TITLE: Inverse of the natural log function $y =\ln x$
QUESTION [8 upvotes]: Of course, it is a well known fact that the inverse of $y=\ln x$ (natural logarithm of x) is $e^x$. 
Assuming we haven't heard of the exponential function at all, how do we prove that the inverse of $\ln x$ i.e ($\ln^{-1} x$ ) is some other function, which indeed is the so called exponential function $e^x$? 

Let me be a little more concrete. If $y = \ln x$, then $x= A^y$, how do I prove that $A= e$?

There's another method by which I tried to arrive at the inverse of natural logarithm of $x$. 
By a theorem of differentiation of inverse functions
$$\left.\dfrac{df}{dx}\right|_{x=a} \cdot\left.\dfrac{df^{-1}}{dx}\right|_{x=f(a)}=1$$ 
I got an equation which looked like
$$\left.\dfrac{dF(x)}{dx}\right|_{x=\ln k}=k$$  where $k$ is a real number. 
How do I actually show that $F(x)$ is that same exponential function $e^x$? 
Is there any way of solving such an equation, or such a problem? Or am I just talking nonsense? I would love to be enlightened by all of you.
Thanks in advance. :)

REPLY [3 votes]: Supposing we have $\ln x$ and not $e^x$, we may as well assume that we've defined $\ln x$ as an anti-derivative of $x^{-1}$, that is:
$$ \ln x = \int_1^x t^{-1} \; dt.$$
This is in fact a standard approach that can be taken to define logarithms and exponentials in Calc I or II (see Stewart's "Gray Pages" in Calculus and elsewhere).
Then we get $(\ln x)' = x^{-1}$.  If $g(x)$ is an inverse for $f(x) = \ln x$, then
$$ g'(x) = \frac{1}{f'(g(x))} = \frac{1}{(g(x))^{-1}} = g(x).$$
In other words, the inverse function of $\ln x$ satisfies the differential equation $g'(x) = g(x)$.  Furthermore, we have $g(0) = 1$ (since $\ln 1 = 0$).  At this point, if we've defined $e^x$ in some other context, then we find that $g(x)=e^x$ does satisfy the initial value problem.  By uniqueness, $g(x) = e^x$ must be the inverse of $\ln x$.
Hope this helps!