# how to test whether an unknown function is logarithmic function or not? can we figure that out by “ the fundamental property of a logarithm”

My question is derived another question:How do we know that ln function (natural log) is a logarithmic function if we start from its calculus definition?

I recently learned the calculus definition of ln, before which I only know that $\ln(a) = \log_e(a)$.

Suppose that we don't know anything about ln function except its calculus definition: $\ln(a)$ equals the integral of $y = \frac{1}{x}$ evaluated on $(1,a)$ ,then we can prove that $\ln(ab) = \ln(a) + \ln(b)$. After this, wikipedia says "this function ($\ln$ function) is a logarithm because it satisfies the fundamental property of a logarithm, $\ln(ab)=ln(a)+ln(b)$"

It can be inferred that there's something called "the fundamental property of a logarithm", which is a criteria to test out whether an unknown function is logarithmic function or not. But why can this property of logarithm test out whether a function is logarithmic or not?

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I think what you are asking for is a proof that if $f(rs)=f(r)+f(s)$ for all positive $r$ and $s$, then $\int_1^x(1/t)\,dt=f(x)$. Is that what you want? –  Gerry Myerson Feb 10 at 1:51
Surely that's false? I can define $f$ arbitrarily at every prime number, and get an $f$ on $\mathbb{Q}$ with the product property. I suppose the trick is that only one of these can be extended continuously to $\mathbb{R}$? –  user7530 Feb 10 at 1:54
@GerryMyerson, I believe you mean $f(x) = a \int_1^x dt / t$ (you only cover $\ln x$, there are certainly more logarithmic functions). –  vonbrand Feb 10 at 2:04
Folks, I know that what I asked for isn't quite true; I wanted to know whether that was what OP wanted. I was prepared to bargain with OP about what one could get, instead. –  Gerry Myerson Feb 10 at 2:11

Note: This depends a lot on exactly how you define "is a logarithm function". In this answer I have assumed that for you a "logarithm function" means any multiple of the natural logarithm, the latter being defined as the integral of $1/x$. Now, rereading the question, I'm not so sure about that anymore, so caveat lector.

The fundamental property of a logarithm function is the formula $$f(ab) = f(a) + f(b)$$ If some function $\mathbb R_+ \to \mathbb R$ satisfies this for all $a$ and $b$, and it is continuous and not identically zero, then it must be a logarithm function to some base, and that base is uniquely the number such that $f(\mathit{base})=1$.

To prove this, I would start by showing that the natural logarithm satisfies the property for base $e$ (this would probably be the definition of $e$, in fact), and that every continuous function $f:\mathbb R_+\to \mathbb R$ such that $f(ab)=f(a)+f(b)$ and $f(e)=1$ must be equal the logarithm at $e^q$ for every rational $q$, and therefore everywhere by continuity.

You can also show that if $f(ab)=f(a)+f(b)$ and $f(x_0)=0$ for some $x_0\ne 1$, then $f$ is zero everywhere. (Again, first prove this for rational powers of $x_0$, and then apply continuity).

Now, if you're given some arbitrary continuous function that satisfies $f(ab)=f(a)+f(b)$ and is not zero everywhere, then in particular $f(e)$ must be nonzero. Therefore $x \mapsto \frac{f(x)}{f(e)}$ satisfies exactly the conditions that we've just see imply that it must be the natural logarithm, and therefore $f(x)=f(e)\ln(x)$ everywhere.

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OP may wish to note that Henning has written log where OP would write ln. –  Gerry Myerson Feb 10 at 2:13
@Gerry: Oops, sorry. Fixed. –  Henning Makholm Feb 10 at 2:27
I'd much rather see OP change to log than see you change to ln, but I think I'll quit while I'm not too far behind. –  Gerry Myerson Feb 10 at 3:04