# Lession 2: Fractional Indices

# Fractional Indices

## introduction

In the previous lesson you learnt about positive indices.In this lesson you will be exposed to another aspect of indices i.e fractional indices.

By the end of this lesson you should be able to
i)state laws of fractional indices ii)evaluate expressions in fractional indices |

## Lesson Content

In this lesson you will be learning about cases when the indices are fractions .

From previous work; 5^{1/2}*5^{1/2}=5^{1/2 + 1/2}

=5^{1}=5(from laws of positive indices)

which means that(5^{1/2})^{2}=5
therefore,5^{1/2}=2√5

now look at the following example

8^{1/3}*8^{1/3}*8^{1/3}=8^{1/3+1/3+1/3}=8^{1}=8

Hence(8^{1/3})^{3}=8

therefore8^{1/3}=3√8

Similarly a^{1/3}=3√a

In general a^{1/n}=n√a

Consider also 8^{2/3}

Now
8^{2/3}*8^{2/3}*8^{2/3}=(8^{2/3})^{3}

=8^{2/3*3/1}=8^{2}

Now if (8^{2/3})^{3}=8^{2}Then 8^{2/3}=
3√8^{2}

(by taking cube roots on both sides) so in general a^{1/n} =n√a

Therefore you have a^{m/n}=n√a^{m}

When m=1, then a^{m/n}=a^{1/n}=n√a^{1}
=n√a

Examples:Evaluate the following

i) 81^{1/4}

ii) 27^{1/3}
iii)(16/25)^{1/2}

Solutions
i) 81^{1/4} =(3^{4)})^{1/4}=3^{1}= 3

ii)27^{1/3}=(3^{3)}^{1/3}=3^{1}=3

iii)(16/25)^{1/2}= (4^{2})^{1/2}/(5^{2})^{1/2}

=4^{1}/5^{1}=4/5

- (81/16)
^{3/4} - Solve following equation:2
^{a2}=16^{a-1}

Maina 15:42, 26 February 2007 (CET)