That counterpart to last week's development of the exponential function is the logarithm. As anyone who ever played with the antique known as a slide rule, the logarithmic function converts multiplication into addition:

Log(a b) = Log(a) + Log(b) , (1)

but that's getting ahead of ourselves.

Assume that (1) is all we know: Consider a real function G defined on at least some part of the real numbers, and having the property

G(a b) = G(a) + G(b) , (2)

for any real numbers a and b. What can we say about this function?

First, take a = 0. Then we have

G(0 b) = G(0) + G(b) , or

G(0) = G(0) + G(b) . (3)

So either G(b) = 0 for all b, or G(0) is undefined. (Yes, we can say that G(0) = ∞, but I'd rather not go there right now.) Since G = 0 is uninteresting, we'll assume that G(0) is undefined. For now, at least, we'll restrict ourselves to x > 0 and so avoid the problem.

If we set a = 1 in (2) we get

G(b) = G(1) + G(b) , which means that

G(1) = 0 , (4)

which implies
G[a (1/a)] = G(a) + G(1/a) = G(1) = 0 , or

G(1/a) = - G(a) . (5)

For any integer M we can repeatedly apply (2) and get

G(a^{M}) = M G(a) . (6)

If we let b = a^{M}, or a = b^{1/M}, then by turning around (6) we get

G(b^{1/M}) = (1/M) G(b) , (7)

and combining the two we get

G(x^{M/N}) = (M/N) G(x) (8)
for any integers M and N.

Furthermore, we can play the same trick as we did last week and extend this to all positive real numbers y:

G(x^{y}) = y G(x) . (9)

What about continuity? I'm going to be a little more physicist-like this week (though last week was probably not properly mathematical anyway), and go through it rather quickly. Let's take two numbers, x > 0 and y > -x. Then we can always write

x + y = x Q , Q = 1+y/x > 0 ,

and we want to study the behavior of

G(x + y) = G(x Q) = G(x) + G(Q) as y → 0 or Q → 1.

When we put it like that, the answer is obvious. Write Q = h^{1/N}. Then as N → ∞, Q → ∞ 1, and

lim_{N → ∞} G(Q) = lim_{N → ∞} G(h)/N → 0 . (10)

Thus

lim_{y→0} |G(x+y) - G(x)| = 0 , (11)

and G(x) is a continuous function for x > 0.

Using the same tricks, let's look at the derivative, if it exists:

G'(x) = lim_{y→∞0} [G(x+y)-G(x)]/y , (12)

With the above substitutions we can write this as

G'(x) = lim_{N→∞} [G(x h^{1/N}) - G(x)]/[x (h^{1/N}-1) , (13)

which reduces to
G'(x) = [G(h)/x]/ [lim_{N→∞} N (h^{1/N} - 1)] . (14)

Last week we showed that the limit in the denominator of (14) is finite, so it follows that G'(x) is well defined for x > 0.

To evaluate the derivative, we do the standard tricks:

d/dy G(x y) = x G'(x y) = d/dy [G(x) + G(y)] = G'(y)

and set y = 1. Then

G'(x) = G'(1)/x , x > 0 . (15)

Thus each and every log function (2) is uniquely defined once we define G'(1). Since G(1) = 0, we can integrate (15) to find

G(x) = G'(1) ∫_{1}^{x} dt / t . (16)

The logical prototype logarithm function (which we could call the, ahem, natural logarithm) is then the function with G'(1) = 1:

ln x = ∫_{1}^{x} dt / t . (17)

Which is exactly how my college calculus course started out.

One more thing: we want to show that ln x and exp x are inverse functions of one another. Write

H(x) = ln(exp x) . (18)

Obviously H(0) = ln(exp 0) = ln(1) = 0. Furthermore,

H'(x) = [d/dx exp x]/exp x = exp x/exp x = 1 , (19)

so H(x) = x. We could go the opposite way as well and show that

exp(ln x) = x . (20)

That concludes this little exploration of the log and exponential functions. Next week we'll take a look at sines and cosines.

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