Definition of logarithm function derived from its useful properties

Question

My understanding of the point of logarithms is that they turn multiplication into addition, and exponentiation into multiplication.

i.e.

$$ \ln cx = \ln c + \ln x $$

$$ \ln x^c = c \ln x $$

Let's call the above two statements about logarithms their "useful" properties.

The above two are somewhat the "point" of logarithms in so far as they were originally invented to simplify calculations: Take the logarithm, do all the calculations in the easier world of "log space" where multiplication is addition, then at the end take the inverse-logarithm (exponentiation) to convert the result back.

I'm aware of various definitions of logarithm, such as:

$$ \ln c = \lim_{h\to 0} \frac{c^h-1}{h} $$

and there are then proofs that this definition leads to what I have called the "useful" properties.

But are there any derivations that go the other way?

i.e. a definition of logarithm that starts with

"Define $\ln x$ as the function that turns multiplication into addition" etc.

and concludes with

"$\ln$ exists and it can be computed by calculating $\lim_{h\to 0} \frac{x^h - 1}{h} $

Answer 1

I do not see why we would need to have both $$ \begin{align} \ln(cx)&=\ln(c)+\ln(x)\\ \ln(x^c)&=c\ \ln(x) \end{align} $$ since we can derive the first directly from the latter writing $c=x^t$ so that $t \ln(x)=\ln(c)$: $$ \ln(cx)=\ln(x^{t+1})=(t+1)\ln(x)=\ln(c)+\ln(x) $$ but it is true that solutions to $f(x^c)=c\ f(x)$ (with $\mathbb R^+$ as domain) are proportional and in fact uniquely determined by say $f(2)$ since any $y\in\mathbb R^+$ can be written in the form $y=2^c$ so that $$ f(y)=c\ f(2) $$ so once $k=f(2)$ is determined all values of $f(y)$ are. But to have $f(\mbox{e})=1$ so that we are talking about the natural logarithm $\ln$ we will first need to define $\mbox{e}$ in some way or another.

Paramanand Singh: Do we explicitly need to define $\log$ or $\exp$ before defining $x^h$ for some given $x\in\mathbb R^+$? If $p/q\in\mathbb Q$ is given ($p,q$ integers) then $$ \left(x^{p/q}\right)=x^p $$ uniquely determines $x^{p/q}>0$. So the we may find a sequence in $q_1,q_2,...\mathbb Q$ converging to $h$ and define $$ x^h=\lim_{n\rightarrow \infty}x^{q_n} $$ But generally I see how the definitions given by the asker are insufficient to establish a connection to the limit $$ \ln(x)=\lim_{h\rightarrow 0}\frac{x^h-1}{h} $$

Answer 2

Let us try with the simple functional equation $f(xy) = f(x) + f(y)$ satisfied by $f(x) = \log x$. We will show that any solution of this equation has to be of the form $k\log x$ where $k$ is a constant. We need to assume the differentiability of $f$. Treating $y$ as constant and differentiating with respect to $x$ we get $yf'(xy) = f'(x)$ and exchanging $x, y$ we get $xf'(xy) = f'(y)$ so that $xf'(x) = yf'(y) = xyf'(xy)$. Since $x, y$ are arbitrary it follows that $xf'(x) = k$ and then $f'(x) = k/x$.

Clearly if $k = 0$ then $f(x)$ is constant and from the function equation this constant must be $0$. If $k \neq 0$ then we can write $f(x) = k\int_{a}^{x}(1/t)\,dt$ where $a, x$ are of same sign. Putting this into the functional equation we will get $a = 1$ so that $f(x) = k\int_{1}^{x}(1/t)\,dt$

Also note that the functional equations can't determine $k$. Hence from the function equation you can't derive $$\ln x = \lim_{h \to 0}\frac{x^{h} - 1}{h}$$ There is another fundamental problem that we can't talk about the above limit unless we define $x^{h}$ for all $h$. This can only be done is a simple way if $\log$ and $\exp$ functions are already defined.