Section 4.2

$$a\times n=\underset{n\text{ amount of times}}{\underbrace{a+a+\dots +a}}$$

$$a^n=\underset{n\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}$$

$$a^n\cdot a^m=\underset{n+m\text{ amount of times}}{\underbrace{\left(\underset{n\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}\right)\cdot \left(\underset{m\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}\right)}}=a^{m+n}$$

$$\begin{array}{rl}{\left(a^m\right)}^n& =a^m\cdot a^m\cdot \dots \cdot a^m\\ & =\underset{m+m+\cdots +m=nm\text{ amount of times}}{\underbrace{\left(\underset{m\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}\right)\cdot \left(\underset{m\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}\right)\cdot \dots \cdot \left(\underset{m\text{ amount of times}}{\underbrace{a\cdot a\cdot \dots \cdot a}}\right)}}\\ & =a^{nm}\end{array}$$

Property 10 (Law of Exponent)

Given the previous equation we can see the following:

$$a^n\cdot a^m=a^{n+m}$$

This implies that ${\left(a^m\right)}^n=a^{m\cdot n}$.

We have seen that $a^{-n}=\frac{1}{a^n}$ and $a^0=1$ (but $a$ cannot be $0$).

This and previous equations imply $\frac{a^m}{a^n}=a^{m-n}$.

Example 69

Show $\frac{a^m}{a^n}=a^{m-n}$.

“Proof:”

$$\begin{array}{rl}\frac{a^m}{a^n}& =a^m\cdot a^{-n}\text{ (since }\frac{1}{a^n}=a^{-n}\text{)}\\ & =a^{m-n}\text{ (since }{a}^m\cdot a^n=a^{m+n}\text{)}\end{array}$$

Something to remember when it comes to the base of the exponent.

$$-2^2\ne (-2{)}^2$$

Property 11

Let $n$ be a natural number.

$$\sqrt[n]{x}=x^{\frac{1}{n}}$$

When it comes to the exponential equation we have the following.

Property 12

The equation $a^m=a^n$ if and only if $m=n$.

We are familiar with linear growth.

$$2,4,6,8,...$$

The first term is $a_1=2$, second term is $a_2=4$, and so on. The $n^{\text{th}}$ term is? The answer would be $a_n=2n$. This would be similar to $f(x)=2x$ where the domain is all real numbers instead of all natural numbers. The function would be called a linear function.

Next, we consider exponential growth.

$$2,4,8,16,32,...$$

The first term is $a_1=2$, second term is $a_2=4$, third term is $a_3=8$, and so on. The $n^{\text{th}}$ term is $a_n=2^n$. This would be similar to $f(x)=2^x$ where the domain is all real numbers instead of all natural numbers. The function would be called an exponential function.

Definition 47

If $a>0$ and $a\ne 1$ then the exponential function base $a$ is

$$f(x)=a^x$$

• The domain of $f$ is the set of all real numbers. (Consider $2^x$ what $x$ value would case $2^x<0$. Answer: there isn’t a real number that would cause this.)

• The range of $f$ is the interval $(0\mathrm{\infty })$.

• The function $f$ is a continuous function over its domain.

• If $a>1$, then $f$ is increasing on its domain.

• If $0<a<1$, then $f$ is decreasing on its domain.

• The function $f$ has a horizontal asymptote $y=0$.

• The graph passes through the points: $(-1,\frac{1}{a})$, $(0,1)$, and $(1,a)$.

The graph of the exponential function changes based on the value of $a$.

The graph of $f(x)=a^x$ when $a>1$ we have:

The graph of $f(x)=a^x$ where $0<a<1$ we have:

Example 70

Solve $2^{x^2-x}=64$.

Solution:

We want to use the fact that $a^m=a^n$ if and only if $m=n$. However, the equation does not satisfy the left-hand-side condition. However, $64=2^6$. That is,

$$2^{x^2-x}=2^6$$

Which implies $x^2-x=6$. Solve this equation we have:

$$\begin{array}{rl}{x}^2-x& =6\\ x^2-x-6& =0\\ (x-3)(x+2)& =0\end{array}$$

Therefore, the solution is $\{3,-2\}$.

It is important to notice that the past equation is only possible to solve since $64$ is a value of $2$ to some power. However, in the future that will not always be the case.

Remember $f(x)=a^x$ is continous everwhere and the range is $(0,\mathrm{\infty })$.

Let $f(x)=2^x$. Notice that $f(2)=2^2=4$ and $f(3)=2^3=8$. By the Intermediate Value Theorem, there exists a $c\in (2,3)$ such that $f(c)=7$. However, the value of $x$ is not common knowledge. This leads to the next section. What value of $x$ causes $2^x=7$?

Definition 48 (Compount Interest)

If $P$ dollars are deposited in an account paying an annual rate of interest $r$ compounded $n$ times per year, then after $t$ years the account will contain $A$ amount of dollars, according to the following formula.

$$A=P{(1+\frac{r}{n})}^{tn}$$

An important base for the exponential equation is the number $e$ defined as:

$$e=\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x$$

If the investment is compounded continuously we have the following formula:

$$A=P{e}^{rt}$$