TAOCP 1.2.2 Exercise 23

Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 23. [M25] Give a geometric proof that $\ln xy = \ln x + \ln y$, based on Fig. 6.

Verified: yes
Solve time: 1m51s

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Solution

Let

$$ L(z) $$

denote the area under the hyperbola

$$ y=\frac1t $$

between $t=1$ and $t=z$. Figure 6 defines the natural logarithm by this area:

$$ L(z)=\ln z. $$

We must prove geometrically that

$$ L(xy)=L(x)+L(y). $$

Consider the region under the hyperbola from $t=1$ to $t=xy$. Split it at $t=x$. Then

$$ L(xy) $$

is the sum of:

1. the area under the curve from $t=1$ to $t=x$, which is $L(x)$;
2. the area under the curve from $t=x$ to $t=xy$.

Thus it remains to show that the second area equals $L(y)$.

Let $R$ be the region

$$ x\le t\le xy,\qquad 0\le u\le \frac1t . $$

Apply the geometric transformation

$$ (t,u)\longmapsto \left(\frac{t}{x},,xu\right). $$

This transformation compresses horizontal distances by a factor $x$ and stretches vertical distances by the same factor $x$. Therefore every area is multiplied by

$$ \frac1x\cdot x=1, $$

so the transformation preserves area.

Under this map:

$$ t=x \quad\longmapsto\quad \frac{t}{x}=1, $$

and

$$ t=xy \quad\longmapsto\quad \frac{t}{x}=y. $$

Furthermore, points on the hyperbola remain on the hyperbola. Indeed, if

$$ u=\frac1t, $$

then after the transformation

$$ u' = xu = \frac{x}{t} = \frac{1}{t/x} = \frac1{t'}. $$

Hence the boundary curve $u=1/t$ is carried onto itself.

Therefore the region $R$ is mapped exactly onto the region under

$$ u=\frac1t $$

from $t=1$ to $t=y$. Since the transformation preserves area, these two regions have equal area. Thus

$$ \text{Area}(R)=L(y). $$

Combining this with the decomposition at $t=x$,

$$ L(xy)=L(x)+\text{Area}(R) =L(x)+L(y). $$

Since $L(z)=\ln z$,

$$ \boxed{\ln(xy)=\ln x+\ln y}. $$

This is the desired geometric proof. ∎