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How to find a continuous function $f(x)$ for $x > 12$, such that $f(f(x))=\ln(x)$? Preferably analytic too.

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    Since x>\exp(\dots\exp(1)\dots) with $n-1$ exponents, we get $f_n\le (\ln(\ln x))^a$. Sorry, no more time now.2012-01-09

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