So i have been thinking about the value of $\ln{(x\ln{(x\ln{(x…)})})}$ for different value of x, and my solution is as follows: $$\ln{(x\ln{(x\ln{(x…)})})}$$ For $x=e$ $$=\ln{(e\ln{(e\ln{(e…)})})}$$ $$=1$$ Since $\ln{(e)}=1$,

$\ln{(a)}>1$ and $\ln{(b)}<1$ where $0<b<e<a$

Therefore, for $x>e$: $$\ln{(x\ln{(x\ln{(x…)})})}$$ $$=\ln{(x\ln{(xa_1)})}$$ $$=\ln{(xa_2)}$$ $$=a_3$$ where $a_n$ are some constants that are larger than $e$

Since $\ln{(x)}\to\infty$ as $x\to\infty$, $a_3$ doesn’t exist and diverges to $\infty$

And for $x<e$ $$\ln{(x\ln{(x\ln{(x…)})})}$$ $$=\ln{(x\ln{(xb_1)})}$$ $$=\ln{(xb_2)}$$ $$=b_3$$

where $b_n$ are some constants that are smaller than $e$

But as $x\to-\infty$, I have no idea what $\ln{(x)}$ approaches…

So $$\ln{(x\ln{(x\ln{(x…)})})} =\begin{cases} ?? & x<e \\ 1 & x=e \\ \infty & x>e \end{cases}$$

And I have 3 questions

  1. Is my solution correct?
  2. What is the expression when $x<e$?
  3. What if $x$ is complex? Does it approaches any value?

It will be appreciated if anyone answers. :)

Edit: the implicit form is $y=\ln{(xy)}$ and I am an absolute idiot who did not notice that.

  • 3
    $\begingroup$ The implicit form is $y=\ln(xy)$ $\endgroup$ May 19, 2023 at 11:56
  • $\begingroup$ Oh i didn’t noticed that… it would have been much easier $\endgroup$
    – YesSpoon3
    May 19, 2023 at 11:57
  • $\begingroup$ That means if the limit $y$ exists, then it satisfies $y = \ln(xy)$. But the existence of a solution to that equation doesn't necessarily mean the sequence $\ln a_n$ or $\ln b_n$ actually approaches $y$. You know $a_n$ is an increasing sequence, but that doesn't necessarily mean it approaches $+\infty$. $\endgroup$
    – aschepler
    May 19, 2023 at 12:06
  • $\begingroup$ The real $\ln$ function is not defined on negative numbers, so if any $b_n<0$, the expression does not have a defined value. For the complex $\ln$ function, you'll need to choose a branch. Usually if not otherwise specified, we use the branch $-\pi < \operatorname{Im}(\ln z) \leq \pi$. $\endgroup$
    – aschepler
    May 19, 2023 at 12:09
  • 1
    $\begingroup$ The starting value matters , so ... is not justified. Log of negative numbers have branches. Im not sure when this actually converges, fixpoints exists but probably cycles too ? $\endgroup$
    – mick
    May 19, 2023 at 17:57

2 Answers 2



We can rewrite the function $f\left( x \right) = \ln\left( x \cdot \ln\left( x \cdot \ln\left( x \cdots \right) \right) \right)$ to something like $f\left( x \right) = \left( g \circ g \circ \cdots \circ g \right)\left( x\right)$ and wrap this equation in a recursive relation of the form: $$ \begin{align*} a_{n + 1} &= \ln\left( x \cdot a_{n} \right)\\ \end{align*} $$

What you are looking for are the $\lim\limits_{n \to \infty}\left[ a_{n} \right]$. Let's ask ourselves what a $\lim\limits_{n \to \infty}\left[ a_{n} \right]$ contextual would mean: If we put in a number $a_{n}$, we get a number $a_{n + 1}$. If $\lim\limits_{n \to \infty}\left[ a_{n} \right]$ is to exist, it must be self-explanatory aka $f\left( \lim\limits_{n \to \infty}\left[ a_{n} \right] \right) = \lim\limits_{n \to \infty}\left[ a_{n} \right]$. So we can define $\lim\limits_{n \to \infty}\left[ a_{n} \right] \equiv z$ and solve for $z$: $$ \begin{align*} a_{n + 1} &= \ln\left( x \cdot a_{n} \right)\\ z &= \ln\left( x \cdot z \right) \quad\mid\quad \exp\left( \cdot \right)\\ \exp\left( z \right) &= \exp\left( \ln\left( x \cdot z \right) \right)\\ \exp\left( z \right) &= x \cdot z\\ \exp\left( z \right) &= x \cdot z \quad\mid\quad \cdot\exp\left( -z \right)\\ \exp\left( z \right) \cdot\exp\left( -z \right) &= x \cdot z \cdot\exp\left( -z \right)\\ 1 &= x \cdot z \cdot\exp\left( -z \right) \quad\mid\quad \div x\\ \left( -1 \right) \cdot \frac{1}{x} &= \left( -1 \right) \cdot z \cdot\exp\left( -z \right) \quad\mid\quad \cdot \left( -1 \right)\\ -\frac{1}{x} &= -z \cdot\exp\left( -z \right) \quad\mid\quad W_{k}\left( \cdot \right)\\ W_{k}\left( -\frac{1}{x} \right) &= W_{k}\left( -z_{k} \cdot\exp\left( -z_{k} \right) \right)\\ W_{k}\left( -\frac{1}{x} \right) &= -z_{k} \quad\mid\quad \cdot \left( -1 \right)\\ \left( -1 \right) \cdot W_{k}\left( -\frac{1}{x} \right) &= -\left( -1 \right) \cdot z_{k}\\ -W_{k}\left( -\frac{1}{x} \right) &= z_{k}\\ \end{align*} $$ where $W_{k}$ is the Lambert W-Function and $k \in \mathbb{Z}$.

$$\fbox{$ \lim\limits_{n \to \infty}\left[ a_{n} \right] = -W_{k}\left( -\frac{1}{x} \right) $}$$

Similar to the complex logarithm, the Lambert W-Function has infinitely many branches $k$ and thus infinitely many different possible complex outcomes.

Plot of $x\left( f \right)$ the branches $1$ and $2$:

enter image description here

Here you can also read directly which values ​​could come out when inserting special $x$ into the equation.

Special Cases

If $x = e$: $$ \begin{align*} f\left( e \right) &= -W_{k}\left( -\frac{1}{e} \right)\\ f\left( e \right) &= -W_{k}\left( -e^{-1} \right)\\ f\left( e \right) &= -W_{k}\left( -1 \cdot e^{-1} \right)\\ f\left( e \right) &= -\left( -1 \right)\\ f\left( e \right) &= 1\\ \end{align*} $$

If $x \to -\infty$: $$ \begin{align*} f\left( -\infty \right) = \lim\limits_{x \to -\infty}\left[ -W_{k}\left( -\frac{1}{x} \right) \right]\\ f\left( -\infty \right) = -W_{k}\left( -\lim\limits_{x \to -\infty}\left[ \frac{1}{x} \right] \right)\\ f\left( -\infty \right) = -W_{k}\left( 0 \right)\\ f\left( -\infty \right) = -W_{0}\left( 0 \right) \wedge -W_{-1}\left( 0 \right)\\ f\left( -\infty \right) = -0 \wedge -\left( -\infty \right)\\ f\left( -\infty \right) = -0 \wedge \infty\\ \end{align*} $$

If $x \to 0$: $$ \begin{align*} f\left( 0 \right) &= \lim\limits_{x \to 0}\left[ -W_{k}\left( -\frac{1}{x} \right) \right]\\ f\left( 0 \right) &= -W_{k}\left( -\lim\limits_{x \to 0}\left[ \frac{1}{x} \right] \right)\\ f\left( 0 \right) &= -W_{k}\left( \infty \cdot e^{\arg\left( \hat{\infty} \right) \cdot i} \right)\\ \end{align*} $$

If $x = \pi^{-1} \cdot i$: $$ \begin{align*} f\left( \pi^{-1} \cdot i \right) &= -W_{k}\left( -\frac{1}{\pi^{-1} \cdot i} \right)\\ f\left( \pi^{-1} \cdot i \right) &= -W_{k}\left( -\frac{i}{\pi^{-1} \cdot i^{2}} \right)\\ f\left( \pi^{-1} \cdot i \right) &= -W_{k}\left( -\frac{i}{-\pi^{-1}} \right)\\ f\left( \pi^{-1} \cdot i \right) &= -W_{k}\left( \frac{i}{\pi^{-1}} \right)\\ f\left( \pi^{-1} \cdot i \right) &= -W_{k}\left( \pi \cdot i \right)\\ f\left( \pi^{-1} \cdot i \right) &= -W_{-1}\left( \pi \cdot i \right)\\ f\left( \pi^{-1} \cdot i \right) &= -\left( -\pi \cdot i \right)\\ f\left( \pi^{-1} \cdot i \right) &= \pi \cdot i\\ \end{align*} $$

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    $\begingroup$ Thank your very much! I just want to ask if you mean if $x\to0$, $f(x)$ is undefined, and if $x\to-\infty$, $f(x)=0$ $\endgroup$
    – YesSpoon3
    May 20, 2023 at 9:16
  • $\begingroup$ @YesSpoon3: Yes: if $x \to -\infty{:}~ f(x) = 0$. And $x \to -\infty{:}~ f(x)$ does not converge to any number defined in complex numbers. However, if we extend the range of complex numbers to include complex infinity ($\hat{\infty}$), then it would already be defined to a certain extent... $\endgroup$ May 20, 2023 at 12:02
  • $\begingroup$ What do you mean by “certain extent”? I have always heard of complex infinity, but i do not know how do define it. Thx $\endgroup$
    – YesSpoon3
    Jun 10, 2023 at 11:56

You have actually defined a recurrence relation, namely $a_{n+1} = \ln(xa_n)$. If it converges, then its limit will be a fixed point of the function $f(t) = \ln(xt)$, hence the equation $t = f(t) = \ln(xt)$, which can be rewritten as $-te^{-t} = -1/x$ and is solved by $t = -W(-1/x)$, where $W$ is the Lambert $W$ function $-$ note that this function possesses an infinite number of branches (like the complex logarithm), hence an infinite number of solutions in the complex plane. In the case $x = e$, we have $a_\infty = -W(-1/e) = 1$, as you have already found.

  • $\begingroup$ It would be clearer to say "the $x=e$" case instead of "the present case". The OP question considers different (real) values of $x$. $\endgroup$
    – aschepler
    May 19, 2023 at 12:14
  • $\begingroup$ @aschepler Done. $\endgroup$
    – Abezhiko
    May 19, 2023 at 13:45

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