How to understand $\ln a + \ln b = \ln(ab)$ looking at the areas defining the three quantities? Im reading the book: What Is Mathematics An Elementary Approach to IDEAS AND METHODS.
There is a description before the proof of $\ln a + \ln b = \ln(ab)$

Intuitively, this formula could be obtained by looking at the areas defining the three quantities lna, lnb and ln(ab). But we prefer to derive it by a reasoning typical of the calculus.... (following the proof in calculus.)

I can't understand this, why the areas defining this is intuitive? thanks.
Edit: I know integrals and the proof in calculus way, what I want to know is just how to understand this quote from the book.
 A: For an intuition, it is enough to consider the case where $a > 1$ and
$b > 1.$  The value of $\ln a$ is equal to the area of the region
$\{(x, y) : 1\leqslant x\leqslant a, \ y\leqslant 1/x \},$ and the
value of $\ln(ab) - \ln b$ is equal to the area of the region
$\{(x^*, y^*) : b\leqslant x^*\leqslant ab, \ y^*\leqslant 1/x^*\}.$
The function $(x, y)\mapsto (bx, y/b)$ is a bijection of the first
set of points to the second.  Intuitively, it preserves areas,
because it preserves the areas of small rectangles: a rectangle
$[x, x + \delta x] \times [y, y + \delta y]$ in the first region
maps to a rectangle $[bx,bx+b\delta x]\times[y/b,y/b+\delta y/b],$
of equal area, in the second region. Therefore, the two regions are
of equal area. That is, $\ln a = \ln(ab) - \ln b.$
A: Mainly too long for a comment:
The book is describing a strategy to prove a statement that might seem appealing to student but will actually be quite messy and hard to follow.
The book has defined $\ln M$ as $\int_1^M \frac 1x dx$.  Geometrically we can think of $\int_a^b f(x) dx$ as the area of the graph below $f(x)$ between $x=a$ and $x = b$.
Therefore to prove $\ln a + \ln b = \ln ab$ would be a matter of proving $\int_1^a\frac 1x dx + \int_1^b \frac 1x dx = \int_1^{ab} \frac 1x dx$.  And we could prove that by comparing the area under $\frac 1x$ between $x=1$ and $x=a$, and the area under $\frac 1x$ between $x=1$ and $x=b$, and the area under $\frac 1x$ between $x=1$ and $x=ab$ and somehow try and see that if we add two of the areas together we will get the third.
The book is warning us not to try to do that.  It is intuitively a valid idea and it does work. But it is not straightforward or clear just how we can manipulate the areas to get the result we want.
(How do we add to areas occupying the same space together? And how do we view them against a third area stretched out? It's not clear how to do, even though it is clear that is what the statement [if it is true] implies.)
So the book suggests a different strategy.  I don't know exactly what the book suggests.  It might be what Sayan Dutta suggests.
The hard part of Sayan Dutta's very good answer is convincing ourselves that if $\int_1^b \frac 1x dx$ is equal to the area under $\frac 1x$ between $1$ and $b$ that that should be equal to the area under $\frac 1x$ between $a$ and $ab$.  And I think trying to convince ourselves geometrically would be very unsatisfactory to the student[1].
So I think the book was probably going to suggest, as Sayan Dutta did, that by substitution and the chain rule that $\int_{x=a}^{x=ab} \frac 1x dx=\int_{t=\frac 1ax;t=1}^{t=b}\frac 1{at}(a\ dt)=\int_1^b \frac 1x dx$.
Or maybe the book was going to do something else.
====
[1] Although it can be done:  The area under the graph of $\frac 1x$ between $x=a$ and $x=ab$ is to divide the segment $[a,ab]$ into $n$ slivers and to take the areas of thin little rectangles that are $\frac{ab-a}n$ at the base and $\frac 1{x}$ in height, and sum up the areas of those rectangles as: $\sum_{k=1}^n (\frac {ab-a}n)\frac 1{x_k}$ where $x_k=a + k\cdot (\frac {ab-a}n)$, and then taking the limit as $n\to \infty$.
But $\sum_{k=1}^n (\frac {ab-a}n)\frac 1{a + k\cdot (\frac {ab-a}n)}=\sum_{k=1}^n(\frac {b-1}n)\frac 1{1+k(\frac {b-1}n)}$ which, when we take the limit, is the approximation of the area under the graph of $\frac 1x$ between $x=1$ and $x=b$.
A: I guess, by "area", they mean using integration.
We can use the fact that
$$\ln a=\int_1^a \frac 1x \;\text{d}x$$
and hence
\begin{align*}
\ln (ab)&=\int_1^{ab} \frac 1x\;\text{d}x\\
&=\int_1^{a} \frac 1x\;\text{d}x + \int_a^{ab} \frac 1x\;\text{d}x\\
&=\int_1^{a} \frac 1x\;\text{d}x + \int_1^{b} \frac 1{at} a\;\text{d}t\\
&=\int_1^{a} \frac 1x\;\text{d}x + \int_1^{b} \frac 1x\;\text{d}x\\
&=\ln a+\ln b
\end{align*}
hence completing the proof.

Edit: I got to know a little bit more about this problem, and it turns out that you don't have to think of areas as integrals, in the sense that it can also be dealt using more elementary notion of area. That is, it is possible to do the calculus done in this proof more intuitively using scaling and dilation. Although Calum and Fleablood did a pretty good job of explaining this in their answers, I am placing this edit mainly to refer you to "Measurements" by Lockhart which is a vast sea of many other proofs of this kind.
