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How is it possible to find the limiting value of Log[x Log[x Log[x Log[...]]]] in Mathematica? Apparently, it should give -ProductLog[-1,-1/x], but I can't seem to replicate that ...

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    $\begingroup$ If u==Log[x Log[x ...]] then u==Log[x u] and FullSimplify[ Reduce[u==Log[x u],u]] will show you the range of possibilities depending on all the values that x might have. $\endgroup$
    – Bill
    Dec 2 '14 at 22:40
  • $\begingroup$ Related: (66085) $\endgroup$
    – Mr.Wizard
    Dec 2 '14 at 22:48
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    $\begingroup$ Letting y=Log[x*Log[x*...]] gives Exp[y]=x*Log[x*...]=x*y. Now solve for y. $\endgroup$ Dec 2 '14 at 23:00
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You can use FixedPoint:

FixedPoint[Log[3.5 #] &, 3.5] == -ProductLog[-1, -1/3.5]
True

In general, if $y = f(f(...f(f(...))))$, then $y = f(y)$. Solving for $y$ will give us the formula for the infinitely nested expression.

In your case, f == Log[x #]&, which gives

sol = Refine[Reduce[y == Log[x y], y], x > 0]

enter image description here

Unfortunately FullSimplify can't prove Im[ProductLog[-1, -(1/x)]] >= -π for x > 0. We can extract the possible solutions for y manually:

Or @@ Cases[sol, y == f_ :> y == FullSimplify[f], Infinity]
y == -ProductLog[-1, -1/x] || y == -ProductLog[-1/x] || y == -ProductLog[1, -1/x]
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  • $\begingroup$ great solution - thank you! :) $\endgroup$
    – martin
    Dec 3 '14 at 9:20

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