We want to show that the Hartley function $\log_2(n)$ is the only function mapping natural numbers to real numbers that satisfies

1. $H(mn) = H(m)+H(n)$ (additivity),
2. $H(m) \leq H(m+1)$ (monotonicity), and
3. $H(2)=1$ (normalization).

Let $f$ be a function on positive integers that satisfies the above three properties. Using the additive property, it is easy to see that the value of $f(1)$ must be zero. So we want to show that $f(n) = \log_2(n)$ for all integers $n\geq 2$ .

From the additive property, we can show that for any integer $n$ and $k$ , \begin{equation} f(n^k) = kf(n). \end{equation}

Let $a>2$ be an integer. Let $t$ be any positive integer. There is a unique integer $s$ determined by $$ a^s \leq 2^t < a^{s+1}. $$ Therefore, $$ s \log_2 a\leq t < (s+1) \log_2 a $$ and $$ \frac{s}{t} \leq \frac{1}{\log_2 a} < \frac{s+1}{t}. $$

On the other hand, by monotonicity, $$ f(a^s) \leq f(2^t) \leq f(a^{s+1}). $$ Using Equation (1) and $f(2)=1$ , we get $$ s f(a) \leq t \leq (s+1) f(a), $$ and $$ \frac{s}{t} \leq \frac{1}{f(a)} \leq \frac{s+1}{t}. $$

Hence, $$ \Big| \frac{1}{f(a)} - \frac{1}{\log_2(a)} \Big| \leq \frac{1}{t}. $$

Since $t$ can be arbitrarily large, the difference on the left hand side of the above inequality must be zero, $$ f(a) = \log_2(a). $$