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I noticed that $\ln(x)-\ln(\ln(x))$ seems to by asymptotic with the Lambert W Function. Is this true?

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By definition:

$$x=W(x)e^{W(x)}$$

It thus follows that

$$e^{W(x)}=\frac x{W(x)}$$

$$W(x)=\ln\left(\frac x{W(x)}\right)$$

By applying the fixed point method, we find that for any initial condition greater than $e$ and $x>e$, we have

$$W(x)=\ln\left(\frac x{W(x)}\right)=\ln\left(\frac x{\ln\left(\frac x{W(x)}\right)}\right)=\dots=\ln\left(\frac x{\ln\left(\frac x{\ln\left(\frac x{\vdots}\right)}\right)}\right)$$

And by expanding with log rules with some obvious bounds on $W(x)$, we find that

$$W(x)\sim\ln(x)-\ln(\ln(x)-\ln(\ln(x)-\ln(\dots)))\sim\ln(x)-\ln(\ln(x))+o(1)$$

Here is a visualization of this with the dotted line being the Lambert W function:

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