $a^x$ is defined as $e^{x \log a}$ for positive $a$. So for positive $x$, we define $x^x$ as $e^{x\log x}$. If you want to use Newton's method, it's a bit easier to take (natural) logarithms at this point. Solving $x^x = c$ for some number $c$ is then equivalent to solving $x \log x = d$ where $d = \log c$, so let's concentrate on solving $x \log x = d$. We are then trying to solve $f(x) = 0$ where $f(x) = x \log x -d$. Notice that the derivative $f^{\prime}(x) = 1 +\log x$.

Newton's method tries to solve $f(x) = 0$ by picking a starting approximation to a solution $x_0$, and then by generating new approximations via the formula $x_{n+1} = x_n - \frac{f(x_n)}{f^{\prime}(x_n)}.$ This is justified by the heuristic that $f(x + h)$ is close to $f(x) + hf^{\prime}(x)$ when $h$ is small, so if we take $h = \frac{-f(x)}{f^{\prime}(x)}$, we should get something close to zero. Newton's method doesn't always converge to a solution, and analysing under which conditions it does is quite complicated.

Anyway, for this choice of $f$, Newton's method tells us to set $x_{n+1} = x_{n} - \frac{x_n \log (x_n) - d}{1+\log (x_n)}$, which we can rewrite as $ x_{n+1} = \frac{x_n + d}{1 + \log( x_n)}$. This, or something like it, may be how your calculator found a solution.

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  • $\begingroup$ Theorem : If $f$ is increasing and convex (then $f$ has a unique zero) and if $x_0$ is greater than the zero of $f$, then Newton's method converges (geometricaly, with explicit bounds). Proof : Just notice that in these conditions, the sequence is decreasing and bounded from below by the zero. $\endgroup$ – user10676 Jul 8 '11 at 17:22
  • $\begingroup$ Thanks. That's helpful for this example, but generally speaking it's not always so easy to know how Newton's method behaves (things like Lipschitz conditions can sometimes be used). $\endgroup$ – Geoff Robinson Jul 8 '11 at 19:17
  • $\begingroup$ @GeoffRobinson Why $e^{xlogx} $ ? If we use e it should be $e^{xlnx} $ right? $\endgroup$ – BugShotGG Aug 27 '13 at 15:48
  • $\begingroup$ Log to an unspecified base means that natural log by most conventions. $\endgroup$ – Geoff Robinson Aug 27 '13 at 20:16

You have already found the answers to your question, so I'll just explain them.

You want to solve $x^x = c$, for some number $c$. The answer $x$ cannot be expressed in terms of the common functions — polynomials, trigonometric functions, exponential, etc. — but you can express it in terms of another function that has been studied and given a name. Specifically, the Lambert W function is defined as the answer to a related question:

If $x = W(z)$, then $xe^x = z$.

The relation is that $$\begin{align}x^x &= c \\ x\ln x &= \ln c \qquad\qquad \text{ taking logarithms}\\ (\ln x)e^{(\ln x)} &= \ln c \qquad\qquad \text{ writing }x=e^{(\ln x)}\\ \ln x &= W(\ln c) \qquad \text{by definition of }W \\ x &= e^{W(\ln c)} \end{align}$$

So the Lambert W function gives a name for what you want. To find the value numerically, you can use something like Newton's method, which is a very general method. For your problem, the method is something like the following (according to Wikipedia):

  • Pick some value $w_0$ as your initial guess.

  • If you know some value $w_j$ (where $j = 0, 1 \dots$ is an index), find the next value $w_{j+1}$ as $$w_{j+1}=w_j-\frac{w_j e^{w_j}-\ln c}{e^{w_j}+w_j e^{w_j}}.$$

  • At each stage (for each $j$), $e^{w_j}$ is an approximation to the true value $x$ you want, and this approximation gets better as $j$ increases (as you perform more steps).

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  • $\begingroup$ I must confess I don't have a very good understanding of the conditions under which Newton's method converges fast and correctly. For instance, applying Newton's method directly to $f(x) = x^x - c$ gives the iteration $\displaystyle x_{j+1} = x_j - \frac{x_j^{x_j} - c}{x_j^{x_j}(1 + \ln x_j)}$ while applying it to $f(x) = x\ln x - \ln c$ gives the iteration $\displaystyle x_{j+1} = x_j - \frac{x_j\ln x_j - \ln c}{1 + \ln x_j} = \frac{x_j + \ln c}{1 + \ln x_j}$. Both of these look simpler than what I quoted from Wikipedia in the answer above, but I don't know if they're "better". $\endgroup$ – ShreevatsaR Jul 8 '11 at 14:05

None of the "elementary functions" from basic mathematics solves this. A "new" function, the Lambert W function, can be introduced, which solves many things not solvable in elementary functions. Including this equation.

Solve $x^x=y$ for $x$ and get: $$ x = \frac{\operatorname{ln} (y)}{\mathrm W \bigl(\operatorname{ln} (y)\bigr)} $$

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    $\begingroup$ Why not take the logarithm to the basis $x$? $\endgroup$ – Rudy the Reindeer Jul 8 '11 at 13:44
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    $\begingroup$ @Matt: $x\log_x x = \log_x y$ so then $x = \log_x y$, but (since $x$ is on both sides) not yet solved for $x$. $\endgroup$ – GEdgar Jul 8 '11 at 13:47
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    $\begingroup$ @Matt: How do you propose to take logarithm to base $x$ without knowing the value of $x$? $\endgroup$ – ShreevatsaR Jul 8 '11 at 13:52
  • $\begingroup$ @ShreevatsaR: that might indeed be a problem! : ) $\endgroup$ – Rudy the Reindeer Jul 8 '11 at 19:56
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    $\begingroup$ @Gedgar If I want to find the value, how do I go about it? What is W(ln(y))? $\endgroup$ – Ragavan N Jul 11 '11 at 12:22

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