Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$ A logarithm of base b for x is defined as the number u such that $b^u=x$. Thus, the logarithm with base $e$ gives us a $u$ such that $e^u=b$. 
In the presentations that I have come across, the author starts with the fundamental property $f(xy) = f(x)+f(y)$ and goes on to construct the natural logarithm as $\ln(x) = \int_1^x \frac{1}{t} dt$. 
It would be suprising if these two definitions ended up the same, as is the case. How do we know that the are? The best that I can think of is that they share property $f(xy) = f(x)+f(y)$, and coincide at certain obvious values (x=0, x=1). This seems weak. Is their a proof?

The following properties uniquely determine the natural log:
1) $f(1) = 0$.
2) $f$ is continuous and differentiable on $(0, \infty)$ with $f'(x) = \frac{1}{x}$.
3) $f(xy) = f(x) + f(y)$

We will show that the function $f(x) = \int_1^x \frac{1}{t} dt$ obeys properties 
1,2, and 3, and is thus the natural log.
1) This is easy, since $f(1) = \int_1^1 \frac{1}{t} dt = 0$.
2) Defining $f(x) = \int_1^x \frac{1}{t} dt$, we note that since $\frac{1}{t}$ is continuous on any interval of the form $[a,b]$, where $0 < a \leq b$, then the Fundamental Theorem of Calculus tells us that $f(x)$ is (continuous and) differentiable with $f'(x) = \frac{1}{x}$ for all $x \in [a,b]$.
3) 
$$\begin{align}
f(xy) = \int_1^{xy} \frac{1}{t}dt &= \int_1^x \frac{1}{t} dt + \int_x^{xy} \frac{1}{t} dt
\\
&= f(x) + \int_{1}^{y} \frac{1}{u} du
\\
&= f(x) + f(y)
\end{align}$$
where in the last step we perform the substitution $t = ux$ (viewing $x$ as constant).  
