Retrieving the Taylor series for $\log$ from its functional equation Consider the unique continuous function $\mathbb{R}^+\to\mathbb{R}$ such that:
$$f(xy)=f(x)+f(y),\qquad f(e)=1$$
where $\displaystyle e=1+\sum_{n=1}^{\infty} \frac{1}{n!}$.
Assuming $f$ has a Taylor series about $x=1$, can we show, using only this information, that: $$|x|<1:\quad f(1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}\,?$$ 
 A: Here is the proof:
In what follows let 
$$ \alpha = f'(1) $$
Clearly from 
$$ f(x) = f(x \times 1 ) = f(x) + f(1)$$
we can conclude that 
$$f(1)=0$$.
Now
$$
f(x+dx) = f(x~(1+dx/x)) = f(x) + f(1+dx/x)) = f(x) + f(1) + f'(1) dx/x + o(dx/x)$$
Hence 
$$ \lim_{dx\rightarrow 0} \frac{f(x+dx)-f(x)}{dx} = f'(1) \frac{1}{x} =\frac{\alpha}{x}$$
i.e.
$$ f'(x) = \frac{\alpha}{x}$$
Differentiating this repeatedly we can get second and higher derivatives and the Taylor series at $1$ gives
$$ f(1+x)=\alpha \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
This series on the right is the expansion for $\log(x)$ and hence
$$ f(x) = \alpha \log(x)$$
Finally, since $f(e) = 1$, $\alpha=1$.
Note: Since $\alpha$ is arbitrary, the function with $\alpha=1$ is considered special and $e$ is often defined as that $x$ for which $f(x)=e$. So, somehow, you have to bring in the $\log$ function.
A: Note that $f(1)=0$, since $f(1)=f(1^2)=f(1)+f(1)=2f(1)$. Also, $0=f(1)=f(t\times\frac1t)=f(t)+f(1/t)$, so $f(1/t)=-f(t)$.
From $f(e)=1$, we get 
$$
f(e^n)=nf(e)=n,\ f(e^{-n})=-n,\ \ mf(e^{1/m})=f(e)=1,
$$
so $f(e^{1/m})=1/m$ and then $f(e^{n/m})=n/m$. By continuity, $f(e^t)=t$ for all $t\in\mathbb R$. Differentiating this equality (we know that $f$ is differentiable at least on $(0,2)$, since it has a power series expansion),
$$\tag{1}
e^tf'(e^t)=1.
$$
Evaluating at $t=0$, we get $f'(1)=1$. 
We can rewrite $(1)$ as 
$$
af'(a)=1,\ \ a>0. 
$$
In particular, assuming $f(1+t)=\sum_{k=0}^\infty b_kt^k$ for $t$ near $0$,
$$
1=(1+t)\,f'(1+t)=(1+t)\,\sum_{k=1}^\infty b_k\,k\,t^{k-1}=(1+t)\,\sum_{k=0}^\infty b_{k+1}(k+1)t^k.
$$
So
$$
\sum_{k=0}^\infty b_{k+1}(k+1)t^k=\frac1{1+t}=\sum_{k=0}^\infty\,(-1)^kt^k.
$$
This shows that 
$$
b_{k+1}=\frac{(-1)^k}{k+1}.
$$
A: Let g(x) = e^(f(x)), then g satisfies the equation: g(x*y) = g(x)*g(y). We show for x > 0, g(x) > 0. Let x = y = x^(1/2), then g(x) = (g(x^1/2))^2 > 0. So let h(x) = ln(g(e^x)), then h is well defined and is continuous. Further, h satisfies the equation: h(x+y) = h(x) + h(y). So h(x) = cx for some c in R. So let e^x = t we have: g(t) = e^(h(ln(t))) = e^(c*ln(t)) = t^c. So e^(f(t)) = t^c, and f(t) = ln(t^c) = c*ln(t). Put t = e we have 1 = f(e) = c. So f(t) = lnt.
