Logarithms and Exponentials

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Last update: Sunday, 2 October 2022.

Goal

Here we'll build the logarithm and exponential from the bottom up. This is possible using only two high-powered theorems: the Fundamental Theorem of Calculus and the Inverse Function Theorem.

Refreshers

First we'll review the necessary theory for building up our functions.

Basic Properties of Integrals

We shall need the following two properties of the definite integral. First, \( \int_a^a f(x)\, dx = 0 \) for every \( f \) and every \( a \). Second, \( \int_b^a f(x)\, dx = -\int_a^b f(x)\, dx \) for every \( a \) and \( b \).

Fundamental Theorem of Calculus

First we recall the Fundamental Theorem of Calculus.

Let \( f \) be a continuous function defined on interval \( I \).

Let \( a \in I \). The function \( F(x) = \int_{t = a}^x f(t)\, dt \) defined on \( I \) is an antiderivative of \( f \).
Let \( a, b \in I \). If \( G \) is any antiderivative of \( f \), then \( \int_a^b f(t)\, dt = G(b) - G(a) \).

Our main use of this theorem will be to make the definition of the logarithm. Recall that the second part of this theorem yields that antiderivatives of \( f \) may differ only by a constant. In particular, for all antiderivatives \( G_1 \) and \( G_2 \) of \( f \) on \( I \) we have \( G_1(x) - G_1(a) = \int_a^x f(x)\, dx = G_2(x) - G_2(a) \); thus \( G_2(x) - G_1(x) = G_2(a) - G_1(a) \), and the right hand side is a constant. We often relabel and express this as \( G_2(x) = G_1(x) + C \) for some constant \( C \).

Injective Functions

Recall that a function is injective (or one-to-one) when \( f(a) = f(b) \) implies \( a = b \) for all \( a \) and \( b \) in the domain of \( f \). Using the Intermediate Value Theorem, the following is relatively easy to establish.

Let \( f \) be a continuous function defined on an interval \( I \). The function \( f \) is injective precisely when it is either strictly increasing or strictly decreasing on \( I \).

This, together with the First Derivative Test means we can use the following simple test for injectivity.

Let \( f \) be a differentiable function on open interval \( I \). If \( f'(x) \neq 0 \) for all \( x \in I \), then \( f \) is injective.

Recall that an injective function \( f \) has an inverse function \( f^{-1} \). The inverse function is determined by \( f(x) = y \) exactly when \( x = f^{-1}(y) \); another way to express this is \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \). This yields that \( \dom(f^{-1}) = \ran(f) \) and \( \ran(f^{-1}) = \dom(f) \).

Inverse Function Theorem

Next we recall the Inverse Function Theorem.

Let \( f \) be a differentiable injective function and suppose \( f(a) = b \). If \( f'(a) \neq 0 \), then the inverse function \( f^{-1} \) is differentiable at \( b \) and satisfies \( (f^{-1})'(b) = \frac{1}{f'(a)} \).

Note that this is sometimes written \( (f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))} \).

Functions \( L(x) \) and \( E(x) \)

To begin, we define a new function \( L(x) = \int_1^x \frac{1}{t}\, dt \) for all \( x > 0 \). Notice that \( f(x) = \frac{1}{x} \) is continuous on the interval \( I = (0, \infty) \), so the function \( L \) is an antiderivative of \( f \) by the Fundamental Theorem of Calculus.

Properties of \( L(x) \)

In this section, we show several interesting properties of \( L(x) \).

Basic Properties

To begin, note \( L(1) = \int_1^1 \frac{1}{x}\, dt = 0 \) by Basic Properties of Integrals. Moreover \( L'(x) = \frac{1}{x} > 0 \) for all \( x > 0 \), so \( L \) is a strictly increasing function.

Range

Next we show that \( L(x) \) has range \( (-\infty, \infty) \). First, we show that if \( r > 0 \), then there is an \( x > 0 \) such that \( L(x) > r \). To do so, recall that \( \int_a^b g(x)\, dx \) represents the area between the graph of \( g \) and the \( x \)-axis on the interval \( [a, b] \). Given \( r > 0 \), we let \( k \) be any positive integer with \( r < \frac{k}{2} \). Now we use \( k \) uneven rectangles to approximate the area beneath \( f(x) = \frac{1}{x} \) on \( [1, 2^{k+1}] \). In particular, the \( i^{\mathrm{th}} \) rectangle sits above the interval \( [2^{i-1}, 2^i] \), and has height \( \frac{1}{2^i} \). Because \( f(x) \) is strictly decreasing and the right hand endpoint \( f(2^i) = \frac{1}{2^i} \), each of these rectangles lies strictly beneath the graph of \( f \) (it's a good idea to draw a picture to convince yourself of this fact). Note that for all \( i \geq 0 \) we have \( \frac{2^{i+1} - 2^i}{2^{i+1}} = \frac{2^i(2 - 1)}{2^{i+1}} = \frac{1}{2} \). In particular, we have constructed \( k \) rectangles beneath the curve each having area \( \frac{1}{2} \); thus we may underestimate the area between the graph of \( f \) and the \( x \)-axis on \( [1, 2^k] \) by \( \frac{k}{2} \).

The above argument shows that \( 0 < r < \frac{k}{2} \leq L(2^k) \). Because \( L \) is differentiable, \( L \) is also continuous. Thus we may apply the Intermediate Value Theorem to obtain a real number \( c \) for which \( L(c) = r \). As this argument works for every \( r > 0 \), we have shown that the range of \( L \) includes \( (0, \infty) \). To see that the range also includes \( (-\infty, 0) \), we can make a similar argument for all \( r < 0 \) on \( [\frac{1}{2^{k+1}}, 1] \) using an integer \( k \) with \( |r| < \frac{k}{2} \). The \( i^{\mathrm{th}} \) rectangle has width \( \frac{1}{2^i} \) and height \( 2^{i-1} \), again yielding an area of \( \frac{1}{2} \) for each rectangle.

Because \( L \) has range \( (-\infty, \infty) \) and is strictly increasing, we may conclude that there is a real number \( e > 1 \) for which \( L(e) = 1 \).

Behaviour on Products

Next we consider and compute \( L(ab) \) for \( a, b > 0 \). Note primarily that for all \( b > 0 \) we have may compute the derivative \[ \ddx{L(xb)} = \frac{1}{x \cdot b} \cdot \ddx{x \cdot b} = \frac{1}{x \cdot b} \cdot b = \frac{1}{x} = \ddx {L(x)} \] Thus \( L(xb) \) and \( L(x) \) are both antiderivatives of \( f(x) = \frac{1}{x} \), and we may express \( L(x \cdot b) = L(x) + C \) for some constant \( C \). Evaluating at \( x = 1 \) we obtain \( L(b) = L(1) + C = 0 + C = C \), so \( L(x \cdot b) = L(x) + L(b) \) for all \( x > 0 \). Relabeling, we may express this as \( L(a \cdot b) = L(a) + L(b) \).

Note that for all \( a > 0 \), we have \( L(\tfrac{1}{a}) + L(a) = L(\tfrac{1}{a} \cdot a) = L(1) = 0 \). Thus \( L(\frac{1}{a}) = -L(a) \) for all \( a > 0 \).

Summary

We can summarize our work above as follows.

Let \( L(x) = \int_1^x \frac{1}{t}\, dt \) for all \( x > 0 \).

The function \( L \) has \( \dom(L) = (0, \infty) \) and \( \ran(L) = (-\infty, \infty) \).
The function \( L \) satisfies \( L(1) = 0 \) and \( L(e) = 1 \).
The derivative \( L'(x) = \frac{1}{x} \).
The function \( L \) is strictly increasing on \( (0, \infty) \).
For all \( a, b > 0 \) we have \( L(a \cdot b) = L(a) + L(b) \).
For all \( a > 0 \) we have \( \ln(\tfrac{1}{a}) = -\ln(a) \).

Inverse Function \( E(x) \)

Because \( L \) is strictly increasing, we obtain that \( L \) is injective. Thus \( L \) has an inverse function; we denote \( E(x) = L^{-1}(x) \). In this section we explore properties of this inverse function.

Basic Properties

Recall that \( L(1) = 0 \) and \( L(e) = 1 \) by our prior work. Thus we have \( E(0) = 1 \) and \( E(1) = e \). Moreover, because \( \dom(L) = (0, \infty) \) and \( \ran(L) = (-\infty, \infty) \) we obtain \( \dom(E) = (-\infty, \infty) \) and \( \ran(E) = (0, \infty) \).

Derivative

We use the Inverse Function Theorem to compute \( E'(x) \) as follows. \[ E'(x) = (L^{-1})'(x) = \frac{1}{L'(L^{-1}(x))} = \frac{1}{1 / L^{-1}(x)} = L^{-1}(x) = E(x) \] Thus \( E'(x) = E(x) \).

Because \( \ran(E) = (0, \infty) \), we see that \( E'(x) > 0 \) for all \( x \); thus \( E \) is a strictly increasing function.

Behaviour on Sums

Let \( a \) and \( b \) be real numbers. Using properties of \( L \) determined earlier, we have \( L(x \cdot y) = L(x) + L(y) \) for all \( x, y > 0 \). As \( L(E(x)) = L(L^{-1}(x)) = x \) for all \( x \), we have \[ L(E(a) \cdot E(b)) = L(E(a)) + L(E(b)) = a + b = L(E(a + b)) . \] Because \( L \) is injective, we have \( E(a) \cdot E(b) = E(a + b) \).

Summary

We can summarize our work above as follows.

Let \( E(x) = L^{-1}(x) \) for all real \( x \).

We have \( \dom(E) = (-\infty, \infty) \) and \( \ran(E) = (0, \infty) \).
We have \( E(0) = 1 \) and \( E(1) = e \).
The derivative \( E'(x) = E(x) \).
The function \( E \) is strictly increasing on \( (-\infty, \infty) \).
For all \( a \) and \( b \) we have \( E(a + b) = E(a)E(b) \).

Remark on Notation

Because these functions are so important, we give them special names. The function \( L(x) \) is the natural logarithm, and is typically denoted \( \ln(x) \). The function \( E(x) \) is the natural exponential, and is typically denoted \( \exp(x) \).

In the rest of this document, I will use the more standard names described in this section.

Other Exponential and Logarithmic Functions

The above theory provides us with a useful framework for a whole new class of functions.

Base \( a \) Exponential Functions

Given \( a > 0 \), we define the function \( a^x = \exp(x \cdot \ln(a)) \) for all real \( x \). The function \( a^x \) is the base \( a \) exponential function. As a special case, the base \( e \) exponential function satisfies \( e^x = \exp(x \ln(e)) = \exp(x) \).

Basic Properties of \( a^x \)

Observe that \( a^x > 0 \) for all real \( x \) because \( \exp(y) > 0 \) for all real \( y \). Moreover, \( a^0 = \exp(\ln(a) \cdot 0) = \exp(0) = 1 \) and \( a^1 = \exp(\ln(a) \cdot 1) = \exp(\ln(a)) = a \) for all \( a \).

Behaviour of \( a^x \) on Sums

By the properties discussed above for \( \exp \), we have \[ a^x \cdot a^y = \exp(x \ln(a)) \cdot \exp(y \ln(a)) = \exp(x \ln(a) + y \ln(a)) = \exp((x + y) \ln(a)) = a^{x + y} . \]

Derivative of \( a^x \)

We compute the derivative of this function using the chain rule: \[ \ddx{a^x} = \ddx{\exp(x \cdot \ln(a))} = \exp(x \cdot \ln(a)) \cdot \ddx{x \cdot \ln(a)} = \exp(x \cdot \ln(a)) \cdot \ln(a) = \ln(a) a^x . \] If \( a < 1 \), then \( \ln(a) < 0 \) and \( a^x \) is a strictly decreasing function by the First Derivative Test. If \( a > 1 \), then \( \ln(a) > 0 \) and \( a^x \) is a strictly increasing function by the First Derivative Test. If \( a = 1 \), then \( \ln(a) = 0 \) and \( a^x \) is a constant function; because \( 1^0 = 1 \) by our previous observations, we see \( 1^x = 1 \) for all \( x \).

Base \( a \) Logarithms

Our work above shows that the base \( a \) exponential function has an inverse function provided \( 0 < a \neq 1 \). Thus, if \( 0 < a \neq 1 \), then \( a^x \) is injective function and has an inverse function; let \( \log_a(x) \) denote the inverse function of \( a^x \). As a special case, note that \( \log_e(x) \) is the inverse function of \( e^x = \exp(x) \), so \( \log_e(x) = \ln(x) \). Note that \( \log_1(x) \) is not a well-defined function, because \( 1^x \) is not injective (and thus has no inverse function)!

Basic Properties of \( \log_a(x) \)

Because \( a^0 = 1 \) and \( a^1 = a \), we have \( \log_a(1) = 0 \) and \( \log_a(a) = 1 \). Moreover, \( \log_a(x) \) has \( \dom(\log_a) = (0, \infty) \) and \( \ran(\log_a) = (-\infty, \infty) \).

Formula for \( \log_a(x) \)

We claim that \( \log_a(x) = \frac{\ln(x)}{\ln(a)} \) for all \( a > 0 \). To see this, let \( M(x) = \frac{\ln(x)}{\ln(a)} \) and compute as follows for all \( x > 0 \) and all real \( y \).

\begin{align*} a^{M(x)} &= a^{\frac{\ln(x)}{\ln(a)}} = \exp\left(\frac{\ln(x)}{\ln(a)} \cdot \ln(a)\right) = \exp(\ln(x)) = x \\ M(a^y) &= \frac{\ln(a^y)}{\ln(a)} = \frac{\ln(\exp(y \cdot \ln(a)))}{\ln(a)} = \frac{y \cdot \ln(a)}{\ln(a)} = y \end{align*}

Hence \( M(x) \) is the inverse function of \( a^x \), and we have shown that \( \log_a(x) = \frac{\ln(x)}{\ln(a)} \) as claimed. This is sometimes called the change of base formula.

Derivative of \( \log_a(x) \)

We can compute the derivative of \( \log_a(x) \) using the formula we just discovered. \[ \ddx{\log_a(x)} = \ddx{\frac{\ln(x)}{\ln(a)}} = \frac{1}{\ln(a)} \cdot \ddx{\ln(x)} = \frac{1}{\ln(a)} \cdot \frac{1}{x} = \frac{1}{x \ln(a)} . \] Recall that \( \ln(a) > 0 \) when \( a > 1 \) and \( \ln(a) < 0 \) when \( a < 1 \). Together with the fact that \( \frac{1}{x} > 0 \) for all \( x > 0 \), we have by the First Derivative Test that \( \log_a(x) \) is strictly increasing if \( a > 1 \) and strictly decreasing if \( a < 1 \). Note that \( \log_a(x) \) is undefined for \( a = 1 \) because \( 1^x \) is not injective!

Exponential Identity for \(log_a(x) \)

As a result of the above work, we can now compute as follows for all \( a, b > 0 \) and real \( r \): \[ \log_a(b^r) = \frac{\ln(b^r)}{\ln(a)} = \frac{\ln(\exp(r \ln(b)))}{\ln(a)} = \frac{r \ln(b)}{\ln(a)} = r \frac{\ln(b)}{\ln(a)} = r \log_a(b) . \] Because \( \ln(x) = \log_e(x) \), we have \( \ln(a^r) = r \ln(a) \) for all \( a > 0 \) and \( \) As a simple result of the identity above, we have the following product identity for exponential functions. \[ a^{rs} = \exp(rs \ln(a)) = \exp(s(r \ln(a))) = \exp(s \ln(a^r)) = (a^r)^s . \]

Summary

We can summarize our work above as follows.

Let \( a > 0 \).

The function \( a^x \) has domain \( (-\infty, \infty) \) and range \( (0, \infty) \).
The function \( \log_a(x) \) has domain \( (0, \infty) \) and range \( (-\infty, \infty) \).
The base \( a \) exponential satisfies the following for all real numbers \( r \) and \( s \).
1. \( a^0 = 1 \) and \( a^1 = a \).
2. If \( a > 1 \), then \( a^x \) is strictly increasing; if \( a < 1 \), then \( a^x \) is strictly decreasing.
3. \( a^{r + s} = a^r \cdot a^s \).
4. \( a^{rs} = (a^r)^s \).
If \( a \neq 1 \), then the base \( a \) logarithm satisfies the following for all real numbers \( r, s > 0 \).
1. \( \log_a(1) = 0 \) and \( \log_a(a) = 1 \).
2. If \( a > 1 \), then \( \log_a(x) \) is strictly increasing; if \( a < 1 \), then \( \log_a(x) \) is strictly decreasing.
3. \( \log_a(rs) = \log_a(r) + \log_a(s) \).
4. \( \log_a(x) = \frac{\ln(x)}{\ln(a)} \).
For all \( a, b > 0 \) and all real \( r \) we have \( \log_a(b^r) = r \log_a(b) \).