Does the sum of $\frac{\pm a}{\log(1+ax)}$ have at most finitely many zeros?

Suppose $a_1, a_2, ..., a_k\in\mathbb{R}$. Is it true that $f(x)=\sum\limits_{1 \leq i \leq k}{\dfrac{\pm a_i}{\log(1+a_ix)}}$ has at most finitely many zeros (on the domain where $1+a_ix>0$ for all $i$) for every sign combination? I have plotted several curves in Mathematica and feel it might be correct. Any ideas to prove it?

-
Thanks for the response. I have deleted my (now useless) answer and edited your question according to your explanation. Please see if I understand you correctly. If not, please feel free to roll back the question to an earlier version. – user1551 Dec 20 '12 at 2:31
@user1551, thanks for the change. Just now I found a solution to this problem. – Strin Dec 20 '12 at 8:14

I find a simple proof for this. For large $x$, we will have $\log(1+a_i x) \sim \log(a_i)+\log(x)$. Let $t = \log(x), c_i = \log(a_i)$, then let's look at the problem whether $g(t) = \sum\limits_{1 \leq i \leq k}{\frac{a_i}{\log(a_i)+t}}$ has infinite vanish points. But it obviously does not, because the equivalent polynomial $\sum\limits_{1 \leq i \leq k}{ a_i \prod\limits_{j \not=i}{ (\log(a_j)+t)}}$ does not infinite vanish points.

On the other hand, it is trivial to show $f(t)-g(t) = o(f(t)) = o(g(t))$, which means if $g(t) > 0$ (or $g(t) < 0$) in the long run, we will also have $f(t) > 0$ (or $f(t) < 0$) in the long run. QED.

-