$\log_j(\log_j(\log_j(x)))=\log(x);\ \ j=?$ $\log_j(\log_j(x))=\log(x)$ has solution $j=x^{\exp-W(\log^2(x))}$ for real $x\neq0$, where $W=$ Lambert W function.
But what is the solution to $\log_j(\log_j(\log_j(x)))=\log(x)$? Mathematica can't do it - can it be done?

We can make things easier on ourselves defining $a = \ln j$ and $y = \ln x$. First let's look at the two-log case:
$$
\log_j(\log_j(x)) = \frac{1}{a}\ln\left(\frac{y}{a}\right) = \frac{\ln y - \ln a}{a} = y\Longrightarrow ay +\ln(ay) = 2\ln y
$$
The key here is that we have found two one-variable functions $f$ and $g$ such that $f(h(a,y)) = g(y)$ for some function $h$. $a$ is now isolated, and can be solved as $a = h^{-1}_a(f^{-1}(g(y)),y)$. Ideally we can choose $f$ and $g$ such that $f$ and $h$ have closed forms (and all the better if $f^{-1}$ and $h^{-1}_a$ do too), but that's not always possible.
Now let's apply this to the three-log case. We get
$$
\log_j(\log_j(\log_j x)) = \frac{1}{a}\ln\left[\frac{1}{a}\ln\left(\frac{y}{a} \right)\right] = \frac{\ln\ln(y/a)-\ln a}{a} = y\\\Longrightarrow \ln\ln(y/a)= ay + \ln a
$$
That double logarithm is going to make it very hard to find a way to isolate $a$ in closed form (or $y$, for that matter). Not saying its impossible, but it's pretty unlikely.
That said, we can get some info about $a(y)$ (and hence $j(x)$) from looking at this plot by WolframAlpha:

*

*The smallest value $x$ can take and have a solution is about $0.744$, with $j\approx 0.506$

*The largest value $j$ can take and have a solution is about $1.663$, with $x\approx5.27$.

*For $x$ values between the minimum value and $1$, there are two solutions. For $j$ values between the maximum value and $1$, there are also two solutions. There are no solutions with $j < 1 < x$ or $x < 1 < j$.

*$a(y)$ looks qualitatively like $W(y)$ for small $y$. However, for large $y$, $W(y)$ is unbounded, whereas it appears $a\rightarrow 0$ as $y\rightarrow \infty$
Anyone bold enough to try to find the exact value of those extremal points?
