Deriving the additive property of natural log from power series? I know the additive property of logarithms, that
$$\ln(x) + \ln(y) = \ln(xy)$$
is easy to prove using the logarithm's nature as the inverse of the exponential function. However, I'm interested in trying to prove it just using the power series for $\ln(1-x)$, namely
$$\ln(1-x) = -\sum_{n=1}^\infty \frac{x^n} n . $$
I've used this definition to try and examine the expression 
$$\log(1-x) + \log(1-y) - \log((1-x)(1-y)) = -\sum_{n=1}^\infty \frac{1}{n} ( x^n + y^n - (x+y-xy)^n)$$
and show it is identically zero, but it has proved too much for me. 
Any suggestions, either on that method or any other approach? 
Thanks 
 A: Using the Taylor series directly is not a very motivated or clean way of showing the additivity of $\log$, but here's one approach. It's clearly sufficient to prove that $\exp(x + y) = (\exp x)(\exp y)$. The function $y =  \exp(x)$ is the unique solution of the differential equation $y' = y$ with $y(0) = 1$. (The uniqueness result here is standard and easy to show, though introductory differential equations courses are often just about finding explicit solutions to certain equations rather than anything more useful.) That's often taken as the definition $\exp$, or it's easy to show directly if one defines $\exp(x) = \sum x^n/n!$. But given a solution $y(x)$ of the equation, clearly $\tilde y(t) = y(x + t)/y(t)$ is also a solution for any fixed $t$ with $y(t)\not =0 $. It follows that $\tilde y(x) = y(x)$; that is, $y(x + t) = y(x)y(t)$. There's a similar argument using the definition $\log x = \int_1^x \frac{1}{t}$, which is equivalent to the one here after noting that $(f^{-1})'(f(x)) = 1/f'(x)$.
A: If I had to prove this result, I'd probably use a less direct way than what seems to be considered in the posted question, but of course that can't be the only way to do it. We have $$ \ln(1-x) = - \sum_{n=1}^\infty \frac{x^n}n. $$  Power series can be differentiated term-by-term within the interiors of their intervals of convergence, so we have
$$
\frac d {dx} \ln(1-x) = -\sum_{n=1}^\infty x^{n-1} = \frac 1 {x-1}.
$$
Then we have $$ \ln(1-x) = \int_0^x \frac{dw}{w-1} $$ (here we have to note that the values of both sides when $x=0$ agree: they're both $0$, so the "constant of integration" has the right value).  This can be written as
$$
\ln u = \int_0^{1-u} \frac{dw}{w-1} = \int_1^u \frac{dv} v
$$
via the substitution
\begin{align}
v & = 1-w, \\
-dv & = dw, \\
-v & = w-1, \\
\text{when } w & = 0 \text{ then }v=1, \\
\text{when } w & = 1-u \text{ then } w=u.
\end{align}
So
$$
\ln(tu) = \int_1^{tu} \frac{dv} v = \int_1^u \frac{dv} v + \int_u^{tu} \frac {dv} v = (\ln u) + \int_u^{tu} \frac{dv} v.
$$
Then deal with that last integral:
$$
\int_u^{tu} \frac{dv}v. \tag 1
$$
Let $y=uv$ so that $dy = u\,dv$ and $\dfrac{dv} v = \dfrac{u\,dv}{uv} = \dfrac{dv} v$. As $v$ goes from $u$ to $tu$, then $y$ goes from $1$ to $t$, and  the integral $(1)$ becomes
$$
\int_1^t \frac {dy} y = \ln t.
$$
However, if you want to do it more directly from working with the power series, maybe you'll have to do something quite different.
