Deriving properties of the logarithm from its integral representation Suppose we define:
$$\ln(x) = \int_{a}^{x} \left[ \frac{1}{r} dr\right]$$
Such that 
$$ \ln(1) = 0, \ln(e) = 1$$
How does one derive all the properties of the logarithm from the properties of the function $$ \frac{1}{x}$$
For example:
If we let $x,y$ be functions of t. Then it follows that 
$$ \frac{1}{x} \frac{dx}{dt} + \frac{1}{y} \frac{dy}{dt} = \frac{1}{xy} \left( y \frac{dx}{dt} + x \frac{dy}{dt} \right) $$ 
Thus 
$$ \ln(x(t)) + \ln(y(t)) = \ln(x(t)y(t)) $$
Arriving at the above statement was artificial. I first assumed the latter property true and then differentiated to find what initial algebraic identitiy had to exist. Is there a more natural motivation for this?
 A: Another way to prove that product formula:
$$\log(ax)=\int_{1}^{ax} \frac{dt}{t} =\int_{1}^x \frac{dt}{t} + \int_{x}^{ax} \frac{dt}{t} = \log(x) + \int_{x}^{ax} \frac{dt}{t}$$
Substituting $t=xu$, we get $$\int_{x}^{ax} \frac{dt}{t} = \int_{1}^{a}\frac{du}{u}=\log(a)$$
By induction from here, you can show $\log(x^n)=n\log(x)$.
If you define $e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$, then:
$$\log(e) =\lim_{n\to\infty} \frac{\log\left(1+\frac{1}{n}\right)}{1/n}$$
The right hand is (essentially) the definition of the deivative of $\log(x)$ at $x=1$, so the limit is $\frac{1}{1}$ by the Fundamental Theorem of Calculus.
A: Define a function $f$ as 
$$f(x) = \int_1^{x} \frac{1}{t} \,dt$$
for $x>0$.  It is easy to see show the following properties.

Property (1):  The function $f$ is increasing.
To show this, assume $x_2>x_1>0$, and form the difference $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \frac{1}{t}\,dt$.  Note that this difference is strictly positive since the integrand is also strictly positive. Thus, $f(x_2)>f(x_1)$ whenever $x_2>x_1$, which implies $f$ is strictly increasing.

Property (2):
The function $f$ is differentiable with $f'(x)=\frac{1}{x}$.
This follows immediately from the fundamental theorem of calculus.  Note that this implies automatically that $f$ is continuous for $x>0$.

Property (3):
The limit $\lim_{x \to \infty} f(x)=+\infty$ 
We can easily verify that $\lim_{x \to \infty} f(x)=+\infty$ since for $n>1$
$$\begin{align}
\int_1^{2^n} \frac{1}{t} dt&=\int_{1}^{2} \frac{1}{t} \,dt+\int_{2}^{4} \frac{1}{t} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{t} \,dt\\\\
&\ge \int_{1}^{2} \frac{1}{2} \,dt+\int_{2}^{4} \frac{1}{4} \,dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{2^n} \,dt\\\\
&=\frac12 (2-1)+\frac14 (4-2) + \cdots + \frac{1}{2^{n}}(2^n-2^{n-1})\\\\
&=\frac{n}{2}
\end{align}$$
which tends to infinity as $n \to \infty$.

Property (4):
The limit $\lim_{x \to 0} f(x)=-\infty$
Examine the integral
$$f(\frac{1}{n})=-\int_{\frac{1}{n}}^1 \frac{1}{t}\,dt$$
and substitute $u=\frac{1}{t}$.  Then, we have
$$-\int_{\frac{1}{n}}^1 \frac{1}{t}\,dt=-\int_1^n \frac{1}{u}\,du$$
But from property (3), this integral tends to $+\infty$ and thus $f(\frac1{n})$ tends to $-\infty$ as $n \to 0$.

Property (5):
The function $f$ satisfies $f(x^n) = nf(x)$.  
We know by definition that
$$f(x^n) = \int_1^{x^n} \frac{1}{t} \,dt$$
Then, by substituting $t=u^n$, with $dt=nu^{n-1}du$, and the limits going from $1$ to $x$ reveals
$$\begin{align}
\log (x^n)  &=\int_1^{x} \frac{nu^{n-1}}{u^n} \,du\\\\
&=\int_1^{x} n\frac{1}u\,du\\\\
&=nf (x)
\end{align}$$


The other well-known properties for the log of a product and quotient follow from Property (5).  For example, in THIS ANSWER, I show that $f(xy)=f(x)+f(y)$.

