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In this answer I established the following characterization of $\exp(x)$:

If $f:\mathbb{R} \to \mathbb{R}$ is a function such that

  • $f(x) \geq 1 + x$ for all $x \in \mathbb{R}$
  • $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$

then $f(x) = \exp(x)$.

Actually I established in the linked answer that $f(0) = 1, f'(0) = 1$ and this implies $f'(x) = f(x)$ so that the characterization of $f$ as the exponential function is complete.

Similarly it can be proved that the following characterization for $\log x$ holds:

If $g: \mathbb{R}^{+} \to \mathbb{R}$ is a function such that

  • $g(x) \leq x - 1$ for all $x \in \mathbb{R}^{+}$
  • $g(xy) = g(x) + g(y)$ for all $x, y \in \mathbb{R}^{+}$

then $g(x) = \log x$.

These characterizations don't mention anything about continuity or differentiability and instead rely on inequalities and the functional equation.

Do we have any other characterizations of exponential and logarithmic functions which don't rely on analytic properties (continuity, derivatives etc) and instead rely on properties which are purely algebraic in nature?

I am thinking of monotone nature. If we augment the functional relation with the requirement that the function is strictly increasing on its domain, will it be sufficient to determine the functions $f, g$ uniquely? I guess it does not work. The function $f(x) = a^{x}$ with $a > 1$ is strictly monotone and satisfies the functional equation, so we still don't get uniqueness. My bad!

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  • $\begingroup$ How does $f(0)=f'(0)=1$ implies $f(x)\equiv f'(x)$? $\endgroup$ – Kenny Lau May 8 '16 at 15:17
  • $\begingroup$ Well, all of the characterizations will ultimately be equivalent, so it'd say there are infinitely many. $\endgroup$ – orion May 8 '16 at 15:17
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    $\begingroup$ @KennyLau: we have $$f'(x) = \lim \frac{f(x + h) - f(x)}{h} = \lim f(x)\frac{f(h) - f(0)}{h} = f(x)f'(0) = f(x)$$ here we have used functional equation to get $f(x + h) = f(x)f(h)$ and $f(x) = f(x)f(0)$. $\endgroup$ – Paramanand Singh May 8 '16 at 15:18
  • $\begingroup$ @orion: agree all the characterizations have to be equivalent because they characterize the same thing, but still it makes sense to know many alternative formulations (focus being to find simpler and simpler ones) so as to get to the essence of the simplest properties which characterize these important elementary functions. $\endgroup$ – Paramanand Singh May 8 '16 at 15:22
  • $\begingroup$ That way every property of a function is its valid characterisation ! $\endgroup$ – Narasimham May 8 '16 at 15:36

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