## Content

### A rigorous approach to logarithms and exponentials

#### What does an exponential mean anyway?

Throughout this module, we've assumed that functions like $$f(x) = 2^x$$ are defined for all real numbers $$x$$. But are they really?

There's no problem defining $$2^x$$ when $$x$$ is a positive integer; this just means repeated multiplication and is certainly well defined.

There's also no problem when $$x$$ is a negative integer, using the index law

$2^{-n} = \dfrac{1}{2^n}.$

For example, $$2^{-3} = \dfrac{1}{8}$$.

Nor is there a problem when $$x$$ is a rational number. Using the index laws again,

$2^\frac{a}{b} = (2^a)^{\frac{1}{b}} = \sqrt[b]{2^a}.$

For example, $$2^\frac{3}{2} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2}$$.

However, it's not so clear what to do when $$x$$ is irrational. What does $$2^{\sqrt{2}}$$ mean? So far in the module, this issue has been quietly suppressed.

One approach we might use is continuity. We could take a sequence of rational numbers $$r_1, r_2, r_3, \dotsc$$ which approach $$\sqrt{2}$$, and consider $$2^{r_1}, 2^{r_2}, 2^{r_3}, \dotsc$$. If these numbers approach a limit, then we can call that limit $$2^{\sqrt{2}}$$.

In any case, it's not a trivial matter to define exponential functions like $$2^x$$ for irrational $$x$$.

One way to avoid all of the difficulties is to develop the entire story a different way, starting with the logarithmic function. As discussed in the Motivation section, this approach may appear less natural but is more rigorous and abstract.

#### The natural logarithm, rigorously

We begin by defining the natural logarithm as an integral.

###### Definition

For any real number $$x>0$$,

$\ln x = \int_1^x \dfrac{1}{t} \; dt.$

This equation was exercise 11. It is now a definition. Definition of the natural logarithm as an integral.
Detailed description

As an aside, note the standard fact that the integral of $$t^n$$ is

$\dfrac{1}{n+1} \, t^{n+1}.$

This is true for any $$n \neq -1$$. The integral above is of $$t^n$$ when $$n = -1$$, precisely the value of $$n$$ for which this standard formula does not apply.

From this definition, we can see immediately that $$\ln 1 = 0$$. We can also see, using the fundamental theorem of calculus, that the derivative of $$\ln x$$ is $$\dfrac{1}{x}$$. (We refer to the module Integration for the details). So the function $$\ln x$$ is increasing, for all $$x > 0$$, as its gradient $$\dfrac{1}{x}$$ is positive. (This can also be seen from the diagram above, where $$\ln x$$ is shown as a signed area.) It now follows that $$\ln x < 0$$, for $$x \in (0,1)$$, and $$\ln x > 0$$, for $$x \in (1, \infty)$$. It is clear that $$\ln x$$ is a continuous function, and it's not too difficult to show that $$\ln x \to +\infty$$ as $$x \to \infty$$, and that $$\ln x \to -\infty$$ as $$x \to 0$$.

Our newly defined function $$\ln x$$ and our previously defined function $$\log_e x$$ both have the same derivative $$\dfrac{1}{x}$$. Since $$\ln 1 = 0 = \log_e 1$$, it follows that $$\ln x$$ and $$\log_e x$$ are in fact the same function — that is, if you manage to overcome all the difficulties with our previous definition of $$\log_e x$$ and arrive at a well-defined function.

From the definition, it's not clear that the new function $$\ln x$$ behaves like a logarithm at all. However, we will now show directly that this new function obeys the logarithm laws. We use a similar method to exercise 10.

Take a positive number $$y$$, considered as a constant, and differentiate the two functions $$f(x) = \ln(xy)$$ and $$g(x) = \ln x + \ln y$$. We obtain

$f'(x) = \dfrac{y}{xy} = \dfrac{1}{x} \qquad \text{and} \qquad g'(x) = \dfrac{1}{x},$ so $$f'(x) = g'(x)$$, and therefore $$f(x), g(x)$$ differ by a constant. As $$f(1) = \ln y = g(1)$$, it follows that $$f(x) = g(x)$$. This shows that $\ln(xy) = \ln x + \ln y,$ and so proves one of the logarithm laws.

Using a similar method, we can show that

$\ln \Bigl( \dfrac{x}{y} \Bigr) = \ln x - \ln y$ and $\ln \Bigl( \dfrac{1}{x} \Bigr) = - \ln x.$

Next, consider the two functions $$f(x) = \ln(x^n)$$ and $$g(x) = n \, \ln x$$, where $$n$$ is any rational number.3

Differentiating these functions gives

$f'(x) = \dfrac{1}{x^n} \cdot n x^{n-1} = \dfrac{n}{x} \qquad \text{and} \qquad g'(x) = \dfrac{n}{x}.$

As $$f$$ and $$g$$ have the same derivative, they must differ by a constant. We further have $$f(1) = 0 = g(1)$$, so $$f(x) = g(x)$$ and we have proved the logarithm law

$\ln ( x^n ) = n \, \ln x.$

As our new function $$\ln x$$ obeys the familiar logarithm laws, we are justified in calling it a logarithm!

#### Exponentials, rigorously

Having established our new version of the natural logarithm function, we now turn to exponentials.

The function $$\ln x$$ mapping $$(0, \infty)$$ to $$\mathbb{R}$$ is a continuous function, strictly increasing from $$-\infty$$ to $$+\infty$$, and hence has an inverse function. This inverse function has domain $$\mathbb{R}$$ and range $$(0, \infty)$$. We will, for the moment, call this inverse function $$\exp x$$. So, by definition,

$\exp x = \ln^{-1}(x).$

We then have

\begin{align*} \exp(\ln x) &= x \quad \text{for all positive $$x$$,} \\ \ln( \exp x) &= x \quad \text{for all real $$x$$.} \end{align*}

We define the number $$e$$ to be $$\exp 1$$. So $$\exp 1 = e$$ and $$\ln e = 1$$. It will turn out that $${\exp x = e^x}$$; however this is not at all clear from the definition.

We can compute the derivative of $$\exp x$$, since it is the inverse of $$\ln x$$, and we know that the derivative of $$\ln x$$ is $$\dfrac{1}{x}$$. Let $$y = \exp x$$. Then $$x = \ln y$$ and we have

$\dfrac{dx}{dy} = \dfrac{1}{y},$ so $\dfrac{d}{dx} \, \exp x = \dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}} = y = \exp x.$

We can also show that $$\exp$$ satisfies the index laws. For instance, we have

\begin{align*} \ln \bigl( \exp x \cdot \exp y \bigr) &= \ln (\exp x) + \ln (\exp y)\\ &= x+y\\ &= \ln \bigl(\exp (x+y)\bigr). \end{align*}

Here we just used a logarithm law and the fact that $$\ln$$ and $$\exp$$ are inverses. Since the function $$\ln$$ is one-to-one and we have just shown that $$\ln \bigl( \exp x \cdot \exp y \bigr) = \ln \bigl(\exp (x+y)\bigr)$$, we conclude the index law

$\exp x \cdot \exp y = \exp(x+y).$

Using a similar method, we can show that $$\exp$$ also obeys the index law

$\exp(x-y) = \dfrac{\exp x}{\exp y}.$

For the remaining index law, take a rational number $$r$$; we will show $$\exp(rx) = (\exp x)^r$$. We observe

\begin{align*} \ln \bigl( (\exp x)^r \bigr) &= r \, \ln ( \exp x)\\ &= rx\\ &= \ln \bigl( \exp(rx) \bigr), \end{align*} where we just used a logarithm law and the fact that $$\ln$$ and $$\exp$$ are inverses. We have $$\ln \bigl( (\exp x)^r \bigr) = \ln \bigl( \exp(rx) \bigr)$$, and cancelling $$\ln$$'s establishes the index law $(\exp x)^r = \exp(rx),$ for any rational number $$r$$. Since $$\exp 1 = e$$, we then have, for any rational number $$r$$, \begin{align*} \exp r = \exp(r \cdot 1) = (\exp 1)^r = e^r. \end{align*}

We can now use this to define irrational powers. We declare for any real number $$x$$, possibly irrational, that

$e^x = \exp x.$

We can go further and use this idea to define any real power $$a^x$$ of any positive number $$a$$. When $$r$$ is rational, we have from the index law above,

$\exp(r \, \ln a) = \bigl(\exp (\ln a ) \bigr)^r = a^r,$ and so for any real number $$x$$ we can define $$a^x$$ as follows.
###### Definition

For any real number $$x$$ and any $$a>0$$,

$a^x = \exp(x \, \ln a).$

It's not difficult to show that $$a^x$$ varies continuously with $$a$$ and $$x$$.

Exercise 20

Let $$\alpha$$ be any real number and let $$f(x) = x^\alpha$$, for $$x > 0$$. Using the above definition for $$x^\alpha$$, prove that $$f'(x) = \alpha x^{\alpha - 1}$$.

In this way, we have rigorously defined the functions $$\ln x$$ and $$e^x$$, found their derivatives, established the index and logarithm laws, and rigorously defined irrational powers.

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