# The value of $\ln{(x\ln{(x\ln{(x…)})})}$

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

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 $$x $$\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 $$\ln{(x\ln{(x\ln{(x…)})})} =\begin{cases} ?? & xe \end{cases}$$

And I have 3 questions

1. Is my solution correct?
2. What is the expression when $$x?
3. What if $$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.

• The implicit form is $y=\ln(xy)$ May 19, 2023 at 11:56
• Oh i didn’t noticed that… it would have been much easier May 19, 2023 at 11:57
• 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 $+\infty$. May 19, 2023 at 12:06
• 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 $-\pi < \operatorname{Im}(\ln z) \leq \pi$. May 19, 2023 at 12:09
• 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 ?
– mick
May 19, 2023 at 17:57

### General

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 $$2$$:

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*}

• Thank your very much! I just want to ask if you mean if $x\to0$, $f(x)$ is undefined, and if $x\to-\infty$, $f(x)=0$ May 20, 2023 at 9:16
• @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... May 20, 2023 at 12:02
• What do you mean by “certain extent”? I have always heard of complex infinity, but i do not know how do define it. Thx 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.

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