How are $x$ and $\ln(\ln(x))$ related?

Question

It is known that the graphs of the functions $e^x$ and $\ln(x)$ are reflections of each other about the line $y = x$. Are the functions $e^{\ln(x)}$ i.e. $x$ and $\ln(\ln(x))$ also symmetric in some similar way? Are they reflections of each other about some nice function(of course not $y = x$).

Answer 1

Let $f_1(x) = \ln(x)$ and $f_2(x) = e^x$. Then $f_1(f_2(x)) = x$ and $f_2(f_1(x))$. This relationship is the "symmetry" (intentionally in quotes) that you see between $\ln(\ln(x))$ and $e^{\ln(x)}$.

For $e^{\ln(x)}$ and $\ln(\ln(x))$, you have simply replaced $x$ with $\ln(x)$.

So, in a certain sense, $\ln(\ln(x))$ is "symmetric" to $e^{\ln(x)}$ across $\ln(x)$.

A plot shows this "symmetry":

The middle green plot is $y = \ln(x)$.

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This same "symmetry" applies to any substitution in $x$, for example:

$f(x) = \sin(x)+2.$

Here is the graph:

You may notice that $\ln(f(x))$ lies below $f(x)$ and $e^{f(x)}$ lies above $f(x)$: this is obviously apparent from the fact that $\ln(x) < x < e^x$ for all $x$ (where $\ln(x)$ is defined, of course).

If you want to play around a bit more with these graphs with different $f$s, here is a desmos calculator with everything set up, just change $f(x)$.