Exponential Functions/Logarithmic Functions

Definition of Exponents Let $a$

be a positive real number, $n$

be a natural number, $m$

be an integer. Then

$a^{0} = 1$
$a^{n} = \overbrace{a\cdot a \cdots a}^{n}$
$a^{-n} = \frac{1}{a^{n}}$
$a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$
Let $\alpha$ be irrational number, $\alpha = \lim_{n \to \infty}\alpha_{n}$ , where $\{\alpha_{m}\}$ is sequence of rational number. Then $a^{\alpha} = \lim_{n \to \infty}a^{\alpha_{n}}$

Definition of Exponents To define the exponents, $a$ has to be positive.
Note that $\frac{1}{a}\cdot a = \frac{a}{a} = 1$ , and $a^{-1}\cdot a = a^{-1+1} = a^0 = 1$ . Thus $a^{-1} = \frac{1}{a}$

NOTE At 5. Consider the case $a \geq 1$ and $a^{\sqrt{2}}$ . First we create an increasing sequence of rational numbers which converges to $\sqrt{2}$ , say $\{\alpha_{n}\}$ . For example,

$\displaystyle \{1.4,1.41,1.414,\ldots\}.$

Then for

, we have $\alpha_{m} < \alpha_{n}$ and

$\displaystyle \frac{a^{\alpha_{n}}}{a^{\alpha_{m}}} = a^{\alpha_{n}-\alpha_{m}} > a^{0} = 1.$

Thus, $\{a^{\alpha_{n}}\}$ is increasing . Also $a^{\alpha_{n}} \leq a^{\sqrt{2}}$ implies that

$\{a^{\alpha_{n}}\}$ is a bounded above increasing sequence. Thus it converges. Let

$\displaystyle a^{\sqrt{2}} = \lim_{n \rightarrow \infty} a^{\alpha_{n}}$

Similarly for $0 < a < 1$

. The proof looks OK. But there are many rational valued sequences which converge to $\sqrt{2}$ . So, we have to show using other rational sequence $b_n$

which converges to $\sqrt{2}$ , $\lim_{n \rightarrow \infty}a^{b_{n}}$ is the same. In mathematics, we call this Uniqueness.

Check
Since , $\alpha_{n} \leq \sqrt{2}$ implies $a^{\alpha_{n}} \leq a^{\sqrt{2}}$ .

Let $\{\alpha_{n}\},\{\beta_{n}\}$ be two increasing rational valued sequences which converege to $\sqrt{2}$ . Since $\alpha_{n} \rightarrow \sqrt{2}, \beta_{n} \rightarrow \sqrt{2}$ , for all $\beta_{n}$ satisfying $\alpha_{1} \leq \beta_{n}$ , we can choose $\alpha_{m},\alpha_{m+1}$ so that $\alpha_{m}\leq \beta_{n} < \alpha_{m+1}$ . $\alpha_{n}, \beta_{n}$ are both increasing and converge $\sqrt{2}$ . Thus we can choose $\beta_{n}$ which is larger that $\alpha_{1}$ . Then we can choose $\alpha_{k}$ which is larger that $\beta_{n}$ . Thus we cha choose $m$ so that $\alpha_{m} \leq \beta_{n} < \beta_{m+1}$ .

For $a > 1$ ,

$\displaystyle a^{\alpha_{m}} \leq a^{\beta_{n}} \leq a^{\alpha_{m+1}}$

and for

$\displaystyle a^{\alpha_{m}} \geq a^{\beta_{n}} > a^{\alpha_{m+1}}$

Now $n \rightarrow \infty$ implies that $m \rightarrow \infty$ and by the squeezing theorem,

$\displaystyle \lim_{m \rightarrow \infty}a^{\alpha_{m}} = \lim_{n \rightarrow \infty}a^{\beta_{n}}$

This shows that $a^{\sqrt{2}}$ is independent from the choice of $\{a_{n}\}$ .

For $a > 0$ and $a \neq 1$ , a function $y = a^{x}$ is called Exponential Function .

The domain is $(-\infty,\infty)$ and the range is $(0, \infty)$ .

Laws of Exponential Function

Theorem 1..13 Suppose that $a > 0$

. Then for every real numbers $x,y$

, we have

$a^{x+y} = a^{x}a^{y}$
$a^{x-y} = \frac{a^{x}}{a^{y}}$
$(a^{x})^{y} = a^{xy}$
$e^{x}$ is continuous on $(-\infty,\infty)$ and strictly increasing function.
$\lim_{x \rightarrow \infty}e^{x} = \infty, \lim_{x \rightarrow -\infty}e^{x} = 0$

NOTE $a^{x}$ for $a > 0$ is defined for irrational number $x$ by considering $a^{\alpha}$ .

$\displaystyle a^{\alpha} = \lim_{n \rightarrow \infty}a^{\alpha_{n}}, \alpha_{n} \rightarrow \alpha,\{\alpha_{n}\}$ sequence of rational numbers $\displaystyle , a > 0$

Then the case 2., we have $a^x = \lim_{n \rightarrow \infty}a^{\alpha_{n}}, a^y = \lim_{n \rightarrow \infty}a^{\beta_{n}}$ and

$\displaystyle a^{x}a^{y} = \lim_{n \rightarrow \infty}a^{\alpha_{n}} \cdot \lim... ...{\beta_{n}} = \lim_{n \rightarrow \infty}a^{\alpha_{n} + \beta_{n}} = a^{x+y}.$

Graph of Exp Among all exponential functions $a^x$ , $e^x$ is the most important function.

Example 1..30 Draw the graph of $y = 2^{x}$ and $y = (\frac{1}{2})^{x}$ .

**Figure 1.47:** Example1-30
$\includegraphics[width=3.5cm]{SOFTFIG-1/exponenntial_gr1.eps}$

NOTE We find values of $x$ and corresponding values of $y$ . Then plot those points and connect by smooth curve.

$\begin{displaymath}\begin{array}{c\vert cccccccccccccccc}\hline x & \cdots & -3 ... ...frac{1}{4} & \cdots & \frac{1}{8} & \cdots\ \hline \end{array}\end{displaymath}$

Exercise 1..30 Suppse that $f(x)$

satisfies

for all

and $f(0) \neq 0$ . Show the followings.
1. $f(0) = 1$

2. $f(-y) = \frac{1}{f(y)}$

The property $f(x+y) = f(x)f(y)$

is generalization of exponetial function.

SOLUTION 1. We can write $f(0) = f(0+0) = f(0)f(0)$ . Then $f(0)(f(0) - 1) = 0$ . Since $f(0) \neq 0$ , we have $f(0) = 1$ $\blacksquare$
2. $1 = f(0) = f(y - y) = f(y)f(-y)$ implies that $f(-y) = \frac{1}{f(y)}$ $\blacksquare$

Definition of Logarithm Let $a$ be a positive real number and $a \neq 1$ . The for every real number $x$ and $y = a^{x}$ , we write $x = \log_{a}{y}$ and call $a$ base of Logarithm.

Table of Logarithm Table of logarithm of base 2. NOTE

$\begin{displaymath}\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...6 & 7\ \hline y & 2 & 4 & 8 & 16 & 32 & 64 & 128 \end{array}\end{displaymath}$

Calculate $8 \times 16$ . Find the number above $8$

and the number above $16$

. Then we have 3 and 4. Now add these two numbers to obtain $7$

. Then the numer below $7$

the result of $8 \times 16$ . Express the number 3 above 8 as $\log_{2}{8}$ . Then we get $\log_{2}(8 \times 16) = \log_{2}{8} + \log_{2}16 = 3 + 4 = 7$ . Next we calclulate $128 \div 32$ . This time we subtract the number above 128 from the number above 32. Then we get $7 - 5 = 2$

. Now the number 4 is the result of $128 \div 32$ . Thus $\log_{2}(128 \div 32) = \log_{2}{128} - \log_{2}{32} = 7 - 5 = 2$ .

Laws of Logarithmics

Theorem 1..14 Let $a > 0$

, $a \neq 1$ , $p > 0, q > 0$

. Then we have the followings.

$\log_{a}{pq} = \log_{a}p + \log_{a}q$
$\log_{a}\frac{p}{q} = \log_{a}p - \log_{a}q$
$\log_{a}\frac{1}{q} = -\log_{a}q$
$\log_{a}q^r = r\log{a}q$

Laws of Logarithms Take logarithms Then
a product becomes an addition,
a quotient becomes a subtraction,
a power becomes a product.

NOTE Consider $y = \log_{a}{x}$ . Then by the definition of logarithm, we have $x = a^{y}$ . Now take the logarithm of both sides with the base $b$ . Then

$\displaystyle \log_{b}x = \log_{b}a^{y} = y\log_{b}a$

From this, we have

$\displaystyle y = \frac{\log_{b}x}{\log_{b}a}.$

We call this change-of-base formula.

Existence of Inverse Function Strictly increasing function is one-to-one. Thus, we have a inverse function. Natural Logarithm We write natural logarithm without the base $b$ .

The domain of $y = a^{x}$ is $(0, \infty)$ , and $y = a^x$ is strictly increasing and continuous function. Thus there exists a unique inverse function and we write $x = f^{-1}(y) = \log_{a}{y}$ . Note $\log_{a}{y}$ is defined and continuous on $(0, \infty)$ . we say $\log_{a}{y}$ logarithmic function with base $a$ .

Laws of Logarithmic Functions

Theorem 1..15 Let $a > 0$

. Then for every real numer $x > 0,y > 0$

, we have the followings.

$\log_{a}{1} = 0, \log{e} = 1$
$\log_{a}(xy) = \log_{a}{x} + \log_{a}{y}$
$\log_{a}\frac{x}{y} = \log_{a}{x} - \log_{a}{y}$
$\log_{a}x^{y} = y\log_{a}{x}$
$\log{x}$ is continuous and monotonically increasing on $(0, \infty)$
$\lim_{x \rightarrow \infty}\log{x} = \infty,\lim_{x \rightarrow 0+}\log{x} = -\infty$

**Figure 1.48:** Graph of Logarithm
$\includegraphics[width=3.5cm]{SOFTFIG-1/logfunc_gr1.eps}$

NOTE To show 2. Let $\alpha = \log_{a}x, \beta = \log_{a}y$ . Then $x = a^{\alpha}, y = \alpha^{\beta}$ and

$\displaystyle xy = a^{\alpha}a^{\beta} = a^{\alpha + \beta}$

Thus,

$\displaystyle \log_{a}x + \log_{a}y = \alpha + \beta = \log_{a}{xy}$

Example 1..31 Find the domain of the following function.

$\displaystyle f(x) = \log{(x\sqrt{1+3x})}$

**Figure 1.49:** Example1-31
$% latex2html id marker 40904 \includegraphics[width=3.5cm]{SOFTFIG-1/logteigiiki_gr1.eps}$

A logarithmic function can not take the values less than 0. SOLUTION Since a logarithmic function can only take positive values, we have $x\sqrt{1+3x} > 0$ . Thus,

$\displaystyle x > 0$ and $\displaystyle \sqrt{1+3x} > 0$ or $\displaystyle x < 0$ and $\displaystyle \sqrt{1+3x} < 0.$

$\sqrt{1+3x} \geq 0$ implies $x > 0$

and $\sqrt{1+3x} > 0$ . Since a square root function can not take negative values, we have $1 + 3x > 0$

. Thus

and $x > -\frac{1}{3}$ . Express using interval, we have $(0,\infty) \cap (-\frac{1}{3},\infty) = (0, \infty)$ $\blacksquare$

Exercise 1..31 Suppose that $f(x)$

satisfies

for

. Then
1. $f(1) = 0$

2. $f(\frac{1}{y}) = -f(y)$

The property $f(xy) = f(x)+ f(y)$

is generalization of logarithm.

SOLUTION 1. $f(1) = f(1\cdot 1) = f(1) + f(1)$ implies $f(1) = 0$ $\blacksquare$
2. $0 = f(1) = f(\frac{y}{y}) = f(y \cdot \frac{1}{y}) = f(y) + \frac{1}{f(y)}$ implies that $f(\frac{1}{y}) = -f(y)$ $\blacksquare$

Hyperbolic Functions

A function below is called hyperbolic function.

$\displaystyle \sinh{x} = \frac{e^{x} - e^{-x}}{2}, \cosh{x} = \frac{e^{x} + e... ... \tanh{x} = \frac{\sinh{x}}{\cosh{x}} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$

The curve described by a uniform chain hanging from two supports in a uniform gravitational field is called a catenary. To exptess catenary, $\cosh{x}$ is used.

Example 1..32 Show $\displaystyle{\lim_{x \rightarrow \infty}(1 + \frac{1}{x})^x = e}$ .

SOLUTION As $x \rightarrow \infty$ , there exists $n$ which satisfies $n < x < n+1$ . Thus Calculation Take the reciprocal of . Then we have $\frac{1}{n} > \frac{1}{x} > \frac{1}{n+1}$ . Now add 1 to both sides of inequality. $\frac{1}{n}+ 1 > \frac{1}{x} + 1 > \frac{1}{n+1} + 1$
Also, $(1 + \frac{1}{n+1})^n < (1 + \frac{1}{n+1})^x < (1 + \frac{1}{x})^x$

$\displaystyle 1 + \frac{1}{n+1} < 1 + \frac{1}{x} < 1 + \frac{1}{n}$

Use

one more time to get

$\displaystyle (1 + \frac{1}{n+1})^n < (1 + \frac{1}{x})^x < (1 + \frac{1}{n})^{n+1}$

Note that

$\displaystyle (1 + \frac{1}{n+1})^n = \frac{(1 + \frac{1}{n+1})^{n+1}}{(1 + \fr... ...}{n+1})}, (1 + \frac{1}{n})^{n+1} = (1 + \frac{1}{n})^{n} (1 + \frac{1}{n}).$

Then

$\displaystyle e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n+1})^n \leq \lim_{... ... + \frac{1}{x})^x \leq \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^{n+1} = e.$

By the squeezing theorem

$\displaystyle \lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e\ensuremath{ \blacksquare}$

We already know $\lim_{n \to \infty}(1 + \frac{1}{n})^n = e$ . So, we extend this to real number $x$ .

Exercise 1..32 Show $\displaystyle{\lim_{x \rightarrow -\infty}(1 + \frac{1}{x})^x = e}$ .

$x \to -\infty$ For $x \to -\infty$ , it is better to put $y = -x$ to avoid negative infinity. Now $y \to \infty$ and by Example1.32

SOLUTION Put $x = -y$ . Then

$\displaystyle \lim_{x \to -\infty}(1 + \frac{1}{x})^x$	$\displaystyle =$	$\displaystyle \lim_{y \to \infty}(1 - \frac{1}{y})^{-y} = \lim_{y \to \infty}(\frac{y-1}{y})^{-y}$
	$\displaystyle =$	$\displaystyle \lim_{y \to \infty}(\frac{y}{y-1})^y = \lim_{y \to \infty}(1 + \frac{1}{y-1})^y$
	$\displaystyle =$	$\displaystyle \lim_{y \to \infty}(1 + \frac{1}{y-1})^{y-1}(1 + \frac{1}{y-1}) = e\ensuremath{ \blacksquare}$

Exercise A

2.: From the table, find the following values:

(a): $\displaystyle{\log{20}}$
(b): $\displaystyle{\log{16}}$
(c): $\displaystyle{\log{3^{4}}}$
(d): $\displaystyle{\log{0.1}}$
(e): $\displaystyle{\log{\sqrt{630}}}$
(f): $\displaystyle{\log{0.4}}$

$\begin{displaymath}\begin{array}{ll\vert ll} n & \log{n} & n & \log{n} \ \hline... ... & 1.39 & 9 & 2.20\\ 5 & 1.61 & 10 & 2.30\ \hline \end{array}\end{displaymath}$

3.

Solve the following equations:

(a) $\displaystyle{\log{x} = 2}$ (b) $\displaystyle{\log{x} = -1}$ (c) $\displaystyle{(2-\log{x})\log{x} = 0}$

(d) $\displaystyle{\log{(2x+1)(x+2)} = 2\log(x+2)}$

Exercise B

1.

Find the limit of the following functions:

(a) $\displaystyle{\lim_{x \rightarrow 0}\frac{\log{(1+x)}}{x}}$ (b) $\displaystyle{\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}}$ (c) $\displaystyle{\lim_{x \rightarrow a}\frac{\log{x} - \log{a}}{x - a}}$

(d) $\displaystyle{\lim_{x \rightarrow 0}\frac{(1+x)^{\alpha} - 1}{x}}$ (e) $\displaystyle{\lim_{x \rightarrow 0+}x^x}$

5.

Given the arithmetic mean $a_{n+1} = \frac{a_{n} + b_{n}}{2}$ and the geometric mean $b_{n+1} = \sqrt{a_{n}b_{n}}$ . Answer the following questions:

(a) Show $\{a_{n}\}$ and $\{b_{n}\}$ converge for $a_{0}, b_{0} \geq 0$ .

(b) Show $\lim_{n \to \infty}a_{n} = \lim_{n \to \infty}b_{n}$ . This limit is denoted by ${\rm agm}(a_{0},b_{0})$