Notation for the solution to $y^y=x$.

Question

Let $x$ be a positive real number.

• The solution to $y+y=x$ is written $$y=x/2.$$
• The solution to $y\cdot y=x$ is written $$y=x^{1/2}.$$
• Is there a notation for the solution to $y^y=x$?

Answer 1

There's a well known function called the Lambert W function, defined to be the inverse of $xe^x$. If $y^y = x$, then $$\ln(y)e^{\ln(y)} = y\ln(y)=\ln(x) \implies$$ $$\ln(y) = W(\ln(x)) \implies$$ $$y = e^{W(\ln(x))}$$ I don't know of any function simply defined to be the inverse of $x^x$, though, but problems like this can often be solved with the W function.

Note: $xe^x$ and $x^x$ aren't injective on $(0,\infty)$, so you have to be careful about the possibilities of multiple solutions to these equations.

Answer 2

Take logs: $$y\ln y=\ln x=\ln y\cdot e^{\ln y}$$ The definition of the Lambert W function satisfies $$W(\ln x)e^{W(\ln x)}=\ln x$$ Therefore $$W(\ln x)=\ln y$$ $$y=e^{W(\ln x)}=\frac{\ln x}{W(\ln x)}$$ Any expression that relates an exponential of a variable with a linear function of the same variable is amenable to solution via Lambert W. Indeed, such expressions pop up in many physics equations, especially those relating to quantum mechanics.

Answer 3

As other's have stated, the $W$ function yields the correct answer to the question you asked. A related sequence is this: $$2+y=x\implies y=x-2$$ $$2\cdot y=x\implies y=x/2$$ $$2^y=2\uparrow y=x\implies y=\log_2(x)$$ $$\underbrace{2^{2^{2^{\cdot^{\cdot^\cdot}}}}}_{y}= 2\uparrow \uparrow y=x\implies y=\text{slog}_2(x)$$ where $\text{slog}$ is a superlog: https://en.wikipedia.org/wiki/Super-logarithm