$f(x)=\log x\Leftrightarrow f^{-1}(x)=e^x$. Why $e=2.73\cdots$? 
$$f(x)=\log x\Leftrightarrow f^{-1}(x)=e^x.$$

Ok, $\log x$ is defined as the function $f(\cdot)$ such that: $f'(x)=\dfrac{1}{x}$. How to get, from this, the inverse of it $f^{-1}(x)$? And why $e=2.73\cdots$?

You also need to impose $\log(1) = 0$; otherwise, it's only determined up to a constant. The inverse $f^{-1}$ has derivative
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = f^{-1}(x)$$
and $f^{-1}(0) = 1$ ; this is exactly the definition (well, one definition) of $e^x$.
