Inflection points with natural logs. Where are the inflection points, if any exist, for $f(x)=10\frac{\ln(\ln(x))}{\ln(x)}$?
I don't think any exist. But could use a little help. My guess would be that with the second derivative, if $x = e^{-2}$ you get 0, but this is not in the domain of the function. Is this correct, or completely off? Any help would be appreciated.
 A: Inflection point must exist.
Indeed we have a maximum at $x=e^e$ and $f(x)\to 0$ as $x\to +\infty$, therefore concavity must change because $f(x)$ has no intersection with $x$-axis after $x=e$.
We have $$f''(x)=10\,\frac{ (\log x+2) \log (\log x)-\log x-3}{x^2 \log ^3x}$$
$f''(x)=0$ if $$(\log x+2) \log (\log x)-\log x-3=0$$
substitute $\log x=u$
$$(u+2)\log u-u-3=0\to u=3.28455$$
Finally we have $x=e^{3.28455}\approx 26.7$
the graph below shows the curve $y=f(x)$, the inflection point and the tangent

$$...$$

A: First $e^{-2}$ doesn't belong to the domain of $f$.
Using Geogebra, you can be convinced that the second derivative does cancel.
$$f''(x)=10\frac{\ln x\ln(\ln x))-\ln x+2\ln(\ln x)-3}{x^2\ln^3x}$$
The numerator can be studied, it's a strictly increasing function with limits $-\infty$ and $+\infty$ at the extremities of the domain. So it has to cancel once, and there you find an inflexion point for the curve of your function.
A: You can check that the solution to $f''(x)=0$ occurs at $e^{e^x}$, where $x \in \mathbb{R}$ is the (only) solution to
$$xe^x-e^x+2x-3=0$$
We may rewrite the equation above as $(x-1)(e^x-2)=1$, but I'm not sure how much better this is actually.
I see no reason to assume that $x$ has a 'nice' symbolical representation
