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?

Put $l(x) = \int_1^x dt/t.$  Then $l'(x) = 1/x$ if $x > 0$ by the fundamental theorem of calculus.  Since $l' > 0$ on $(0,\infty)$, $l$ is 1-1 and therefore has an inverse.  Denote its inverse by $m$.  We have $l(m(x))= x;$ differentiate to see that
$$1 = l'(m(x))m'(x) = m'(x)/m(x).$$
Multiply to get $m'(x) = m(x).$  It is not at all hard to see that $m(0) = 1.$ Therefore, $m(x) = e^x$.
