Numerical approximation, or is there a better way? $x=1+\ln(1+\ln(1+\ln(x)))$

Question

Don't have enough reputation to post on this but is there any way of calculating the following?

$x=1+\ln(1+\ln(1+\ln(x)))$

Answer 1

Consider $$f(x)=x-(1+\ln(1+\ln(1+\ln(x)))).$$ It is

$$f'(x)=1-\frac{1}{1+\ln(1+\ln(x))}\cdot \frac{1}{1+\ln (x)}\cdot \frac 1x=\frac{(1+\ln(1+\ln(x)))(1+\ln x)x-1}{(1+\ln(1+\ln(x)))(1+\ln x)x}.$$

If $x\in(e^{1/e-1},1)$ then $f'(x)<0$ (that is, $f$ is strictly decreasing in $(e^{1/e-1},1)$) and if $x\in(1,\infty)$ it is $f'(x)>0$ (that is, $f$ is strictly increasing in $(1,\infty)).$ Since $f(1)=0$ we can conclude that $x=1$ is the only solution of the equation.