LogCeption

The argument of $\ln(x)$ should be positive :

$\displaystyle => \ln(\ln(\ln(\ln(x)))) > 0$

$e^x$ is strictly increasing :

$\displaystyle => e^{\ln(\ln(\ln(\ln(x))))} > e^0$

$\displaystyle => \ln(\ln(\ln(x))) > 1$

Repeating the process:

$\displaystyle => e^{\ln(\ln(\ln(x)))} > e^1$

$\displaystyle => \ln(\ln(x)) > e$

Again:

$\displaystyle => ln(x) > e^e$

$\displaystyle =>\boxed{ x > e^{e^e} }$

I just reverse-engineered it starting from the limiting case:

$\begin{aligned} \ln{(0)} & = \ln{(\ln{(1)})} \\ & = \ln{(\ln{(\ln{(e)})})} \\ & = \ln{(\ln{(\ln{(\ln{(e^e)})})})} \\ & = \ln{(\ln{(\ln{(\ln{(\ln{(e^{e^e})})})})} )} \end{aligned}$ $\Rightarrow a = e^{e^e}$

The smallest value of x in $\ln(x)$ is \color \green{0} .

Thus \ln(\ln(x)), \ln(x)=0\Longrightarrow x=e^{\color \green{0}}=\color \red{1}

\ln(\ln(\ln(x)))\Longrightarrow \ln(\ln(x))=0 \Longrightarrow \ln(x)=1 \Longrightarrow x=e^{\color \red{1}}=e

Notice the pattern: e to the power of the answer to the previous line (shown by the colors). When there are 3 natural logs. We have $x=e$ . When there are 4, $x=e^e$

Thus when there are 5. Our answer will be e^e^e