Coefficients of series Suppose that i have a function $f(x)=\sum_{i=0}^{\infty}a_ix^i$ with radius of convergence $r_f>0$ and that i want to write $f$ in a form $f(x)={e^{g(x)}}$ where $e$ is natural logarithm base and $g(x)=\sum_{i=0}^{\infty}b_ix^i$.
Also, suppose that inside the domain of convergence of $f$ there exists open set $D$ such that on $D$ we have $f(x\in D)>0$ (we need this for $g$ to exist because $e^x>0, \forall x \in\mathbb R$).
Now, i want to express coefficients $b_i$ as a function of the coefficients $a_0,a_1,...,a_n,...$ or, in other direction, to express $a_i$ as a function of $b_0,b_1,...,b_n,...$ when these are known and this pisses me off because i cannot find a way to do that, is there someone who know how to solve this problem(even if it boils down to a system of infinite number of equations with infinite number of unknowns, at least something is known about such systems)?
 A: Let $\alpha_k = a_k/a_0$ for $k \ge 1$, we have
$$b_0 + \sum_{k=1}^{\infty}b_k x^k = g(x) = \log f(x) = \log a_0 + \log\left(1 + \sum_{k=1}^{\infty} \alpha_k x^k \right)$$
It is clear $b_0 = \log a_0$. Now differential with $x$ on both sides, we get:
$$\sum_{k=1}k b_k x^{k-1} = \frac{\sum_{k=1}^{\infty} k \alpha_k x^{k-1}}{1 + \sum_{k=1}^{\infty} \alpha_k x^k}
\;\;\iff\;\;\left(\sum_{k=1}k b_k x^{k-1}\right)\left(1 + \sum_{k=1}^{\infty} \alpha_k x^k\right) = 
\sum_{k=1}^{\infty} k \alpha_k x^{k-1}$$
Comparing coefficients of $x^{n-1}$, we obtain a recurrence relation of $b_n$ for $n \ge 1$.
$$n b_n + (n-1)b_{n-1}\alpha_1 + (n-2)b_{n-2}\alpha_2 +\cdots+ b_1 \alpha_{n-1} = n \alpha_n$$
One can use these two equivalent set of relations to compute $\alpha_k$ from $b_k$ or
$b_k$ from $\alpha_k$:
$$
     \alpha_n = b_n + \frac{1}{n}\sum_{k=1}^{n-1} k b_k \alpha_{n-k}
\quad\iff\quad b_n = \alpha_n - \frac{1}{n}\sum_{k=1}^{n-1} k b_k \alpha_{n-k}$$
When one expand these out, one can express $\alpha_n$ as a sum of products of $b_1, b_2, \ldots, b_n$ over the set of representation of $n$ as a sum of positive integers $\le n$. Similarly, we can expand $b_n$ as a sum of products of $\alpha_1, \alpha_2, \ldots, \alpha_n$.
What this mean is foreach n-tuple $(e_1,e_2,\ldots,e_n) \in \mathbb{N}^n$ such that $\sum_{k=1}^n k e_k = n$, one can assign two numbers $\lambda_{e_1e_2\ldots e_n}$ 
and $\mu_{e_1e_2\ldots e_n}$independent of choice of $\alpha_k$ and $b_n$ such that:
$$
\alpha_n = \sum_{\sum_{k=1}^{n} ke_k = n } \lambda_{e_1e_2\ldots e_n} \prod_{k=1}^n b_k^{e_k}\quad\text{ and }\quad
b_n = \sum_{\sum_{k=1}^{n} ke_k = n } \mu_{e_1e_2\ldots e_n} \prod_{k=1}^n \alpha_k^{e_k}
$$
If I'm not mistaken, 
$$\begin{align}
\lambda_{e_1e_2\ldots e_n} = \frac{1}{\prod_{k=1}^n e_k!}
\quad & \implies\quad 
\alpha_n = \sum_{\sum_{k=1}^{n} ke_k = n }\left( \prod_{k=1}^n \frac{b_k^{e_k}}{e_k!}\right)\\
\\
\mu_{e_1e_2\ldots e_n} = \frac{(-1)^{m-1} (m-1)!}{\prod_{k=1}^n e_k!}
\quad & \implies\quad
b_n    = \sum_{\sum_{k=1}^{n} ke_k = n}
\left(
(-1)^{m-1}(m-1)!
\prod_{k=1}^n \frac{\alpha_k^{e_k}}{e_k!}\right)
\end{align}$$
where $m$ is the short hand for $\sum_{k=1}^n e_k$.
A: The Maple commands f := sum(a[i]*x^i, i = 0 .. infinity):
series(log(f), x, 5);
produce 
ln(a[0])+a[1]*x/a[0]+(a[2]/a[0]-(1/2)*a[1]^2/a[0]^2)*x^2+(- 
(1/3)*a[1]*a[2]/a[0]^2+a[3]/a[0]-(1/3)*(2*a[0]*a[2]-
a[1]^2)*a[1]/a[0]^3)*x^3+(-(1/4)*a[1]*a[3]/a[0]^2+a[4]/a[0]-(1/4)*
(2*a[0]*a[2]-a[1]^2)*a[2]/a[0]^3-(1/4)*
(3*a[0]^2*a[3]-3*a[0]*a[1]*a[2]+a[1]^3)*a[1]/a[0]^4)*x^4+O(x^5). 

I don't see any general formula for expressing $b_i$ through $a_i$ (in your notation).
