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

The other answer has not answered your question, “Why is $e$ equal to $2.718281828\cdots$?”.
Let’s form the number $E=\lim_n(1+\frac1n)^n$, and evaluate it knowing the continuity of the log function and what its derivative is. Of course this number $E$ is computable, even if slowly, directly by hand. And if you take $n$ large enough, you will indeed find that your result is close to the number I quoted above.
We have:
$$
\log(E)=\log\left[\lim_n(1+\frac1n)^n\right]=\lim_n\frac{\log(1+\frac1n)-\log(1)}{1/n}=\log'(1)=1/1\,,
$$
just using the definition of the derivative and applying it to the logarithm. So $\log(E)=1$, and I think that’s what you wanted to know.
