Intuitively, how would you explain that the logarithm and the number $e$ are related to the function $\frac{1}{x}$ in particular? Why does the integral of the function $\frac{1}{x}$ from 1 to $e$ have to be equal to 1 ?
Why does it mathematically make sense?
How come a number related to instantaneous, continuous growth has to appear under the curve of $\frac{1}{x}$ in particular?
Why is the area under this curve related to exponentiation?
Thank you.
EDIT : I do not mind formal mathematics as long as the answer conveys the intuition that would help answering these questions.
 A: It depends on how you define $e$.  One definition, the one that Mohammad alludes to, defines $F(x) = \ln x$ as the antiderivative of $f(x) = \frac{1}{x}$ which satisfies $F(1) = 0$.  Then, because $F$ is increasing, it must assume the value $1$ at some point, so we define $e$ to be that number such that $F(e) = 1$.  This doesn't help with intuition, though, in my opinion.
Another definition starts from the fact that any continuous function $g(x)$ satisfying $g(x+y) = g(x)g(y)$ must also satisfy $g(x) = g(1)^x$.  So functions which “turn addition into multiplication” are exponential functions.  This includes any function of time that exhibits constant relative growth.  
If the function is differentiable at $0$, it's differentiable anywhere and $g'(x) = g'(0) g(x)$.  That's also notable; the derivative of such a function is a constant multiple of itself.  By experimentation, this constant is less than one if the “base” $g(1)$ is $2$, and it's greater than one if $g(1) = 3$.  So assuming some kind of continuity in the base, there should be a number $e$ such that if $g(x) = e^x$, then $g'(0) = 1$.  In other words, $e$ is defined by the equation $\lim_{h\to 0} \frac{e^h -1}{h} = 1$.
Now assume $g(x) = e^x$.  Its derivative is positive, so it's increasing and has a differentiable inverse $G(x)$, which we call $\ln x$.  By the inverse function theorem, 
$$
    G'(x) = \frac{1}{g'(G(x))} = \frac{1}{g'(0) g(G(x))} = \frac{1}{1 \cdot x} = \frac{1}{x}
$$
So the derivative of $\ln x$ must be $\frac{1}{x}$.
That's the basics of the story (borrowed heavily from Michael Spivak's Calculus).  I don't know if any of it's intuitive, but other than the inverse function theorem I think it stems from the fact that $g(x+y) = g(x)g(y)$ for all $x$ and $y$ implies that $g(x) = g(1)^x$.
A: Looking at the area under the curve $y = \frac 1x$
Compare the shape of the region from $1$ to $b,$ and the shape of the region from $a$ to $ab$

The region from $a$ to $ab$ has been compressed vertically and dilated horizontally by the same factor.
The areas are the same
$\int_1^{b} \frac {1}{x} dx = \int_a^{ab} \frac {1}{x} dx$
And this can be done "analytically" with a u-substitution.  But, it is nice to see what is going on geometrically.
If we define some function: 
$f(a) = \int_1^{b} \frac {1}{x} dx$
and $f(ab) = f(a) + f(b)$ and $f(a^n) = nf(a)$  (which you probably should prove)
Then $f(a)$ behaves like a logarithm. 
If it behaves like a logarithm, then the function IS a logarithm.  The question is what is its base? 
We don't know!  We define $e$ to be the base of this logarithm! If it feels all a bit circular, that is because it is.  
It won't be until near the end of of the course that you will have enough calculus to show that this $e$ is the same $e$ that you learned about in Algebra class when discussing compound interest calculations.
A: I think, it's easier to introduce the constant $e$ first.  This constant provides solutions to the ordinary differential equation:
$${dy \over dt} = y.$$
In other words, it helps answer the question about invariants (https://en.wikipedia.org/wiki/Invariant_(mathematics)): Which functions are their own derivatives?
Using the power series approach, $y(t) = \sum_{k \geq 0} a_{k} t^{k}$, we arrive at $a_{k} = 1 / (k!)$.
Now, since it turns out that the function $f(x) = e^{x}$ is a bijection from the real line to the real strictly positive half-line $y > 0$, one can ask:  What is the inverse of $f(x)$?  That's how we arrive at the natural logarithm $\ln(y)$.
Finally, what is the rate of change of $\ln(y)$ with respect to $y$?  Well, by the above,
$$
1 = {dy \over dy} = {d \over dy} (e^{\ln y}) = e^{\ln y} {d \over dy} \ln(y).
$$
(The last equality is the Chain Rule.)
That gives the rate relations you were asking about.
The definite integral $\int_{1}^{e} {dx \over x}$ can now be computed using the above and the Fundamental Theorem of Calculus.
A: This really, really depends on how you define the natural logarithm and Euler's constant.  We can start by defining the logarithm as the area under the hyperbola:

Definition: The natural logarithm is the function $\log : (0,\infty) \to \mathbb{R}$ defined by the formula
$$ \log(x) := \int_{1}^{x} \frac{1}{t} \,\mathrm{d}t. $$

There are other ways of defining a logarithm.  In lower level classes, for example, we define logarithms as the inverse functions of exponential functions.  We might also define the logarithm by its power series expansion.  However, this definition is quite nice, and the one that I would use if I were the king of mathematics.
Next, I would like to define the Euler's constant via a power series.  The following theorem gives us what we need.

Theorem: The function $\exp : \mathbb{R} \to \mathbb{R}$ defined by the power series
$$ \exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$
is analytic on $\mathbb{R}$.  Via term-by-term integration, we can see that
$$ \frac{\mathrm{d}}{\mathrm{d}x} \exp(x) = \exp(x).$$
Define Euler's constant to be
$$ \mathrm{e} = \exp(1). $$

We could also define the exponential function to be the unique solution to the initial value problem $u'(t) = u(t)$ subject to $u(0) = 1$, but there are questions about whether or not such a solution even exists that I would prefer not to deal with.  In any event, the fact that $\exp$ is its own derivative is hugely important here.
At first blush, it does not look like $\log$ and $\exp$ have anything to do with each other.  However, we can show that they are inverses!  First, we should note that $\log$ is monotonically increasing—as $x$ gets larger, so too does $\log(x)$.  But then $\log$ must be injective, which implies that it must have an inverse.  I claim that $\exp$ is the inverse of $\log$!
To see this, we can compute as follows:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} \log(\exp(x))
&= \frac{\mathrm{d}}{\mathrm{d}x} \int_{1}^{\exp(x)} \frac{1}{t} \,\mathrm{d}t \\
&= \frac{1}{\exp(x)} \frac{\mathrm{d}}{\mathrm{d}x} \exp(x) & \tag{FTC and chain rule} \\
&= \frac{\exp(x)}{\exp(x)} & \tag{definition of $\exp$} \\
&= 1.
\end{align}
Since we have $(\log(\exp(x))' = 1$, we can integrate on each side to obtain
$$ \log(\exp(x)) = x + C $$
for some constant $C$.  But when $x=0$, we get
$$ C = \log(\exp(0)) = \log(1) = \int_{1}^{1} \frac{1}{t} \,\mathrm{d}t = 0. $$
But then $\log(\exp(x)) = x$, which implies that $\log$ is the inverse of $\exp$, as claimed.
Finally, why does $\int_{1}^{\mathrm{e}} \frac{1}{t}\,\mathrm{d}t = 1$?  This follows from the work done above.  Let $I$ denote the value of this integral; that is, define
$$ I := \int_{1}^{\mathrm{e}} \frac{1}{t}\,\mathrm{d}t = \log(\mathrm{e}). $$
Since $\log$ and $\exp$ are inverses, it follows that
$$ \exp(I) = \mathrm{e}. $$
But we defined $\mathrm{e}$ to be $\exp(1)$, and $\exp$ is injective, hence
$$ \exp(I) = \exp(1) \iff I = 1, $$
which is what we wanted to show.
A: The function $$f:t\mapsto \int_{1}^t\frac{ds}{s}$$ is multiplicative $f(tt')=f(t)+f(t')$, and this follows immediately from a change of variables: $$f(tt')=\int_{1}^{tt'}\frac{ds}{s}=\int_{1}^t\frac{ds}{s}+\int_{t}^{tt'}\frac{ds}{s}=\int_{1}^t\frac{ds}{s}+\int_{1}^{t'}\frac{ds}{s}=f(t)+t(t'),$$ substituting in the second integral the variable $s$ by $s/t$. We define the logarithm of a number $t$ by declaring it to be equal to $f(t)$. By definition, also, the exponential is the inverse of the logarithm. There is no intuition involved in all this: the exponential is related to that integral simply because we define it to be related to it.
The technical explanation for this is that $ds/s$ is the Haar measure for the multiplicative group $\mathbb R^\times$, that is, the unique (up to scalars) translation invariant measure on that group. 
