$\log_j(\log_j(\log_j(x)))=\log(x);\ \ j=?$

Question

$\log_j(\log_j(x))=\log(x)$ 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?

Answer 1

I presume that the only possible solutions corresponds to $j=\frac 1k$ to make the equation $$\frac{\log \left(-\frac{\log \left(-\frac{\log (x)}{\log (k)}\right)}{\log (k)}\right)}{\log (k)}+\log (x)=0$$ Numerically solving for $x$ for a given $k$, the following (not very exciting) results $$\left( \begin{array}{cc} k & x_k \\ 2 & 0.743810 \\ 3 & 0.773972 \\ 4 & 0.806977 \\ 5 & 0.831510 \\ 6 & 0.849777 \\ 7 & 0.863823 \\ 8 & 0.874964 \\ 9 & 0.884035 \\ 10 & 0.891583 \\ 20 & 0.930106 \\ 30 & 0.945785 \\ 40 & 0.954705 \\ 50 & 0.960604 \\ 60 & 0.964859 \\ 70 & 0.968106 \\ 80 & 0.970684 \\ 90 & 0.972792 \\ 100 & 0.974555 \\ 1000 & 0.994683 \end{array} \right)$$

Answer 2

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?