Give geometric proof of $\ln(xy) = \ln x + \ln y$ based on below diagram and please explain it 
I've been reading The Art Of Computer Programming Volume 1, and I am having difficulty in solving this problem. Since Maths is Greek to me, can you please help?
At the back, there are hints to the solutions, for this problem below is the picture.Hint to the solution
 A: Assume $y,x>1$. 
The shaded region has area $\ln x$ (if we define $\ln x=\int_1^x\frac1x\,\mathrm dx$). 
Note that if we scale the hyperbola horizontally by a factor $a$ and vertically by a factor $\frac1a$, we end up with the original hyperbola.
Thus if we scale our shaded region vertically by $\frac1y$ and horizontally by $y$, we obtain a region of the same area, bounded above by the same hyperbola, bounded below by the $x$-axis, bounded on the right by the vertical line segment from $(xy,0)$ to $(xy,\frac1{xy})$, and on the left by the vertical line from $(y,0)$ to $(y,\frac1y)$. Together with the region defining $\ln y$ (i.e., ananlogous as in the image, but with all $x$ replaced by $y$), we obtain the region defining $\ln xy$. We conclude $$\ln xy=\ln x+\ln y. $$
A: I can't exactly draw pictures very easily right now, so I'll try to give a verbal explanation as clear as I can.
The provided picture demonstrates that $\ln x$ is the area under the curve $f(t)=1/t$ between $1$ and $x$.
Assume, mostly for geometric clarity, that $x$ and $y$ are both greater than $1$.
Thus, $\ln(xy)$ is the area under $f(t)=1/t$ between $1$ and $xy$. Let's split this region into two pieces by cutting it at $t=x$. This gives us two regions. The first is the area under $f(t)=1/t$ between $1$ and $x$. This has area $\ln x$. The second region is the area under $f(t)=1/t$ between $x$ and $xy$. In order for the geometric argument to work out, this should have area $\ln y$, since then we have split a region with a total area of $\ln(xy)$ into two pieces with areas $\ln x$ and $\ln y$.
So why does the region under $f(t)=1/t$ between $x$ and $xy$ have area $\ln y$?
Well, let's try rescaling it. If we scale it in the $x$ direction by a factor of $1/y$, the region will go from $t=1$ to $t=y$, under the graph $f(t)=1/t$, and we'll have the area divided by a factor of $y$. To cancel this division out, let's also scale the region vertically by a factor of $y$. Now the new region has the same area as our original region, and better yet, it has become the region under the graph $f(t)=1/t$ between $t=1$ and $t=y$. We know this region has area $\ln y$ though! Thus we are done.
I should point out that we're using a special property of the graph $f(t)=1/t$. Namely that if you transform a point $(x,y)$ lying on the graph of this function by the rule $$(x,y)\mapsto (x/c,yc)$$ then it will still lie on the graph, which you can see, since
if we know
$$\frac{1}{x}=y,$$
then we also know
$$\frac{1}{x/c} = \frac{c}{x} = cy.$$
This is why our transformation where we squash $x$-values and stretch $y$-values sends the region under $f(t)=1/t$ to another region which is bounded at the top by $f(t)=1/t$.
A: 
Source: Why is the integral of $\frac{1}{x}$ equal to the natural logarithm of x
One essential feature of logarithms is that they make a multiplication 
  problem equivalent to an addition problem, by which I mean
$$\ln(ab) = \ln(a) + \ln(b)$$
  Meanwhile, $\int\frac{1}{x}\mathrm{d}x$ is usually thought of geometrically as the area underneath a curve. The problem, then, is to try to see visually what an area under a curve has to do with turning multiplication into addition.
Here’s a graph of $\frac{1}{x}$, and we’re finding, as an example, the area under it from $1$ to $2$.

Let’s say now that we multiply the limits of integration by two, so we’re now finding the area from $2$ to $4$. Here’s what that looks like.

The two portions are actually very similar to each other in their overall shape. The orange one is twice is wide as the green one, but also half as tall. Here they are overlaid.

If you take the green shape and first squash it down vertically by a factor of two then stretch it out horizontally by a factor of two, you get the orange shape exactly. (If you don’t believe this, convince yourself it works!) This means the areas of these shapes are exactly the same, even though we don’t know what that area is.
Show for yourself that this result is general. The area under $\frac{1}{x}$ from $a$ to $b$ is the same as that from $ac$ to $bc$.
What, then, is the area from $1$ to $6$? We can break it into two parts – the area from $1$ to $2$ and the area from $2$ to $6$. But the area from $2$ to $6$ is the same as the area from $1$ to $3$, by the above reasoning.
Thus, the area from $1$ to $6$ is the same as the sum of the areas from $1$ to $2$ and from $1$ to $3$. Note that $6 = 3\cdot 2$. Again, this is general. The area under $\frac{1}{x}$ from $1$ to $ab$ is the same as the sum of the areas from  $1$ to $a$ and from $1$ to $b$.
That’s pretty good motivation for the definition
$$\ln(x) = \int_1^x\frac{1}{t}\mathrm{d}t$$
  Note that this is being taken as a definition of the natural logarithm, not a proof of the relationship. Our argument about the integral of $\frac{1}{x}$ now translates to the statement
$$\ln(ab) = \ln(a) + \ln(b)$$

