Peter Alfeld, --- Department of Mathematics, --- College of Science --- University of Utah

Why Logarithms?

It turns out that the only (non-zero differentiable) functions with the key property

$\displaystyle f(xy) = f(x) + f(y)\qquad(*)$

are logarithmic functions. To see this we derive a differential equation for $ f$ whose solution is a logarithm. Differentiating in $ (*)$ with respect to $ x$ gives

$\displaystyle yf'(xy) = f'(x)\quad\Longrightarrow\quad f'(xy) = \frac{f'(x)}{y}.$

Similarly, differentiating with respect to $ y$ gives

$\displaystyle xf'(xy) = f'(y)\quad\Longrightarrow\quad f'(xy) = \frac{f'(y)}{x}.$


$\displaystyle \frac{f'(x)}{y} = \frac{f'(y)}{x}.$

In the special case $ y=1$ this turns into

$\displaystyle f'(x) = \frac{f'(1)}{x}.$

From $ (*)$ if follows that $ f(1) = 0$. The solution of this initial value problem is

$\displaystyle f(x) = C\ln x = \log_b x$

where $ b = e^{1/C}$ and $ C= f'(1)$ is an arbitrary constant.

So only logarithms work. However, if you want to settle for properties that are more complicated than $ (*)$ there are other possibilities. For example, you may want to think about how to construct a slide rule based on the identity

$\displaystyle \sin x \sin y =
\frac{1}{2}[\cos(x-y) - \cos(x+y)].$