## Section5.4The logarithm and the exponential

Note: 1 lecture (optional, requires the optional sections Section 3.5, Section 3.6, Section 4.4)

We now have the tools required to properly define the exponential and the logarithm that you know from calculus so well. We start with exponentiation. If $$n$$ is a positive integer, it is obvious to define

\begin{equation*} x^n := \underbrace{x \cdot x \cdot \cdots \cdot x}_{n \text{ times}} . \end{equation*}

It makes sense to define $$x^0 := 1\text{.}$$ For negative integers, let $$x^{-n} := \nicefrac{1}{x^n}\text{.}$$ If $$x > 0\text{,}$$ define $$x^{1/n}$$ as the unique positive $$n$$th root. Finally, for a rational number $$\nicefrac{n}{m}$$ (in lowest terms), define

\begin{equation*} x^{n/m} := {\bigl(x^{1/m}\bigr)}^n . \end{equation*}

It is not difficult to show we get the same number no matter what representation of $$\nicefrac{n}{m}$$ we use, so we do not need to use lowest terms.

However, what do we mean by $$\sqrt{2}^{\sqrt{2}}\text{?}$$ Or $$x^y$$ in general? In particular, what is $$e^x$$ for all $$x\text{?}$$ And how do we solve $$y=e^x$$ for $$x\text{?}$$ This section answers these questions and more.

### Subsection5.4.1The logarithm

It is convenient to define the logarithm first. Let us show that a unique function with the right properties exists, and only then will we call it the logarithm.

#### Proof.

To prove existence, we define a candidate and show it satisfies all the properties. Let

\begin{equation*} L(x) := \int_1^x \frac{1}{t}\,dt . \end{equation*}

Obviously, i holds. Property ii holds via the second form of the fundamental theorem of calculus (Theorem 5.3.3).

To prove property iv, we change variables $$u=yt$$ to obtain

\begin{equation*} L(x) = \int_1^{x} \frac{1}{t}\,dt = \int_y^{xy} \frac{1}{u}\,du = \int_1^{xy} \frac{1}{u}\,du - \int_1^{y} \frac{1}{u}\,du = L(xy)-L(y) . \end{equation*}

Let us prove iii. Property ii together with the fact that $$L'(x) = \nicefrac{1}{x} > 0$$ for $$x > 0\text{,}$$ implies that $$L$$ is strictly increasing and hence one-to-one. Let us show $$L$$ is onto. As $$\nicefrac{1}{t} \geq \nicefrac{1}{2}$$ when $$t \in [1,2]\text{,}$$

\begin{equation*} L(2) = \int_1^2 \frac{1}{t} \,dt \geq \nicefrac{1}{2} . \end{equation*}

By induction, iv implies that for $$n \in \N$$

\begin{equation*} L(2^n) = L(2) + L(2) + \cdots + L(2) = n L(2) . \end{equation*}

Given $$y > 0\text{,}$$ by the Archimedean property of the real numbers (notice $$L(2) > 0$$), there is an $$n \in \N$$ such that $$L(2^n) > y\text{.}$$ By the intermediate value theorem there is an $$x_1 \in (1,2^n)$$ such that $$L(x_1) = y\text{.}$$ We get $$(0,\infty)$$ is in the image of $$L\text{.}$$ As $$L$$ is increasing, $$L(x) > y$$ for all $$x > 2^n\text{,}$$ and so

\begin{equation*} \lim_{x\to\infty} L(x) = \infty . \end{equation*}

Next $$0 = L(\nicefrac{x}{x}) = L(x) + L(\nicefrac{1}{x})\text{,}$$ and so $$L(x) = - L(\nicefrac{1}{x})\text{.}$$ Using $$x=2^{-n}\text{,}$$ we obtain as above that $$L$$ achieves all negative numbers. And

\begin{equation*} \lim_{x \to 0} L(x) = \lim_{x \to 0} -L(\nicefrac{1}{x}) = \lim_{x \to \infty} -L(x) = - \infty . \end{equation*}

In the limits, note that only $$x > 0$$ are in the domain of $$L\text{.}$$

Let us prove v. Fix $$x > 0\text{.}$$ As above, iv implies $$L(x^n) = n L(x)$$ for all $$n \in \N\text{.}$$ We already found that $$L(x) = - L(\nicefrac{1}{x})\text{,}$$ so $$L(x^{-n}) = - L(x^n) = -n L(x)\text{.}$$ Then for $$m \in \N$$

\begin{equation*} L(x) = L\Bigl({(x^{1/m})}^m\Bigr) = m L\bigl(x^{1/m}\bigr) . \end{equation*}

Putting everything together for $$n \in \Z$$ and $$m \in \N\text{,}$$ we have $$L(x^{n/m}) = n L(x^{1/m}) = (\nicefrac{n}{m}) L(x)\text{.}$$

Uniqueness follows using properties i and ii. Via the first form of the fundamental theorem of calculus (Theorem 5.3.1),

\begin{equation*} L(x) = \int_1^x \frac{1}{t}\,dt \end{equation*}

is the unique function such that $$L(1) = 0$$ and $$L'(x) = \nicefrac{1}{x}\text{.}$$

Having proved that there is a unique function with these properties, we simply define the logarithm or sometimes called the natural logarithm:

\begin{equation*} \ln(x) := L(x) . \end{equation*}

Mathematicians usually write $$\log(x)$$ instead of $$\ln(x)\text{,}$$ which is more familiar to calculus students. For all practical purposes, there is only one logarithm: the natural logarithm. See Exercise 5.4.2.

### Subsection5.4.2The exponential

Just as with the logarithm we define the exponential via a list of properties.

#### Proof.

Again, we prove existence of such a function by defining a candidate and proving that it satisfies all the properties. The $$L = \ln$$ defined above is invertible. Let $$E$$ be the inverse function of $$L\text{.}$$ Property i is immediate.

Property ii follows via the inverse function theorem, in particular Lemma 4.4.1: $$L$$ satisfies all the hypotheses of the lemma, and hence

\begin{equation*} E'(x) = \frac{1}{L'\bigl(E(x)\bigr)} = E(x) . \end{equation*}

Let us look at property iii. The function $$E$$ is strictly increasing since $$E'(x) = E(x) > 0\text{.}$$ As $$E$$ is the inverse of $$L\text{,}$$ it must also be bijective. To find the limits, we use that $$E$$ is strictly increasing and onto $$(0,\infty)\text{.}$$ For every $$M > 0\text{,}$$ there is an $$x_0$$ such that $$E(x_0) = M$$ and $$E(x) \geq M$$ for all $$x \geq x_0\text{.}$$ Similarly, for every $$\epsilon > 0\text{,}$$ there is an $$x_0$$ such that $$E(x_0) = \epsilon$$ and $$E(x) < \epsilon$$ for all $$x < x_0\text{.}$$ Therefore,

\begin{equation*} \lim_{n\to -\infty} E(x) = 0 , \qquad \text{and} \qquad \lim_{n\to \infty} E(x) = \infty . \end{equation*}

To prove property iv, we use the corresponding property for the logarithm. Take $$x, y \in \R\text{.}$$ As $$L$$ is bijective, find $$a$$ and $$b$$ such that $$x = L(a)$$ and $$y = L(b)\text{.}$$ Then

\begin{equation*} E(x+y) = E\bigl(L(a)+L(b)\bigr) = E\bigl(L(ab)\bigr) = ab = E(x)E(y) . \end{equation*}

Property v also follows from the corresponding property of $$L\text{.}$$ Given $$x \in \R\text{,}$$ let $$a$$ be such that $$x = L(a)$$ and

\begin{equation*} E(qx) = E\bigl(qL(a)\bigr) = E\bigl(L(a^q)\bigr) = a^q = {E(x)}^q . \end{equation*}

Uniqueness follows from i and ii. Let $$E$$ and $$F$$ be two functions satisfying i and ii.

\begin{equation*} \frac{d}{dx} \Bigl( F(x)E(-x) \Bigr) = F'(x)E(-x) - E'(-x)F(x) = F(x)E(-x) - E(-x)F(x) = 0 . \end{equation*}

Therefore, by Proposition 4.2.6, $$F(x)E(-x) = F(0)E(-0) = 1$$ for all $$x \in \R\text{.}$$ Next, $$1 = E(0) = E(x-x) = E(x)E(-x)\text{.}$$ Then

\begin{equation*} 0 = 1-1 = F(x)E(-x) - E(x)E(-x) = \bigl(F(x)-E(x)\bigr) E(-x) . \end{equation*}

Finally, $$E(-x) \not= 0$$ 1  for all $$x \in \R\text{.}$$ So $$F(x)-E(x) = 0$$ for all $$x\text{,}$$ and we are done.

Having proved $$E$$ is unique, we define the exponential function as

\begin{equation*} \exp(x) := E(x) . \end{equation*}

If $$y \in \Q$$ and $$x > 0\text{,}$$ then

\begin{equation*} x^y = \exp\bigl(\ln(x^y)\bigr) = \exp\bigl(y\ln(x)\bigr) . \end{equation*}

We can now make sense of exponentiation $$x^y$$ for arbitrary $$y \in \R\text{;}$$ if $$x > 0$$ and $$y$$ is irrational, define

\begin{equation*} x^y := \exp\bigl(y\ln(x)\bigr) . \end{equation*}

As $$\exp$$ is continuous, then $$x^y$$ is a continuous function of $$y\text{.}$$ Therefore, we would obtain the same result had we taken a sequence of rational numbers $$\{ y_n \}$$ approaching $$y$$ and defined $$x^y = \lim\, x^{y_n}\text{.}$$

Define the number $$e\text{,}$$ sometimes called Euler's number or the base of the natural logarithm, as

\begin{equation*} e := \exp(1) . \end{equation*}

Let us justify the notation $$e^x$$ for $$\exp(x)\text{:}$$

\begin{equation*} e^x = \exp\bigl(x \ln(e) \bigr) = \exp(x) . \end{equation*}

The properties of the logarithm and the exponential extend to irrational powers. The proof is immediate.

#### Remark5.4.4.

There are other equivalent ways to define the exponential and the logarithm. A common way is to define $$E$$ as the solution to the differential equation $$E'(x) = E(x)\text{,}$$ $$E(0) = 1\text{.}$$ See Example 6.3.3, for a sketch of that approach. Yet another approach is to define the exponential function by power series, see Example 6.2.14.

#### Remark5.4.5.

We proved the uniqueness of the functions $$L$$ and $$E$$ from just the properties $$L(1)=0\text{,}$$ $$L'(x) = \nicefrac{1}{x}$$ and the equivalent condition for the exponential $$E'(x) = E(x)\text{,}$$ $$E(0) = 1\text{.}$$ Existence also follows from just these properties. Alternatively, uniqueness also follows from the laws of exponents, see the exercises.

### Subsection5.4.3Exercises

#### Exercise5.4.1.

Given a real number $$y$$ and $$b > 0\text{,}$$ define $$f \colon (0,\infty) \to \R$$ and $$g \colon \R \to \R$$ as $$f(x) := x^y$$ and $$g(x) := b^x\text{.}$$ Show that $$f$$ and $$g$$ are differentiable and find their derivative.

#### Exercise5.4.2.

Let $$b > 0\text{,}$$ $$b\neq 1$$ be given.

1. Show that for every $$y > 0\text{,}$$ there exists a unique number $$x$$ such that $$y = b^x\text{.}$$ Define the logarithm base $$b$$, $$\log_b \colon (0,\infty) \to \R\text{,}$$ by $$\log_b(y) := x\text{.}$$

2. Show that $$\log_b(x) = \frac{\ln(x)}{\ln(b)}\text{.}$$

3. Prove that if $$c > 0\text{,}$$ $$c \neq 1\text{,}$$ then $$\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\text{.}$$

4. Prove $$\log_b(xy) = \log_b(x)+\log_b(y)\text{,}$$ and $$\log_b(x^y) = y \log_b(x)\text{.}$$

#### Exercise5.4.3.

(requires Section 4.3)   Use Taylor's theorem to study the remainder term and show that for all $$x \in \R$$

\begin{equation*} e^x = \sum_{n=0}^\infty \frac{x^n}{n!} . \end{equation*}

Hint: Do not differentiate the series term by term (unless you would prove that it works).

#### Exercise5.4.4.

Use the geometric sum formula to show (for $$t\not= -1$$)

\begin{equation*} 1-t+t^2-\cdots+{(-1)}^n t^n = \frac{1}{1+t} - \frac{{(-1)}^{n+1}t^{n+1}}{1+t}. \end{equation*}

Using this fact show

\begin{equation*} \ln (1+x) = \sum_{n=1}^\infty \frac{{(-1)}^{n+1}x^n}{n} \end{equation*}

for all $$x \in (-1,1]$$ (note that $$x=1$$ is included). Finally, find the limit of the alternating harmonic series

\begin{equation*} \sum_{n=1}^\infty \frac{{(-1)}^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \end{equation*}

#### Exercise5.4.5.

Show

\begin{equation*} e^x = \lim_{n\to\infty} {\left( 1 + \frac{x}{n} \right)}^n . \end{equation*}

Hint: Take the logarithm.
Note: The expression $${\left( 1 + \frac{x}{n} \right)}^n$$ arises in compound interest calculations. It is the amount of money in a bank account after 1 year if 1 dollar was deposited initially at interest $$x$$ and the interest was compounded $$n$$ times during the year. The exponential $$e^x$$ is the result of continuous compounding.

#### Exercise5.4.6.

1. Prove that for $$n \in \N\text{,}$$

\begin{equation*} \sum_{k=2}^{n} \frac{1}{k} \leq \ln (n) \leq \sum_{k=1}^{n-1} \frac{1}{k} . \end{equation*}
2. Prove that the limit

\begin{equation*} \gamma := \lim_{n\to\infty} \left( \left( \sum_{k=1}^{n} \frac{1}{k} \right) - \ln (n) \right) \end{equation*}

exists. This constant is known as the Euler–Mascheroni constant 2 . It is not known if this constant is rational or not. It is approximately $$\gamma \approx 0.5772\text{.}$$

#### Exercise5.4.7.

Show

\begin{equation*} \lim_{x\to\infty} \frac{\ln(x)}{x} = 0 . \end{equation*}

#### Exercise5.4.8.

Show that $$e^x$$ is convex, in other words, show that if $$a \leq x \leq b\text{,}$$ then $$e^x \leq e^a \frac{b-x}{b-a} + e^b \frac{x-a}{b-a}\text{.}$$

#### Exercise5.4.9.

Using the logarithm find

\begin{equation*} \lim_{n\to\infty} n^{1/n} . \end{equation*}

#### Exercise5.4.10.

Show that $$E(x) = e^x$$ is the unique continuous function such that $$E(x+y) = E(x)E(y)$$ and $$E(1) = e\text{.}$$ Similarly, prove that $$L(x) = \ln(x)$$ is the unique continuous function defined on positive $$x$$ such that $$L(xy) = L(x)+L(y)$$ and $$L(e) = 1\text{.}$$

#### Exercise5.4.11.

(requires Section 4.3)   Since $$(e^x)' = e^x\text{,}$$ it is easy to see that $$e^x$$ is infinitely differentiable (has derivatives of all orders). Define the function $$f \colon \R \to \R\text{.}$$

\begin{equation*} f(x) := \begin{cases} e^{-1/x} & \text{if } x > 0, \\ 0 & \text{if } x \leq 0. \end{cases} \end{equation*}
1. Prove that for every $$m \in \N\text{,}$$

\begin{equation*} \lim_{x \to 0^+} \frac{e^{-1/x}}{x^m} = 0 . \end{equation*}
2. Prove that $$f$$ is infinitely differentiable.

3. Compute the Taylor series for $$f$$ at the origin, that is,

\begin{equation*} \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k . \end{equation*}

Show that it converges, but show that it does not converge to $$f(x)$$ for any given $$x > 0\text{.}$$

$$E$$ is a function into $$(0,\infty)$$ after all. However, $$E(-x) \neq 0$$ also follows from $$E(x)E(-x) = 1\text{.}$$ Therefore, we can prove uniqueness of $$E$$ given i and ii, even for functions $$E \colon \R \to \R\text{.}$$
Named for the Swiss mathematician Leonhard Paul Euler 3  (1707–1783) and the Italian mathematician Lorenzo Mascheroni 4  (1750–1800).
https://en.wikipedia.org/wiki/Leonhard_Euler
https://en.wikipedia.org/wiki/Lorenzo_Mascheroni
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf