Is there any other function than $log$ for which $f(ab) = f(a)+f(b)$ and $f$ is monotonic?

Question

Is there any other function than $log_{x}$ for which $f(ab) = f(a)+f(b)$ and $f$ is monotonic?

If the answer to this question is yes, how can I find such functions?

Answer 1

Not if we assume that $f$ is differentiable and defined on $\mathbb R_{>0}$ (which is where $\log$ is defined), since then $$ y \, f'(xy) = \frac{\text {d}}{\text {d} x} f(xy) = \frac{\text {d}}{\text {d} x} (f(x) + f(y)) = f'(x).$$ Setting $y=\frac{1}{x}$ and $f'(1)=a$, $$ \frac{f'(1)}{x} = \frac{a}{x} = f'(x),$$ we integrate to get $$f(x) = f(1) +\int_1^x \frac{a}{x'} \text {d} x' = f(1) + a \log (x),$$ and since $f(1) = f(1 \cdot1) = f(1)+f(1)$, we necessarily have $f(1)=0$. Thus, all the differentiable solutions to $f(xy) = f(x)+f(y)$ are of the form $a \log (x)$.

Answer 2

We begin with the provided functional equation:

$$\mathcal {f} \left( x y \right) = \mathcal {f} \left( x \right) + \mathcal {f} \left( y \right).$$

Rewrite the equation:

$$\mathcal {f} \left( x y \right) - \mathcal {f} \left( x \right) = \mathcal {f} \left( y \right).$$

See

$$\frac {\mathcal {f} \left( x y \right) - \mathcal {f} \left( x \right)}{x y - x} = \frac {1}{x} \cdot \frac {\mathcal {f} \left( y \right) - 0}{y - 1}.$$

Return for now to the original equation. Letting $x = y = 1$, we see $\mathcal {f} \left( 1 \right) = 0$.

Using this result, we continue and write

$$\frac {\mathcal {f} \left( x y \right) - \mathcal {f} \left( x \right)}{x y - x} = \frac {1}{x} \cdot \frac {\mathcal {f} \left( y \right) - \mathcal {f} \left( 1 \right)}{y - 1}.$$

Let $y \to 1$. Assuming differentiability, by definition,

$$\mathcal {f}' \left( x \right) = \frac {1}{x} \cdot \mathcal {f}' \left( 1 \right),$$

so

$$\mathcal {f} \left( x \right) = \mathcal {f}' \left( 1 \right) \ln \left( x \right).$$

In summary, assuming differentiability, then $\mathcal {f} \left( x \right) = a \ln \left( x \right)$ for all $a \in \mathbb {R}$. A special case is when $a = 0$. Then we have a zero function, which is trivial.

Answer 3

It's easy to see that $f(1)=0$, as $f(1 \cdot 1) = f(1) + f(1)$ :-)

Next thing to do, is choose another value.
Normally you take a value for $x$ and set the function value, but here we do it the other way around: for which $k$ is $f(k) = 1$?

You'll easily see that $f(x)= \log_k(x)$, proving that indeed logarithmic functions (or the constant zero function) are the only solutions.