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.

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.
