Uniqueness of log function with relaxed conditions? Question
If:
$$f(a) + f(b) = f(ab)$$
$$ f(1) = 0 $$
$$ a<b \implies f(a) < f(b)  \forall  a,b \in N  $$
where $N$ is the set of natural numbers.
Prove or disprove $f$ must be the $\log$ function.
Background
I was recently wondering about the uniqueness of a function given:
$$f(a) + f(b) = f(ab)$$
$$ f(1) = 0 $$
$$ a<b \implies f(a) < f(b)  \forall  a,b \in R^+  $$
where $R^+$ is the set of positive real numbers.
All of these imply it must be the $\log$ function. I was wondering however what would be the consequence of relaxing the third condition 
$$a<b \implies f(a)<f(b)  \forall  a,b \in N $$
where $N$ is the set of natural numbers.
This would allow $f(x)$ to be a combination of the $\log$ and number theoretic functions such as $A(x)$. Where 
$$A(x) = \text{number of prime factors of $x$}$$
We note,
$$ A(x) + A(y) = A(xy)$$
$$ A(1) =0 $$
 A: It can be shown by induction on $p$ that $f(a^p) = pf(a)$ for any integers $a,p \geq 1$.
We will prove that $f(n) = \log_b n$ for some $b > 1$. After multiplying $f$ by a positive constant, we may assume $f(10) = 1$.
Let $n$ be fixed. We will prove that $f(n) = \log n = \log_{10} n$. Otherwise, we have either $f(n) < \log n$ or $f(n) > \log n$. 
Assume $f(n) < \log n$. Then there exists a rational number $r/s$ such that $f(n) < r/s < \log n$. Then 
$$f(n^s) = sf(n) < r = rf(10) = f(10^r),$$
so $n^s < 10^r$. It follows that $n < (10^r)^{1/s} = 10^{r/s} < 10^{\log n} = n$, a contradiction. Thus the hypothesis $f(n) < \log n$ was absurd. 
The condition $f(n) > \log n$ can similarly be proved absurd. Therefore $f(n) = \log n$.
A: I think this result will follow if you assume that given primes $p_1,p_2$ and $\epsilon>0$ there exists integers $n_1,n_2,n_1',n_2'$ such that
\begin{equation}
1-\epsilon<\frac{n'_1\log p_1}{n'_2\log p_2}<1<\frac{n_1\log p_1}{n_2\log p_2}<1+\epsilon
\end{equation}
With this approximation, you can argue that $f(p_1)/f(p_2)$ is arbitrarily close to $\log p_1/\log p_2$ and conclude with the proof.
I don't have a reference for this property of primes but intuitively it makes a lot of sense.
