How to motivate logarithm by breaking down $1/x$, into increasingly large segments, and proving that the area under the curve stayed the same? James Stewart's Calculus 7e 2011 (not Early Transcendentals) presents the two ways to define a logarithm. First, on p 404 in § 6.3, Stewart defines $\log_ax$ as the inverse of $a^y$. Then on p 414 in § 6.4, he proves that $\frac d{dx} ln|x| = \frac 1x \iff \ln|x| + C = \int \frac 1x dx $.
Second, on p 421 in §6.2*, he first defines $\ln x = \int^x_1 \frac 1t \; dt, x > 0 $. Then on p 429 in §6.3*, he defines $e^x = y \iff \ln y = x$.
In either case, how does this comment add anything to the motivation or intuition behind $\log$?

I'm going to go back in time to the development of the logarithm. The log was a way of performing multiplication (in some base) by simply using addition. Meaning, every time you add one number, your output is proportional to the last output.
At some point in time shortly afterwards, mathematicians were working on integration techniques. They noticed that the basic methods of finding anti derivatives worked for everything except for 1/x. However it was realized, that if you break down the graph of 1/x, into increasingly large segments (1-2, 2-4, 4-8, 8-16...) That the area under the curve stayed the same. this is actually very easily provable! But now look what they found..... They found that the area under the curve increases by simple addition as you double the interval. (Meaning adding the area interval (1-2) + (2-4) + (4-8), is simply the same area added together 3 times).
This should stick out to you, because this is exactly how we defined a logarithm to work. Addition carrying out multiplication with some base number.
Because this logarithm seemed to be so "natural" (it just came about all by itself), it was called the "natural logarithm". So the relation was known that ln(x) = integral of 1/x, before it was known what that base number was. Euler was actually able to calculate the number, and the proof for that is very interesting as well! (See my paper for the proof).

 A: One of the most important properties of logarithms is that
$$
\log a+ \log b = \log ab \, .
$$
In fact, if we try to solve the functional equation
$$
f(a)+f(b) = f(ab) \tag{*}\label{*} \, ,
$$
then the solution is $f(x)=c\log x$, where $c$ is an arbitrary constant and $\log$ is the natural logarithm. Now consider the task of finding the area under the curve of $1/x$. As Stewart notes, 'if you break down the graph of $1/x$ into increasingly large segments ... the area under the curve stayed the same'. Symbolically,
$$
\int_{1}^{2} \frac{1}{t} \, dt \, = \int_{2}^{4}\frac{1}{t} \, dt = \int_{4}^{8}\frac{1}{t} \, dt = \ldots
$$
If we play with this equation a little, we find an intriguing way of computing areas. For example,
\begin{align}
\int_{1}^{4} \frac{1}{t} \, dt &= \int_{1}^{2}\frac{1}{t} \, dt +\int_{2}^{4}\frac{1}{t} \, dt \\[6pt]
&= \int_{1}^{2}\frac{1}{t} \, dt + \int_{1}^{2}\frac{1}{t} \, dt
\end{align}
More generally, let $n$ be a positive integer. Then,
\begin{align}
\int_{1}^{2^n} \frac{1}{t} \, dt &= \int_{1}^{2} \frac{1}{t} \, dt + \int_{2}^{4} \frac{1}{t} \, dt + \int_{4}^{8} \frac{1}{t} \, dt + \ldots + \int_{2^{n-1}}^{2^n} \frac{1}{t} \, dt \\[6pt]
&= \underbrace{\int_{1}^{2} \frac{1}{t} \, dt + \int_{1}^{2} \frac{1}{t} \, dt + \int_{1}^{2} \frac{1}{t} \, dt + \ldots + \int_{1}^{2} \frac{1}{t} \, dt}_{\text{$n$ terms}}
\end{align}
What does this have to with logarithms? Well, if we let
$$
f(x) = \int_{1}^{x}\frac{1}{t} \, dt \, ,
$$
then the above equation translates to
$$
f(2^n) = \underbrace{f(2) + f(2) + f(2) + \ldots + f(2)}_{\text{$n$ terms}}
$$
meaning that it appears that these integrals do have the property
$$
f(a) + f(b) = f(ab) \, !
$$
Further investigation suggests that this property is not unique to powers of $2$—it works for powers of $3$, and all other powers for that matter. It is therefore sensible to conjecture that
$$
f(a) + f(b) = f(ab) \, ,
$$
in other words that the integral
$$
\int_{1}^{x}\frac{1}{t} \, dt
$$
is a logarithm. And we can prove this in the following way:
$$
\int_{1}^{ab}\frac{1}{t} \, dt = \int_{1}^{a}\frac{1}{t} \, dt + \int_{a}^{ab}\frac{1}{t} \, dt
$$
Let $z=t/a$, meaning that $dz=\frac{1}{a}dt=\frac{z}{t}dt$. Then we have
\begin{align}
\int_{1}^{ab}\frac{1}{t} \, dt &= \int_{1}^{a}\frac{1}{t} \, dt + \int_{1}^{b}\frac{1}{z} \, dz \\[4pt]
f(ab) &= f(a) + f(b) \, .
\end{align}
Since the solution to $\eqref{*}$ is $f(x)=c\log x$,
$$
\int_{1}^{x}\frac{1}{t} \, dt = c\log x
$$
for some constant $c$. It is only a matter of time before we discover that $c=1$, providing us with the motivation behind defining
$$
\boxed{
\;\\[4pt]
\quad \log x = \int_{1}^{x} \frac{1}{t} \, dt \, . \quad
\\
}
$$
As indicated by the length of this post, the easiest way to make sense of this definition is to pick another starting point, and then discover that
$$
\log x = \int_{1}^{x} \frac{1}{t} \, dt
$$
as a theorem. Once we have done that, it makes sense to use the integral as an alternative way of defining the exponential function.
A: That comment you linked to is not a particularly meaningful one.  The reason is because any function of the form $f(x) = c/x$ for some constant $c > 0$ obeys the same property, namely that for $a > 0$, $$\int_{x=a}^{2a} \frac{c}{x} \, dx = \int_{x=2a}^{4a} \frac{c}{x} \, dx.$$  So this property in itself does not make the choice $c = 1$ "special" or "natural," although it is the choice for which the number $e$ satisfies $$\log_e x = \int_{t=1}^x \frac{1}{t} \, dt, \quad x > 0.$$
The question basically boils down to how $e$, the base of the natural logarithm, is defined.  And the "explanation" in the comment is inadequate motivation for the reason I described.
In reality, there are several equivalent definitions for $e$, each provable from any one of the others, as the Wikipedia article about $e$ states (see "Alternative characterizations").  Consequently, how one motivates the existence of this constant depends on what properties are of interest.
