Logarithm Series: Symbol Manipulation Proof that $\log(x) + \log(y) = \log (xy)$ Let $R$ be a ring with 1. Define a formal power series $$\log(x)=\sum_{m=1}^\infty (-1)^{m+1}\frac{(x-1)^m}{m}.$$ I would like to show using only manipulations of the power series (pretending we know nothing of exp) that for commuting $x$,$y$, we have $\log(xy)=\log(x)+\log(y)$.
(1) For sanity, this is true, yes?
(2) Assuming it's true, is a symbol manipulation proof reasonably tractable? If so, would anyone be kind enough to reproduce it here?
Thanks.
 A: This can be done formally. For details, and more, there is a nice classical reference on formal power series:

Ivan Niven. Formal Power Series. The American Mathematical Monthly, 76 (8), (1969), 871-889.

Here is how Niven proceeds: 
We verify that $\log(1+t)+\log(1+s)=\log(1+(t+s+ts))$.
For this, let $D$ be the formal derivative operator, so $D(a+bx+cx^2+\dots)=b+2cx+\dots$ and we have $D(f^n)=nf^{n-1}D(f)$ and $D(fg)=fD(g)+gD(f)$ for commuting $f,g$, etc., and $D(c)=0$ iff $c$ is a scalar (a "constant").
Using $D$, one shows by direct manipulation of series:


*

*$D(\log(1+t))=(1+t)^{-1}D(t)$.

*$D(\log((1+t)(1+s)))=D(\log(1+t))+D(\log(1+s))$.


And, from these, the result follows.
A: Use formal differentiation to show that 
$$
\frac{d}{dx}\log (xy)=\frac{d}{dx}\log x.
$$
These are easy computations which involve Neumann series, provided they make sense. Which is true in your case, I believe, since you deal with a subgroup of $GL_n(\mathbb{R})$.
This shows that the $x$ power series $\log(xy)$ and $\log x$ are equal modulo a constant. Then $y=1$ yields the constant.
