Geometric interpretation of ln(c) = ln(xy)= ln(x)+ln(y)

Question

For $c,x,y>0$, and $c$ a constant, factoring $c$ into a product can be interpreted as taking a square of area $c$ and morphing it into a rectangle with sides of length $x$ and $y$ with equal area.

Is there a similarly transparent direct geometric interpretation of $\ln(c)=\ln(xy)=\ln(x)+\ln(y)$ perhaps in hyperbolic geometry?

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It's really quite an amazing invariance when looking at the infinite series

$$ \ln(x) = \ln[1-(1-x)] = -\sum_{k>0}\frac{(1-x)^k}{k}.$$

Then, e.g., with $x=y=1/2$,

$$\ln(1/4) = \ln(1/2) + \ln(1/2)$$

implies

$$ \sum_{k>0}\frac{(3/4)^k}{k} = \sum_{k>0}\frac{2(1/2)^k}{k}. $$

No way. Yet ... .

Answer 1

Provided the log is defined via $$\ln(x)=\int_1^x\frac1t\,dt,$$ we first have for positive number $c$: $$\int_1^x\frac1t\,dt=\int_{c}^{cx}\frac1t\,dt.$$ Felix Klein once gave this explanation:

This means that the area between the ordinates $1$ and $x$ is the same as that between the ordinates $c$ and $cx$ which are $c$ times as far from the origin. We can make this clear geometrically by observing that the area remains the same when we slide it along the $x$-axis under the curve provided we stretch the width in the same ratio as we shrink the height. From this the addition theorem follows at once: $$\begin{align} \int_1^x\frac1t\,dt+\int_1^y\frac1t\,dt&= \int_1^x\frac1t\,dt+\int_x^{xy}\frac1t\,dt\\ &= \int_1^{xy}\frac1t\,dt. \end{align}$$

Elementary Mathematics from an advanced standpoint, p.156