Solving $y^y = x$ for large $x$ I was playing around with recurrence relations and noticed that $\sqrt x$ has the fun property that
$$\frac{x}{f(x)} = f(x)$$
($\sqrt{x}$ and its negation are the only functions $f(x)$ that satisfy this it).
That got me thinking about what functions satisfy
$$\sqrt[f(x)]{x} = f(x).$$
These functions need to satisfy
$$x = f(x)^{f(x)}.$$
If we let $y = f(x)$, this boils down to solving
$$y^y = x.$$
I am having trouble seeing how to solve this. My initial thought was to take the log of both sides, giving
$$y \log y = \log x,$$
and then tried seeing if the change-of-basis formula would help, since the above statement implies that
$$y = \log_y x,$$
but this didn't seem to offer any clarity.
Is there a nice way to solve this equation? Or is there a known name for a function of $x$ that's specifically designed to have this property?
 A: The Lambert W function $W$ is defined to be the inverse of the map $z \mapsto z e^z$ (or more precisely, its restriction to $[0, \infty)$). A little algebra shows that a solution $x$ to $x^x = y$ is (for suitable $y$)
$$x = e^{W(\log y)} = \frac{\log y}{W(\log y)} .$$
A: No, there is no nice way. (Think about it this way: how would you solve $x*x=y$, if you didn't know about the square root function? The square root was invented to solve $x*x=y$; you need another special function to solve $x^x = y$ (and as Travis mentions above, a very similar such function is called the Lambert $W$-function.)
What if you want to find an approximate solution? This is a great place to try Newton's method. If you want to solve $f(x)=y$, and you have some initial guess $x_0$, Newton's method lets you find a better guess (usually) using the formula
$$x_1 = x_0 - [f(x_0)-y]/f'(x_0)$$
where $f'$ is the derivative. For the case $f(x) = x \log x$ and $\log y$ we have
$$x_1 = x_0 - (x_0\log x_0 - \log y)/(\log x_0 + 1)$$
(if you haven't taken Calculus yet, you will have to take it for granted for now how I got that term in the denominator.)
And we can repeat this process to get better and better guesses:
$$x_{i+1} = x_i - (x_i\log x_i - \log y)/(\log x_i + 1).$$
We can try this with $y=1000$. Since $x\log x \approx x$, we will pick $x_0 = \log y = 6.9$ as our initial guess. Plugging into the formula repeatedly I get
$$\begin{array}{lc}x_0 & 6.9\\x_1 & 4.7101\\x_2 & 4.55654\\ x_3 & 4.55554 \\ x_4 & 4.55554\end{array}$$
The numbers have stopped changing, so the method has finished: I check
$$4.55554^{4.55554} = 1000.01$$
so I have a pretty good estimate of the right value of $x$.
Interestingly, this same approach was used to calculate $\sqrt{x}$ by the ancient Babylonians, before calculators (and before Newton invented the general method).
A: Since the title of this question is "Solving $y^y=x$ for large $x$" let us do just this.
We do not resort to using the properties of special functions.
We begin by rewriting the relationship between $y$ and $x$:
\begin{align*}
y^y &= x \\
y\log y &= \log x \\
\log y + \log\log y &= \log\log x.
\end{align*}
Note that for $y\gg 1$ we have that $\log y\gg \log \log y$.
Thus, for large $x$, $\log y \approx \log\log x$ and so $y = \log x$ should solve this problem approximately.
We now obtain successively better approximate solutions by solving
$$y=\frac{\log x}{\log y}$$
by recursive approximation:
\begin{align*}
y_0 &= \log x \\
y_1 &= \frac{\log x}{\log y_0} = \frac{\log x}{\log\log x} \\
y_2 &= \frac{\log x}{\log y_1} 
= \frac{\log x}{\log \frac{\log x}{\log\log x}} 
= \frac{\log x}{\log \log x - \log \log \log x} \\
&\cdots 
\end{align*}
The result for $y_2$ above is equivalent to that obtained by applying the well-known asymptotic expansion for the Lambert $W$ function,
$W(x)\approx \log x-\log \log x$,
to the solution found by @TravisWillse.

Figure 1. Plot of $y$ (in red, using the Lambert $W$ function) and $y_k$ for $k=0,1,\ldots,6$ (in gray, darker shades corresponding to larger values of $k$) for $x\in[0,10^6]$.
Addendum: Recursive asymptotic approximations for the Lambert $W$ function
The technique above can be used to find recursive asymptotic approximations for the Lambert $W$ function.
We have
\begin{align*}
W(x) e^{W(x)} &= x \\
W(x) &= \log x-\log W(x).
\end{align*}
We can see that $W(x)\approx \log x$ for large $x$.
(Corrections will be of order $\log\log x$.)
Thus,
\begin{align*}
W_0(x) &= \log x \\
W_1(x) &= \log x-\log W_0(x) \\
&= \log x-\log\log x \\
W_2(x) &= \log x-\log W_1(x) \\
&= \log x-\log(\log x-\log\log x) \\
&\cdots 
\end{align*}

Figure 2. Plot of $W(x)$ (in red) and $W_k$ for $k=0,1,2$ (in gray, darker shades corresponding to larger values of $k$) for $x\in[0,10^6]$.
The graph of $W(x)$ and $W_2(x)$ are almost indistinguishable on this scale.
