Laws of indices

2 Laws of indices

There is a set of rules which enable us to manipulate expressions involving indices. These rules are known as the laws of indices , and they occur so commonly that it is worthwhile to memorise them.

Key Point 5

Laws of Indices

The laws of indices state:

First law : $a^{m} \times a^{n} = a^{m + n}$ add indices when multiplying numbers with the same base

Second law : $\frac{a^{m}}{a^{n}} = a^{m - n}$ subtract indices when dividing numbers with the same base

Third law : ${(a^{m})}^{n} = a^{m n}$ multiply indices together when raising a number to a power

Example 16

Simplify

$a^{5} \times a^{4}$ ,
$2 x^{5} (x^{3})$ .

Solution

In each case we are required to multiply expressions involving indices. The bases are the same and we use the first law of indices.

The indices must be added, thus $a^{5} \times a^{4} = a^{5 + 4} = a^{9}$ .
Because of the associativity of multiplication we can write
$2 x^{5} (x^{3}) = 2 (x^{5} x^{3}) = 2 x^{5 + 3} = 2 x^{8}$

The first law of indices (Key Point 5) extends in an obvious way when more terms are involved:

$\frac{8^{4}}{8^{2}}$ ,
$x^{18} \div x^{7}$ .

Solution

In each case we are required to divide expressions involving indices. The bases are the same and we use the second law of indices (Key Point 5).

The indices must be subtracted, thus $\frac{8^{4}}{8^{2}} = 8^{4 - 2} = 8^{2} = 64$ .
Again the indices are subtracted, and so $x^{18} \div x^{7} = x^{18 - 7} = x^{11} .$

Task!

Example 19

Simplify

${(8^{2})}^{3}$ ,
${(z^{3})}^{4}$ .

Solution

We use the third law of indices (Key Point 5).

${(8^{2})}^{3} = 8^{2 \times 3} = 8^{6}$
${(z^{3})}^{4} = z^{3 \times 4} = z^{12}$ .

Task!

Two important results which can be derived from the laws of indices state:

Key Point 6

Any non-zero number raised to the power 0 has the value 1, that is

a^{0} = 1

Any number raised to power 1 is itself, that is

a^{1} = a

A generalisation of the third law of indices states:

Key Point 7

{(a^{m} b^{n})}^{k} = a^{m k} b^{n k}

Example 20

Remove the brackets from

${(3 x)}^{2}$ ,
${(x^{3} y^{7})}^{4}$ .

Solution

Noting that $3 = 3^{1}$ and $x = x^{1}$ then ${(3 x)}^{2} = {(3^{1} x^{1})}^{2} = 3^{2} x^{2} = 9 x^{2}$

$or, alternatively$ ${(3 x)}^{2} = (3 x) \times (3 x) = 9 x^{2}$
${(x^{3} y^{7})}^{4} = x^{3 \times 4} y^{7 \times 4} = x^{12} y^{28}$

Exercises

Show that ${(- x y)}^{2}$ is equivalent to $x^{2} y^{2}$ whereas ${(- x y)}^{3}$ is equivalent to $- x^{3} y^{3}$ .

2 Laws of indices

Key Point 5

Example 16

Solution

Example 17

Solution

Task!

Example 18

Solution

Task!

Example 19

Solution

Task!

Key Point 6

Key Point 7

Example 20

Solution

Exercises

Interactive Exercises

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