# Logarithmic and exponential functions

## Introduction

The logarithmic and exponential functions are called transcendental functions. A transcendental function is a function that "transcends" algebra in the sense that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction. In this document we will first define these functions for rational exponents and then we extend their definition to the real and complex numbers.

## Rational exponents

Let $f(x)=a^n, n \in \mathbb{Q}, \; a>0$ be the exponential function and its inverse $\phi(y)=\log_a y$ with $y>0$ the logarithm function. The exponential function is defined for $n \in \mathbb{N}$ as follows: $a^n \triangleq \underbrace{a\, a \, a \,\cdots \,a \, a}_{n\ times}$ From this it is easy to discover the following properties: $a^ma^m=a^{m+n}\Rightarrow f(m+n)=f(m)f(n)$ $(a^m)^n=a^{mn}$ If we assume this property it immediately follows: $a^0a=a^1=a \Rightarrow a^0=1$ Then along the same line of reasoning we have also for negative integers: $a^{-1}a=a^0=1 \Rightarrow a^{-1}=\frac{1}{a}$ and thus $a^{-n}=\frac{1}{a^n}, \; n \in \mathbb{Z}$ We can extend the same reasoning to rational exponents. Let $x \in \mathbb{Q}$ then $x=\dfrac L M$ with $L,M \in \mathbb{Z}$ and $a^x=a^\frac{L}{M}=\left(a^\frac{1}{M}\right)^L$ Notice: $\left(a^\frac{1}{M}\right)^M=1$ We define $a^\frac{1}{M}$ as the $Mth$ root of $a$ also written as: $\sqrt[M]{a}$ However for irrational exponents this algebraic approach fails and we must find a more sophisticated approach.

## Real exponents

We seek now for a generalization of the rational exponents into the real numbers.

Let $f(x)=a^x, \; x\in \mathbb{R}$ be the exponential function, with its inverse the logarithm function: $\phi(y)=\log_a y, \; y>0$.

$f(x)$ should have the following general property: $$f(m+b)=f(m).f(n) \; \forall m,n \text{ and } f(0)=1$$ We find its derivative: \begin{align*} f^\prime(x) &= \lim_{\delta \rightarrow 0}\frac{f(x+\delta)-f(x)}{\delta} \\ &=\lim_{\delta \rightarrow 0}\frac{f(x)f(\delta)-f(x)}{\delta} \\ &=f(x)\lim_{\delta \rightarrow 0}\frac{f(\delta)-1}{\delta} \\ &=f(x)\lim_{\delta \rightarrow 0}\frac{a^\delta-1}{\delta} \\ &=f(x)f'(0) \\ \end{align*} Let us assume for the moment that $f'(0)$ exists, so the function $f(x)$ is differentiable at $x=0$, and denote its value with $\alpha$. $$f^\prime(x)=\alpha f(x) \label{eq01}$$ Let us now investigate the derivative of its inverse by using the following propostion.

### Proposition

If $f$ is differentiable in $x=c$, and $f$ has inverse $\phi(y)$, and $f^\prime(c) \neq 0$ then also $\phi$ is differentiable in $C=f(c)$ and: $\phi^\prime(C)=\frac{1}{f^\prime(c)}=\frac{1}{f^\prime(\phi(C))}$

### Proof

We apply this proposition with $\phi(x)=\log_a x$ with $x>0$ and $f(x)=a^x$ with $x \in \mathbb{R}$ and using the found relation between $f$ and $f^\prime$: \begin{align} \phi^\prime(x)&=\frac{1}{f^\prime(\phi(x))} \notag \\ &=\frac{1}{\alpha f(\phi(x))} \notag \\ &=\frac{1}{\alpha x} \label{eq02} \end{align}

Based on integral calculus we know that for any continuous function $f$ a function $F$ exists defined by: $F(x)=\int_a^x f(t) dt \text{ and } F^\prime(x)=f(x)$ As $f:x \mapsto x^{-1}$ is a continuous function for $x \neq 0$ it is natural to define based on the Fundamental Theorem of Calculus the following function definition for the logarithm function:

Logarithm
For $(x \neq 0)$ and $a>0$: $$\log_a(x)=\frac{1}{\alpha} \int_1^{|x|} \frac{dt}{t}$$

Geometrically, our logarithmic function means the area shown in the figure, bounded above by the rectangular hyperbola $y=\dfrac{1}{t}$ , below by the $x$-axis, and on the sides by the lines $t=1$ and $t=x$. This area is to be reckoned positive, if $x>1$, negative if $x \less 1$. For $x = 1$, the area vanishes.

The problem is still this constant $\alpha$ which depends on $a$. Before we further investigate $\alpha$ we introduce the natural number $e$. The notation $e$ for this number is due to Euler and is also called Euler's number.

There are a number of ways to introduce the natural number $e$. Let us introduce it here by definition:

Number $e$
The number $e$ is defined by the integral: $$1=\int_1^e \frac{dt}{t}$$

But then easily follows: $\log_a(e)=\frac{1}{\alpha} \int_1^e\frac{dt}{t}=\frac{1}{\alpha}$ wich means the same as: $a^{\tfrac{1}{\alpha}} = e$ but then if $a=e$ we find: $e^{\tfrac{1}{\alpha}}=e \to \alpha =1$

We have earlier found: $\alpha=\lim_{\delta \to 0}\frac{a^\delta-1}{\delta}$ Substitue $a=e$ we get $\alpha = \lim_{\delta\to 0} \frac{e^\delta-1}{\delta} = 1$ From this expression, we have, as $\delta\to 0$, \begin{align*} e^\delta - 1 & \to \delta \Rightarrow \\ e^\delta & \to 1+\delta \Rightarrow \\ e & \to (1+\delta)^{1/\delta} \end{align*} or $e = \lim_{\delta\to 0} \left(1+\delta \right)^{1/\delta}$ alternatively if we substitute $\delta=1/x$ and $h \to 0$ by $|x| \to \infty$ we get the usual definition of $e$:

Number $e$
$e = \lim_{x \to \pm \infty} \left(1+\frac{1}{x}\right)^x$

Formally one has to proof that this limit exists, but we have shown already that $\alpha =1$ given $a=e$ based on the first definition, and that must imply that above limit also exists because it is derived from $\alpha=1$.

$e$ is a transcendental number (a type of irrational number), so its decimal expansion never repeats. The initial decimal expansion of $e$ is given by $e = 2.7182818284590452353602874713526624977572470937\ldots$

We now define the special exponential and logarithm functions for base $e$.

Natural exponential and logarithm function
The natural logarithm function is defined by: $\ln(x)=\log_e(x), \; (x \neq 0)$ The natural exponential function is defined by: $\exp(x)=e^x$

From $\alpha=1$ and $\ref{eq01}$ resp $\ref{eq02}$ follows \begin{align*} \exp^\prime(x)&=\exp(x) \\ \ln^\prime(x)&=\frac{1}{x} \end{align*}

### Corollaries

$\ln a^n = n \ln a$
Take $\ln a^3 = 3 \ln a$ and via mathematical induction the proof is made for n.
$\ln \frac{1}{a}=-\ln a$
If $ab=1$ and $a>0$ then $b=a^{-1}$ and $\ln ab = \ln a + \ln a^{-1}=0$
$\ln \frac{a}{b}=\ln a - \ln b$
$\ln \frac{a}{b}= \ln a + \ln \frac{1}{b} = \ln a - \ln b$
$\ln x$ is monotonic increasing on $(0,\to)$
Let $x_1 > x_2 > 0$ then $\dfrac{x_1}{x_2}>1$ and thus $\ln \dfrac{x_1}{x_2} >0$ and $\ln x_1 > \ln x_2$
$\lim_{x\rightarrow \infty} \ln x = \infty$
Take any number $n \in \mathbb{N}$ and $x \geq 2^n$ then $\ln x \geq n \ln 2$.
$\lim_{x \downarrow 0}\ln x=-\infty$
Take any number $n \in \mathbb{N}$ then for $n^{-1}$ and $x \leq 2^{-n}$ we have $\ln x \leq - n \ln 2$.

## Imaginary exponents

A Taylor series expansion is a polynomial (possibly of infinitely high order), and polynomials involve only addition, multiplication, and division. Since these elementary operations are also defined for complex numbers, any smooth function of a real variable, $f(x)$, may be generalized to a function of a complex variable $z=\alpha+i\beta$ by simply substituting the complex variable for the real variable $x$ in the Taylor series expansion of $f(x). The Taylor serie expansion of$f(x)=e^x$around$x=0$is given by: $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}++\frac{x^3}{3!}+....$ Now$f(x)$can be generalized to a function of complex variables$f(z)$, and it makes perfectly sense to define following Taylor expansion of$f(z)$around$x=0: $e^{i\theta}\triangleq \sum_{n=0}^\infty \frac{(i\theta)^n}{n!}=1+i\theta-\frac{\theta^2}{2}-i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+i\frac{\theta^5}{5!}++$ Note: \begin{align*} \mbox{re}\{e^{i\theta}\}&=1-\theta^2/2+\theta^4/4!-.... \\ \mbox{im}\{e^{i\theta}\}&=\theta-\theta^3/3!+\theta^5/5!-.... \end{align*} Let us compare this with the Taylor expansion of\sin \theta$and$\cos \theta$around$\theta=0. First recall: \begin{align*} \frac{d}{d\theta}\cos\theta&=-\sin\theta \\ \frac{d}{d\theta}\cos\theta&=\cos\theta \end{align*} so that the Taylor expansion around\theta=0is: \begin{eqnarray*} \cos \theta = \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+.... \\ \sin \theta = \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n+1}}{(2n+1)!}=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+.... \end{eqnarray*} but then follows: \begin{align*} e^{i\theta}&=\mbox{re}\{e^{i\theta}\}+i\; \mbox{im}\{e^{i\theta}\} \\ &=\frac{(-1)^n\theta^{2n}}{(2n)!}+i\frac{(-1)^n\theta^{2n+1}}{(2n+1)!} \\ &=\cos\theta + i\sin\theta \end{align*} This last equation is called Euler's Indentity. Euler's indentity also gives us a new way to express complex numbers. A complex number can be represented in polar form as: \begin{align*} z&=r(\cos\theta + i\sin\theta) \\ &= re^{i\theta} \end{align*} Polar form often simplifies algebraic manipulations of complex numbers, especially when they are multiplied together. Simple rules of exponents can often be used in place of messier trigonometric identities If we set\theta=\pi$in Euler's identity we get what some have called the most beautiful formula in mathematics'' due to the extremely simple form in which the fundamental constants$e,i,\pi,1,0$together with the elementary operations of addition, multiplication, exponentiation, and equality, all appear exactly once: $e^{i\pi}+1=0$ Furthermore wwe can derive formulas for$\cos$in terms of$e^{i\theta}: \begin{align*} e^{i\theta}+\overline{e^{i\theta}}&=e^{i\theta}+e^{-i\theta} \\ &=(\cos\theta + i\sin\theta)+(\cos\theta - i\sin\theta) \\ &= 2 \cos\theta \end{align*} From which follows: $\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ In the samne way we can derive a formula for\sin$in terms of$e^{i\theta}\$: \begin{align*} e^{i\theta}-\overline{e^{i\theta}}&=e^{i\theta}-e^{-i\theta} \\ &=(\cos\theta + i\sin\theta)-(\cos\theta - i\sin\theta) \\ &= 2 i\sin\theta \end{align*} form which follows: $\sin\theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}$