General Exponential Functions

For fixed a, we want the exponential function a^{x} to have at least the properties:

(i) a^{x+y} = a^{x}a^{y}; (ii) a^{0} = 1; (iii) a^{1} = a; (iv) a^{x} is continuous in x. (1)

Many other properties follow easily from these. If we put y = −x in (i), we get 1 = a^{0} = a^{x+(−x)} = a^{x}a^{−x}, so that

a^{−x} = \frac{1}{a^{x}} , in particular, a^{−1} = \frac{1}{a^{1}} = \frac{1}{a} .  (2)

If we iterate (i), we get for any positive integer n

a^{nx} = a^{x+x+...+x} = a^{x}a^{x} … a^{x} = (a^{x})^{n},  (3)

the nth power of a^{x} in the ordinary sense. By taking x = 1, we see that a^{n} is the nth power of a as usually understood.

Existence and uniqueness We cannot expect to define a^{x} for all real x and a. We note that axiom (i) gives a^{x} = (a^{x/2})^{2} ≥ 0. Also, equation (2) requires a^{x} ≠ 0, so that we must have a^{x} > 0. Then (iii) gives a = a^{1} > 0.

Theorem 4 Given a > 0, there exists a unique function a^{x} that satisfies the axioms equation (1). It satisfies a^{x} > 0 for all x.

For a rational number x = p/q, with p and q integers and q > 0, equation (3) gives (a^{x})^{q} = a^{qx} = a^{p}, which determines a^{x} uniquely. This, with continuity, is enough to determine a^{x} uniquely for all real x. We defer the proof of existence of a^{x}.

Applications We make some immediate applications of Theorem 4. The method in each case is to show that a certain function satisfies the axioms equation (1) for a^{x} for some a, and therefore by the uniqueness in Theorem 4 must coincide with a^{x}.

Our first application is the elementary fact that 1^{x} = 1 for all x. (In detail, define f(x) = 1 for all x; then f(x+y) = 1 = f(x)f(y), f(0) = 1 = f(1), and the constant function f is clearly continuous; hence f(x) = 1^{x}.)

Our second application generalizes equation (3).

Theorem 5 Given a > 0, we have a^{zx} = (a^{z})^{x} for any real z and x.

Proof Fix a and z and define g(x) = a^{zx} for all x. Then g(x+y) = a^{z(x+y)} = a^{zx+zy} = a^{zx}a^{zy} = g(x)g(y), g(0) = a^{0} = 1, g(1) = a^{z}, and g is continuous. Thus g(x) satisfies the axioms for (a^{z})^{x} and must coincide with it.

Our third application shows that a^{x} is multiplicative in a.

Theorem 6 Given a > 0 and b > 0, we have (ab)^{x} = a^{x}b^{x} for all x.

Proof We put h(x) = a^{x}b^{x} for all x. Then h(x+y) = a^{x+y}b^{x+y} = a^{x}a^{y}b^{x}b^{y} = h(x)h(y), h(0) = 1, h(1) = a^{1}b^{1} = ab, and h is continuous. Thus h satisfies the axioms for (ab)^{x}, and so must be (ab)^{x}.

Differentiation In order to differentiate a^{x}, we must take the limit of

\frac{a^{x+h} − a^{x}}{h} = \frac{a^{x}a^{h} − a^{x}}{h} = a^{x} \frac{a^{h} − 1}{h}

as h → 0. As this limit is not obvious, we give it a name.

Definition 7 Given a > 0, we define the natural logarithm log a (or ln a) of a as the limit

log a = ln a = lim_{h→0} \frac{a^{h} − 1}{h} .  (8)

When we plug this in, we obtain the answer

\frac{d}{dx}(a^{x}) = a^{x} log a .  (9)

Theorem 10 We have log a > 0 whenever a > 1, log 1 = 0, and log a < 0 whenever 0 < a < 1.

Proof As a^{x} > 0 always, equation (9) shows that a^{x} is an increasing, decreasing, or constant function according as to whether log a > 0, log a < 0, or log a = 0. Comparison of a^{0} = 1 and a^{1} = a shows which case applies.

The main property of the function log x follows easily from equation (9).

Theorem 11 We have log ab = log a + log b for any a > 0 and b > 0.

Proof When we differentiate in Theorem 6, the product rule gives

(ab)^{x} log ab = a^{x}(log a)b^{x} + a^{x}b^{x} log b,

in which we put x = 0.

Similarly, we can differentiate in Theorem 5.

Theorem 12 We have log(a^{z}) = z log a for any a > 0 and any z.

Proof Differentiation of a^{zx} = (a^{z})^{x} with respect to x gives

a^{zx}z log a = (a^{z})^{x} log(a^{z}) .

Again we put x = 0.