The problem asks to show that the corresponding approximation to area under $y=1/x$ from $b$ to $ab$ by $n$ rectangles has exactly the same area. I'm not sure where to start. Then Deduce by the method of exhaustion, that the areas under $y = 1/x$ from $1$ to a and from b to ab are equal. And finally deduce the definition of $log$, that $log (ab) = log (a) + log (b)$.
Any help would be appreciated, thanks
On
Part 1
We show that corresponding rectangles have equal areas.
Figure 1 demonstrates how to construct the product $ab$ via straightedge and compass. It is important to note that the lines $CB$ and $AD$ are parallel. (You could choose the line through $O$ perpendicular to $OA$, and then visualise the rectangular approximations concurrently on the full graph of $y=1/x$).
$\tag{Fig. 1}$
Next, consider any partition $P_1=(x_0,...,x_n)$ of the interval $[1,a]$, where $1=x_0\lt x_1\lt ... \lt x_n=a$. From Figure 1 we can construct the partition $P_2=(y_0,...,y_n)$ of $[b,ab]$ by the same parallel projection as was used to construct the line segment $OD$. Figure 2 demonstrates the case $n=4$ .
$\tag{Fig. 2}$
The value of this construction is that we can plainly see, via similar triangles, a simple relation between corresponding points of $P_1$ and $P_2$: $$\tag{1}\frac{x_i}{y_i}=\frac{1}{b}.$$
Hence, as areas of rectangles $A\left(\square_{a_i}\right)$ and $A\left(\square_{b_i}\right)$ are simply the widths $\Delta x=x_i -x_{i-1}$, $\Delta y=y_i -y_{i-1}$ multiplied by the heights $f(x_{i-1})=\frac{1}{x_{i-1}}$ and $f(y_{i-1})=\frac{1}{y_{i-1}}$, we get $$\tag{2} A\left(\square_{a_i}\right)=\frac{x_i-x_{i-1}}{x_{i-1}} =\frac{\frac{y_i}{b}-\frac{y_{i-1}}{b}}{\frac{y_{i-1}}{b}}=\frac{y_i-y_{i-1}}{y_{i-1}}=A\left(\square_{b_i}\right).$$
Part 2
In modern parlance The Method of Exhaustion was essentially the demonstration that you could approximate a given figure by shapes with known areas arbitrarily closely. In order to show that the two areas are indeed equal, we take a single rectangle in the partition $P_1$ as in Figure 3.
$\tag{Fig. 3}$
Next, we consider what happens to the shaded area $A\left(\color{grey}{\blacksquare}\right)$ after we subdivide the interval $\Delta x$ in two:
$\tag{Fig. 4}$
It is easy to show what this picture suggests, which is that the sum of the shaded areas in Figure 4 is exactly $$\tag{3}\frac{1}{2}A\left(\color{grey}{\blacksquare}\right).$$ In general, after $n$ such subdivisions the sum of the $2n$ shaded regions is $$\tag{4}\frac{1}{2n}A\left(\color{grey}{\blacksquare}\right)$$
and since we know $\frac{1}{n} \rightarrow 0 $ as $n \rightarrow \infty$ (obviously the Greeks would not have considered a limit like this) the differences between the rectangles inscribed under and circumscribed over the graph of $y$ can be made arbitrarily small. Since the graph of $y=1/x$ is contained in the shaded regions an equivalent statement is that the areas of the rectangles get arbitrarily close to the area under $y$ as $n$ increases. By Part 1 we can conclude $\log{a}$ is equal to the area $\square$ under the graph of $y$ between $b$ and $ab$.
Part 3
By definition $\square$ is equal to the difference between $\log{ab}$ and $\log{b}$. Hence we can conclude $$\tag{4} \log{a}= \log{ab} - \log{b}\implies \log{a} + \log{b}= \log{ab}.$$ (Alternatively, you could note that the same equality holds between $\log{b}$ and the area under the graph between $a$ and $ab$. You can then visualise the sum on the graph by marking off $a$, $b$, and $ab$ and considering the overlaps of shaded regions).
First I suggest you make a graph to visualize the approximation.
You could try to write the first approximation (with $x$ from $1$ to $a$).
You have $n$ rectangles, all of them having width equal to $\frac{a-1}{n}$.And the $k$-th rectangle has height $\frac{1}{1+k\frac{a-1}{n}}$. So the first approximation is $$S_1 = \sum_{k=1}^n\frac{a-1}{n}\times\frac{1}{1+k\frac{a-1}{n}}=\sum_{k=1}^n\frac{a-1}{n+k(a-1)}$$
The second approximation has $n$ rectangles of width $\frac{ba-b}{n}=\frac{b(a-1)}{n}$ and height $\frac{1}{b+k\frac{b(a-1)}{n}}$.
Thus, $$S_2=\sum_{k=1}^n \frac{b(a-1)}{n}\times\frac{1}{b+k\frac{b(a-1)}{n}}$$ The $b$'s simplify and we see that $S_1=S_2$.
Remark: this goes with the fact that $\log(ab) - \log(b)=\log(a)+\log(b)-\log(b) = \log(a) = \log(a)-\log(1)$.