Many real functions support a $\log$-series representation along with their usual power series representation. An example is the canonical
$$ x = 1 + \frac{1}{1!}\ln(x) + \frac{1}{2!}\ln(x)^2 + \frac{1}{3!}\ln(x)^3 + ... $$
To find the $\log$-series one has two strategies effectively but (the question being asked here) sometimes they conflict and it's not clear why.
The substitution and then taylor method...
You start with some $f(x)$ that you want a log expansion for. You then consider the function $f(e^u)$ and compute its taylor series centered at 0 the normal way in terms of powers of $u$, you then make the substitution $u = \ln(x)$ to get back where you started. Now sometimes you might encounter divergent sums as a result, if that happens, you can do some totally non rigorous ramanujan summation / other renormalization and still get away with it.
Example: to find the log series $x$ we consider $e^u$ and then look at it's taylor series
If $f(x) = x$ $$ e^u = 1 + u + \frac{1}{2!}u^2 + \frac{1}{3!}u^3 + ... $$ $$ x = 1 + \log(x) + \frac{1}{2!}\log(x)^2 + ... $$
If $f(x) = \frac{1}{1-x}$ $$ \frac{1}{1-e^u} = -\frac{1}{u} + \frac{1}{2} - \frac{1}{12}u + \frac{1}{720}u^3 ... $$
$$ \frac{1}{1-x} = -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 ... $$
The taylor then substitution method...
You can also go the other way. You start with a power series, replace each instance of $x^n$ with $e^{n \ln(x)}$ and now you expand these things into a massive power series and collect terms with common powers of $\ln(x)$. As usual if you encounter divergent series you ramanujan sum them (as long as they respect indices which is a handwavy idea i vaguely understand that probably has a rigorous interpretation)
if $ f(x) = x $
$$ f(x) = e^{\ln(x)} = 1 + \frac{1}{1!} \ln(x) + \frac{1}{2!}\ln(x)^2 + ... $$
if $ f(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 ... $
$$ f(x) = 1 + e^{\ln(x)} + e^{2 \ln(x)} + e^{3 \ln(x)} + ... $$
$$ f(x) = 1 + \frac{1}{0!} \left(1 + 1 + 1 + ... \right) + \frac{1}{1!} \left( 1 + 2 + 3 + .... \right) \ln(x) + \frac{1}{2!} \left(1^2 + 2^2 + 3^2 + ... \right)\ln(x)^2 + ... $$
Now if you know anything about divergent sums then you realize this is saying
$$ f(x) = 1 - \frac{1}{2} - \frac{1}{12} \ln(x) ... $$
But obviously
$$ \frac{1}{2} - \frac{1}{12} \ln(x) + \frac{1}{720}\ln(x)^3 ... \ne -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 .. $$
Theres a missing laurent term of $\frac{1}{\ln(x)}$. Why did that happen?
Inspired by some weirdness found in my answer here: Why does $1+2+3+\cdots = -\frac{1}{12}$?