# 1.1 Exponents  (Page 2/2)

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${a}^{0}=1$ , $\left(a\ne 0\right)$

For example, ${x}^{0}=1\text{and}\left(\text{1,000,000}{\right)}^{0}=1$ .

Note that the base must be a non-zero value. 0 0 is called an indeterminate number, and has no value. This is because 0 0 = 0/0. If one considers 0 = 0 × n (where n can be any number) then it follows that 0/0 = n , where n can be any number – meaning the value of 0/0 cannot be determined.

## Examples: application using exponential law 1

1. ${\text{16}}^{0}=1$
2. $\text{16}{a}^{0}=\text{16}$
3. $\left(\text{16}+a{\right)}^{0}=1$
4. $\left(-\text{16}{\right)}^{0}=1$
5. $-{\text{16}}^{0}=-1$

## Exponential law 2

Our definition of exponential notation shows that:

${a}^{m}×{a}^{n}={a}^{m+n}$

That is:

${a}^{m}\cdot {a}^{n}=1\cdot a\cdot \dots \cdot a$  ( m times) $\cdot 1\cdot a\cdot \dots \cdot \text{a}$   ( n times)

$=1\cdot a\cdot \dots \cdot a\text{}$     ( m + n times)

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

For example:

$\begin{array}{}{2}^{7}\cdot {2}^{3}=\left(2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\right)\text{}\left(2\cdot 2\cdot 2\right)\\ ={2}^{\text{10}}\\ ={2}^{7+3}\end{array}$

This simple law illustrates the reason exponentials were originally invented. In the days before calculators, all multiplication had to be done by hand with a pencil and a pad of paper. Multiplication takes a very long time to do and is very tedious. Adding numbers, however, is easy and quick. This law says that adding the exponents of two exponential numbers (of the same base) is the same as multiplying the two numbers together. This means that, for certain numbers, there is no need to actually multiply the numbers together in order to find their multiple. This saved mathematicians a lot of time.

## Examples: application using exponential law 2

1. ${x}^{2}\cdot {x}^{5}={\text{x}}^{7}$
2. ${2x}^{3}y\cdot {5x}^{2}{y}^{7}=\text{10}{x}^{5}{y}^{8}$
3. ${2}^{3}\cdot {2}^{4}={2}^{7}$    (Note that the base (2) stays the same.)
4. $3\cdot {3}^{2a}\cdot {3}^{2}={3}^{2a+3}$

## Exponential law 3

${a}^{m}÷{a}^{n}={a}^{m-n}$

We know from Law 2 that ${a}^{m+n}$ is base a multiplied by itself m times plus a multiplied by itself n times. Law 3 extends this to the case where an exponent is negative.

$\frac{{a}^{m}}{{a}^{n}}=\frac{a\cdot a\cdot a\cdots \cdot a}{a\cdot a\cdot a\cdots \cdot a}$ $\frac{\left(m\text{times}\right)}{\left(n\text{times}\right)}$

By factoring out ${a}^{n}$ from both numerator and denominator, we are left with

$=\frac{a\cdot a\cdot a\cdots \cdot a}{a\cdot a\cdot a\cdots \cdot a}$ $\frac{\left(m\text{times}\right)}{\left(n\text{times}\right)}$ $\frac{-a\cdot a\cdot a\cdots \cdot a}{-a\cdot a\cdot a\cdots \cdot a}$ $\frac{\left(n\text{times}\right)}{\left(n\text{times}\right)}$

$=a\cdot a\cdot a\cdots \cdot a$    ( m n times)

$={a}^{m-n}$

For example,

$\begin{array}{}{2}^{7}÷{2}^{3}=\frac{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}{2\cdot 2\cdot 2}\\ =2\cdot 2\cdot 2\cdot 2\\ ={2}^{4}\\ ={2}^{7-3}\end{array}$

## Examples: exponential law 3

1. $\frac{{a}^{6}}{{a}^{2}}={a}^{6-2}={a}^{4}$
2. $\frac{{3}^{2}}{{3}^{6}}={3}^{2-6}={3}^{-4}=\frac{1}{{3}^{4}}$    (Always give the final answer with a positive index)
3. $\frac{\text{32}{a}^{2}}{{4a}^{8}}={8a}^{-6}=\frac{8}{{a}^{6}}$
4. $\frac{{a}^{3x}}{{a}^{4}}={a}^{3x-4}$

## Exponential law 4

${a}^{-n}=\frac{1}{{a}^{n}},a\ne 0$

Our definition of exponential notation for a negative exponent shows that

${a}^{-n}=1÷a÷\cdots ÷a$    ( n times)

$=\frac{1}{1\cdot a\cdot \cdots \cdot a}$ $\frac{}{\left(n\text{times}\right)}$

$=\frac{1}{{a}^{n}}$

The minus sign in the exponent is just another way of writing that the whole exponential number is to be divided instead of multiplied.

For example, starting with Law 3, take the case of ${a}^{m-n}$ , but where  n>m :

$\begin{array}{}{2}^{2-9}=\frac{{2}^{2}}{{2}^{9}}\\ =\frac{2\cdot 2}{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}\\ =\frac{1}{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}\\ =\frac{1}{{2}^{7}}\\ ={2}^{-7}\end{array}$

## Examples: exponential law 4

1. ${2}^{-2}=\frac{1}{{2}^{2}}=\frac{1}{4}$
2. $\frac{{2}^{-2}}{{3}^{2}}=\frac{1}{{2}^{2}\cdot {3}^{2}}=\frac{1}{\text{36}}$
3. ${\left(\frac{2}{3}\right)}^{-3}={\left(\frac{3}{2}\right)}^{3}=\frac{\text{27}}{8}$
4. $\frac{m}{{n}^{-4}}={\text{mn}}^{4}$
5. $\frac{{a}^{-3}\cdot {x}^{4}}{{a}^{5}\cdot {x}^{-2}}=\frac{{x}^{4}\cdot {x}^{2}}{{a}^{3}\cdot {a}^{5}}=\frac{{x}^{6}}{{a}^{8}}$

## Exponential law 5

$\left(\text{ab}{\right)}^{n}={a}^{n}{b}^{n}$

The order in which two real numbers are multiplied together does not matter.

Therefore,

$\left(\text{ab}{\right)}^{n}=a\cdot b\cdot a\cdot b\cdot a\cdot b\cdot \cdots \cdot a\cdot b$     ( n times)

$=a\cdot a\cdot \dots \cdot a$  ( n times) $\cdot b\cdot b\cdot \dots \cdot b$  ( n times)

$={a}^{n}{b}^{n}$

For example:

$\begin{array}{}2\cdot {3}^{4}=\left(2\cdot 3\right)\cdot \left(2\cdot 3\right)\cdot \left(2\cdot 3\right)\cdot \left(2\cdot 3\right)\\ =\left(2\cdot 2\cdot 2\cdot 2\right)\cdot \left(3\cdot 3\cdot 3\cdot 3\right)\\ ={2}^{4}\cdot {3}^{4}\\ ={2}^{4}{3}^{4}\end{array}$

## Examples: exponential law 5

1. $\left({2x}^{2}y{\right)}^{3}={2}^{3}{x}^{2×3}{y}^{3}={8x}^{6}{y}^{3}$
2. $\left(\frac{7a}{{b}^{3}}\right){}^{2}=\frac{\text{49}{a}^{2}}{{b}^{6}}$
3. $\left({5a}^{n-4}{\right)}^{3}=\text{125}{a}^{3n-\text{12}}$

## Exponential law 6

$\left({a}^{m}{\right)}^{n}={a}^{\text{mn}}$

We can find the exponential of an exponential just as well as we can for a number, because an exponential is a real number.

$\left({a}^{m}{\right)}^{n}={a}^{m}\cdot {a}^{m}\cdot {a}^{m}\cdot \dots \cdot {a}^{m}$     ( n times)

$=a\cdot a\cdot \dots \cdot \text{a}$       ( m × n times)

$={\text{a}}^{\text{mn}}$

For example:

$\begin{array}{}\left({2}^{2}{\right)}^{3}=\left({2}^{2}\right)\cdot \left({2}^{2}\right)\cdot \left({2}^{2}\right)\\ =\left(2\cdot 2\right)\cdot \left(2\cdot 2\right)\cdot \left(2\cdot 2\right)\\ ={2}^{6}\\ ={2}^{2×3}\end{array}$

## Examples: exponential law 6

1. $\left({x}^{3}{\right)}^{4}={x}^{\text{12}}$
2. $\left[\left({a}^{4}{\right)}^{3}{\right]}^{2}={a}^{\text{24}}$
3. $\left({3}^{n+3}{\right)}^{2}={3}^{2n+6}$

## Module review exercises

Write the following examples using exponential notation.

$4\cdot 4$

4 2

$\text{12}\cdot \text{12}$

12 2

$9\cdot 9\cdot 9\cdot 9$

9 4

$\text{10}\cdot \text{10}\cdot \text{10}\cdot \text{10}\cdot \text{10}\cdot \text{10}$

10 6

$\text{826}\cdot \text{826}\cdot \text{826}$

826 3

$\text{3021}\cdot \text{3021}\cdot \text{3021}\cdot \text{3021}$

3021 4

$6\cdot 6\cdot 6\cdot \cdots \cdot 6$     (85 factors of 6).

6 85

$2\cdot 2\cdot 2\cdot \cdots \cdot 2$     (112 factors of 2).

2 112

For the next examples, expand the terms. (Do not find the actual values).

5 3

5 · 5 · 5

15 2

15 · 15

117 5

117 · 117 · 117 · 117 · 117

7 4

7 · 7 · 7 · 7

Determine the value of each of the powers.

3 2

9

1 2

1

15 2

225

3 4

81

1 4

1

7 3

343

Simplify as far as possible.

(2x) 3

2 3 · x 3 = 8x 3

302 0

1

(-2x) 3

(-2) 3 · x 3 = -8x 3

2 3 · 2 4

2 3+4 = 2 7

$\frac{{x}^{8}}{{x}^{3}}$

${x}^{8-3}={x}^{5}$

$\frac{\text{25}{x}^{2}}{{5x}^{8}}$

$\frac{\text{25}}{5}{x}^{2-8}={5x}^{-6}=\frac{5}{{x}^{6}}$

(3 -1 +2 -1 ) -1

$\frac{1}{{3}^{-1}+{2}^{-1}}=\frac{1}{\frac{1}{3}+\frac{1}{2}}=1\cdot \left(\frac{3}{1}+\frac{2}{1}\right)=3+2=5$

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