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?

Let $f(x) = e^{\int_1^x \frac{1}{t} dt}$.
Then $f'(x)=\frac{1}{x}e^{\int_1^x \frac{1}{t} dt}=\frac{f(x)}{x}$.
hence
$$x f'(x)-f(x)=0 \Rightarrow \frac{x f'(x)-f(x)}{x^2}=0 \Rightarrow \left(\frac{f(x)}{x}\right)'=0 \,.$$
