## ILS1.1 Indices

What is index notation? When a number such as 16 is written in the form 42 (which means 4 x 4) we say that it is written as an exponential, or in index notation.

There are laws about multiplying and dividing indices as well as how to deal with negative indices.

This module introduces rules for multiplying and dividing expressions using index notation. For example how to simplify expressions like $$4a^{3}b\times3ab^{5}$$ or $$9a^{3}b^{2}c\div3ab^{5}$$. We do not consider fractional indices which are covered in a different module. The plural of index is indices.

Here's a video showing how the rules work:

Transcript (RTF)

### Index Notation

Consider the following examples:

\begin{align*} 3^{4} & =3\times3\times3\times3=81\\ 5^{3} & =5\times5\times5=125\\ 2^{7} & =2\times2\times2\times2\times2\times2\times2=128 \end{align*}

In general: $a^{n}=\underbrace{a\times a\times a\times\ldots\times a}_{n\textrm{ factors}}$ The letter $$n$$ in $$a^{n}$$ is referred to in one of three ways:

1. $$n$$ is the index in $$a^{n}$$ with $$a$$ known as the base.

2. $$n$$ is the exponent or power to which the base $$a$$ is raised.

3. $$n$$ is the logarithm, with $$a$$ as the base. (see the Logarithms module)

When a number such as $$125$$ is written in the form $$5^{3}$$ we say it is written as an exponential or in index notation. Multiplication and division of numbers or expressions written in index notation is achieved using index laws.

### Index Laws

This section states and gives examples of universal index laws.

#### First Index Law

To multiply index expressions you add the indices. For example: \begin{align*} 2^{3}\times2^{2} & =\left(2\times2\times2\right)\times\left(2\times2\right)\\ & =2\times2\times2\times2\times2\\ & =2^{5} \end{align*} Therefore $$2^{3}\times2^{2}=2^{3+2}=2^{5}$$. In general:

First Index Law: $a^{m}\times a^{n}=a^{m+n}$

#### Second Index Law

To divide expressions subtract the indices. For example: \begin{align*} 3^{5}\div3^{3} & =\frac{3^{5}}{3^{3}}\\ & =\frac{3\times3\times3\times3\times3}{3\times3\times3}\\ & =\frac{3\times3}{1}\quad\textrm{cancelling three lots of 3}\\ & =3^{2} \end{align*} Therefore $$3^{5}\div3^{3}=\frac{3^{5}}{3^{3}}=3^{5-3}=3^{2}.$$ In general:

Second Index Law: $a^{m}\div a^{n}=\frac{a^{m}}{a^{n}}=a^{m-n},\quad a\neq0$

Note that expressions in index form can only be multiplied or divided if they have the same base.

#### Third Index Law

To raise an expression in index form to a power, multiply the indices. For example: \begin{align*} \left(5^{2}\right)^{3} & =5^{2}\times5^{2}\times5^{2}\\ & =5^{2+2+2}\quad\textrm{using the first index law}\\ & =5^{6} \end{align*} Therefore $$\left(5^{2}\right)^{3}=5^{2\times3}=5^{6}.$$ In general:

Third Index Law: $\left(a^{m}\right)^{n}=a^{m\times n}$

This also leads to the expression: \begin{align*} (a^{m}b^{p})^{n} & =a^{mn}b^{pn} \end{align*}

Be careful as this is true for multiplication and division only , not addition or subtraction , so that $$(a+b)^{n}\neq a^{n}+b^{n}$$

### Examples of Index Laws

1. Simplify $$x^{5}\times x^{6}$$.
Solution: \begin{align*} x^{5}\times x^{6} & =x^{5+6}\quad\textrm{by the first law}\\ & =x^{11} \end{align*}

2. Simplify $$a^{5}\div a^{3}.$$
Solution: \begin{align*} a^{5}\div a^{3} & =\frac{a^{5}}{a^{3}}=a^{5-3}\quad\textrm{by the second law}\\ & =a^{2} \end{align*}

3. Simplify $$\left(c^{3}\right)^{4}$$.
Solution: \begin{align*} \left(c^{3}\right)^{4} & =c^{3\times4}\quad\textrm{by the third law}\\ & =c^{12} \end{align*}

4. Simplify $$\left(2x^{2}\right)^{3}$$.
Solution: \begin{align*} \left(2x^{2}\right)^{3} & =2^{3}\left(x^{2}\right)^{3}\\ & =8x^{2\times3}\quad\textrm{by the third law}\\ & =8x^{6} \end{align*}

Note that terms with different bases must be considered seperately when using the index laws , such as $$(2a^{3}b^{2})^{4}=2^{4}a^{12}b^{8}$$

Exercise 1 provides practice for these laws.

### Zero Index

So far we have only considered expressions in which each index is a positive whole number1 Whole numbers are called integers and positive whole numbers are called the positive integers.. The index laws also apply if the index is zero, negative or a fraction (fractional indices will be dealt with in another module).

Consider $$2^{3}\div2^{3}=\frac{2^{3}}{2^{3}}=\frac{2\times2\times2}{2\times2\times2}=\frac{8}{8}=8\div8=1$$. Using the second law, $$2^{3}\div2^{3}=2^{3-3}=2^{0}$$

therefore $$1=2^{0}$$. In general any expression with a zero index is equal to 1. Also note that $$0^{0}$$ is ambiguous and so we don’t allow $$a=0$$ in this law.

Zero law of indices: $a^{0}=1,\qquad a\neq0$

#### Examples of the Zero Index Law

$\begin{array}{lllllll} 7^{0}=1 & & \left(xy\right)^{0}=1 & & \left(\frac{1}{2}\right)^{0}=1 & & \left(28x^{2}\right)^{0}=1\\ \end{array}$

### Negative Indices

Consider $$2^{0}\div2^{4}$$. \begin{align*} 2^{0}\div2^{4} & =\frac{2^{0}}{2^{4}}\quad\textrm{remember that 2^{0}=1 }\\ & =\frac{1}{2^{4}}. \end{align*} But \begin{align*} 2^{0}\div2^{4} & =2^{0-4}\quad\textrm{using the second law}\\ & =2^{-4}. \end{align*} So $$2^{0}\div2^{4}=2^{-4}$$ and $$2^{0}\div2^{4}=\frac{1}{2^{4}}$$

therefore $$2^{-4}=\frac{1}{2^{4}}$$. In general2 Note that $$1/0$$ is undefined and so $$a\neq0$$ in the law below.,

$a^{-n}=\frac{1}{a^{n}}\textrm{ which also leads to \frac{1}{a^{-n}}=a^{n},\qquad a\neq0 }$

#### Examples of Negative Indices

$\begin{array}{lllll} 1. \ \ 2^{-3}=\frac{1}{8} & & 2. \ \ \frac{1}{x}=x^{-1} & & 3. \ \ 2y^{-1}=\frac{2}{y}\\ & & & & \\ 4. \ \ \frac{1}{3x^{-2}}=\frac{x^{2}}{3} & & 5. \ \ \frac{1}{\left(-2a\right)^{-3}}=\left(-2a\right)^{3} & & 6. \ \ 5ab^{-4}=\frac{5a}{b^{4}}\\ \end{array}$

Exercise 2 provides practice on zero and negative indices.

### Summary of Index Laws

The following laws should be remembered.

Summary of Index Laws

1. $$\ a^{m}\times a^{n}=a^{m+n}$$

2. $$\ a^{m}\div a^{n}=a^{m-n}$$ , $$\qquad a\neq0$$

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

4. $$\ a^{0}=1$$ , $$\qquad a\neq0$$

5. $$\ a^{-n}=\frac{1}{a^{n}}$$ , $$\qquad a\neq0$$

### Combing of Index Laws

Index laws may be used to simplify complex expressions.

#### Examples

1. Simplify $$\left(4a^{2}b\right)^{3}\div b^{2}.$$
Solution: \begin{align*} \left(4a^{2}b\right)^{3}\div b^{2} & =\left(4^{3}a^{6}b^{3}\right)\div b^{2}\quad\textrm{using law 3}\\ & =4^{3}a^{6}b^{1}\quad\textrm{using law 2}\\ & =4^{3}a^{6}b\\ & =64a^{6}b \end{align*}

2. Simplify $$\left(\frac{3a^{3}b}{c^{2}}\right)^{2}\div\left(\frac{ab}{3c^{-2}}\right)^{-3}$$

remember that $$a\div\frac{b}{c}=a\times\frac{c}{b}$$

Solution: \begin{align*} \left(\frac{3a^{3}b}{c^{2}}\right)^{2}\div\left(\frac{ab}{3c^{-2}}\right)^{-3} & =\frac{3^{2}a^{6}b^{2}}{c^{4}}\div\frac{a^{-3}b^{-3}}{3^{-3}c^{6}}\quad\textrm{by law 3}\\ & =\frac{3^{2}a^{6}b^{2}}{c^{4}}\times\frac{3^{-3}c^{6}}{a^{-3}b^{-3}}\quad\textrm{inverting the last term and multiplying}\\ & =\frac{3^{2-3}a^{6}b^{2}c^{6}}{c^{4}a^{-3}b^{-3}}\quad\textrm{by law 1}\\ & =3^{-1}a^{6-\left(-3\right)}b^{2-\left(-3\right)}c^{6-4}\quad\textrm{by law 2}\\ & =3^{-1}a^{9}b^{5}c^{2}\quad\textrm{simplifying}\\ & =\frac{a^{9}b^{5}c^{2}}{3}\quad\textrm{by negative index law} \end{align*}

1. Write $$x^{-1}+x^{2}$$ as a single fraction.
Solution: \begin{align*} x^{-1}+x^{2} & =\frac{1}{x}+x^{2}\quad\textrm{by negative index law}\\ & =\frac{1}{x}+\frac{xx^{2}}{x}\\ & =\frac{1+xx^{2}}{x}\quad\textrm{using a common denominator}\\ & =\frac{1+x^{3}}{x}\quad\textrm{using law 1} \end{align*}

### Exercise 1

Simplify the following:

$\begin{array}{lllll} a). \ c^{5}\times c^{3}\times c^{7} & & b). \ 3\times2^{2}\times2^{3} & & c). \ a^{3}\times a^{2}b^{3}\times ab^{4}\\ & & & & \\ d). \ 3^{6}\div3^{4} & & e). \ a^{8}\div a^{3} & & f). \ x^{4}y^{6}\div x^{2}y^{3}\\ & & & & \\ g). \ \left(x^{3}\right)^{4} & & h). \ \left(x^{m}y^{n}\right)^{5} & & \\ \end{array}$

$\begin{array}{llll} a)\:c^{15} & b)\;3\times2^{5}=96 & c)\;a^{6}b^{7} & d)\;3^{2}=9\\ e)\;a^{5} & f)\;x^{2}y^{3} & g)x^{12} & h)\;x^{5m}y^{5n} \end{array}$

### Exercise 2

Write with positive indices and evaluate if possible:

$\begin{array}{lllllll} a). \ x^{-6} & & b). \ 250^{0} & & c). \ 3ab^{-5} & & d). \ \left(pq\right)^{-2}\\ & & & & & & \\ e). \ \left(5xy\right)^{-3} & & f). \ \frac{2y}{z^{-5}} & & g). \ 2^{-5} & & h).\ (-2)^{-3}\\ & & & & & & \\ i).\ -(3^{-2}) & & j).\ 2\times(-5)^{-2} & & & & \\ \end{array}$

$\begin{array}{lllll} a)\;\frac{1}{x^{6}} & b)\;1 & c)\;\frac{3a}{b^{5}}. & d)\;\frac{1}{(pq)^{2}} & e)\;\frac{1}{(5xy)^{3}}=\frac{1}{125x^{3}y^{3}}\\ \\ f)\;2yz^{5} & g)\;\frac{1}{32} & h)\;-\frac{1}{8} & i)-\frac{1}{9} & j)\;\frac{2}{25} \end{array}$

### Exercise 3

Simplify the following:

$\begin{array}{lllll} a). \ 2a^{3}b^{2}\times a^{-1}\times b^{3} & & b). \ (5x^{-2}y)^{-3} & & c). \ (3x^{3}y^{-1})^{5}\\ & & & & \\ d). \ (a^{-4}b^{-5})^{-2} & & e). \frac{a^{2}b^{3}c^{-4}}{a^{4}bc^{5}} & & f). \frac{a^{7}\times a^{8}\times a^{3}}{a^{2}\times a^{5}}\\ & & & & \\ g). \ x\left(x-x^{-1}\right) & & h). \ \frac{(2^{4})^{n}}{2^{3}} & & i).\ \frac{15a^{2}b}{3a^{4}b}\times\frac{4a^{5}b^{2}}{5a^{3}b^{4}}\\ & & & & \\ j).\ 2^{4}-2^{3} & & & & \\ \end{array}$

$\begin{array}{lllllc} a)\;2a^{2}b^{5} & b)\;\frac{x^{6}}{5^{3}y^{3}} & c)\;\frac{3^{5}x^{15}}{y^{5}} & d)\;a^{8}b^{10} & e)\;\frac{b^{2}}{a^{2}c^{9}}\\ \\ f)\;a^{11} & g)\;x^{2}-1 & h)\;2^{4n-3} & i)\;\frac{4a^{7}b^{3}}{a^{7}b^{5}}=4b^{-2}=\frac{4}{b^{2}} & j)\;2^{3}(2^{1}-2^{0})=2^{3}=8 \end{array}$