If ln is given paricular times can we find least value for which it is defined?

Question

I was just doing some time-pass with my calculator but then I observe something.I don't know whether it is senseful to ask.So here's my question. ln ln (1) is not defined but for all values greater than 1 it is defined.So then I try to find values for which ln ln ln(x) is defined,then I get to know that it get's defined from 2.72.If ln is taken 4 times it's start giving values from 15.2.So my question is if ln is given particular times how I can come to know the infimum of values for which it is defined?

Answer 1

$\ln (x)$ is defined for $x>0$

$\ln (\color{blue}{\ln (x)})$ will be defined for $\color{blue}{\ln (x)}>0 \implies x >1$

$\ln (\color{blue}{\ln ( \ln (x))})$ is defined for $\color{blue}{\ln ( \ln (x))}>0 \implies \color{blue}{ \ln (x)}>1 \implies x>e$

You see the pattern now?

$$0, e^0, e^1, e^e, e^{e^e} \ldots$$

Answer 2

Denote the minimum value for which $\underbrace{\ln\ln...\ln}_{n \text{ times}} (x)$ is defined to be $y_n$. Since $\ln x$ is defined for $x>0$, this means that $$\underbrace{\ln\ln...\ln}_{n \text{ times}} (y_n)=\ln 0\implies\underbrace{\ln\ln...\ln}_{n-1 \text{ times}} (y_n)=0\\\implies\underbrace{\ln\ln...\ln}_{n-2 \text{ times}} (y_n)=1\\\implies\underbrace{\ln\ln...\ln}_{n-3 \text{ times}} (y_n)=e\\...\\\implies y_n=\underbrace{e^{e^{.^{.^{.^{e}}}}}}_{n-2\text{ times}}$$ for $n>2$. (And $y_1=0, y_2=1$)