How to understand $ln a + ln b = ln(ab)$ looking at the areas defining the three quantities?

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 $x=b$.

How to understand $\ln a + \ln b = \ln(ab)$ looking at the areas defining the three quantities?