**Introduction**

We define the notation 2^{2} as “two squared” or the square of two” or “two raised to the power of 2” and it means 2 x 2.

i.e. 2^{2} = 2 x 2 (i.e. 2 multiplied by itself in 2 places ) also 2^{3} = 2 x 2 x 2 ( i.e. 2 multiplied by itself in 3 places) and so on.

This definition is true for any number whatsoever and for any index or power.

4^{3} = 4 x 4 x 4, 100^{5} = 100 x 100 x 100 x 100 x 100.

The number of places that will multiply the number by itself is called the POWER or the INDEX of the base 2 or 4 or 100 as shown above. For example in the notation 7^{4}; 4 is the power or index while 7 is called the base. In mathematics the words “power” and “index” are used interchangeably i.e. they are synonymous, the mean the same thing.

13^{2} is read as 13 squared or 13 raised to power of 2, or 13 index 2.

13^{2} is an index number, it is an index notation for 169.

Since 13^{2} = 13 x 13 = 169, then the index form of the number 169 is 13^{2}. While the plural of “power” and “index” are “powers” and “indices” respectively and since we usually treat several cases of powers or indices we call this section “**indices” **and not index. Indices can apply to integers, fractions, positive/negative or mixed numbers. Indices can be integral or fractional, positive or negative.

Examples

Write 1/27 in the index forms

Solution

1/27 = 1/3 x 1/3 x 1/3 = (1/3)^{3} this is an index form for the number 1/27 which can also be written as 27^{-1}.

We use the index form usually with 10 as the base for every large numbers or very small numbers e.g. 10^{8} instead of 10,000,000. Indices play an important role in logarithm and vise versa because the concepts are related since the characteristics of logarithm can be compared with the indices. We shall discuss logarithm later.

**Reciprocals: **If the product of any two numbers equals 1 then the two numbers are called reciprocals of each other.

e.g. a. 5 and 1/5 are reciprocals since 5 x 1/5 = 1.

- 1/20 and 20 are reciprocals since 1/20 x 20 = 1.
- Suppose
*a*stands for any number, then*a*and 1/*a*are reciprocals since*a*x 1/*a*= 1.

Reciprocals are obtained by dividing one by the number. This holds whether the number is integral, fractional, mixed number, positive or negative number. The index of the reciprocal of any number *a* is *a*^{-1}.

We simplify what we have after dividing 1 by the number e.g.

- The reciprocal of 4 = ¼
- The reciprocal of 5/8 = 1/5/8 = 1/1 x 8/5 =8/5.
- The reciprocal of 2
^{1}/_{3}= 1¸ 2^{1}/_{3}=^{1}/_{1 }x^{3}/_{7}=^{3}/_{7} - The reciprocal of -5 =
^{1}/_{-5 }= -(^{1}/_{5})

Also reciprocals are numbers also known as the multiplicative inverses. This is because their product gives the multiplicative identity which is 1, an identity is that number which leaves another unchanged after the operation (addition and multiplication). We must note that a and –a are additive inverses because their sum gives additive identity 0.

**Laws of Indices**

In order to solve problems involving more easily, the following laws governing the operations with indices are provided and are true for any numbers *a*, *b* and *N* where *N *¹ 0.

**Multiplication (Products)**N^{a }x N^{b}

By definition *N ^{a}* =

*N*x

*N*x

*N*x …. to

*a*place

And *N ^{b}* =

*N*x

*N*x

*N*x …. to

*b*places

\ *N ^{a}* x

*N*= (

^{b}*N*x

*N*x

*N*x …. to

*a*places) x (

*N*x

*N*x

*N*x …. To be

*b*places)

= (*N* x *N* x *N* x …. To (*a* +*b)* places

\ For multiplication we add the indices

Example 1

- y
^{3}x y^{7}= (y x y x y) x (y x y x y x y x y x y x y)

= y x y x y x y x y x y x y x y x y x y

= y^{10} or y^{7+3}

2** Division (Quotient)** *N ^{a}* x

*N*

^{b}*N ^{a}* ¸

*N*=

^{b}*N*x

*N*x

*N*x …. to

*a*places/

*N*x

*N*x

*N*x …. to

*a*places

Some of the Ns will cancel out, leaving N x N x N x…. to (a + b) places. If a > b, (a – b) is positive, if a < b, (a – b) is negative. We have treated both positive and negative indices and know what they mean. If *a* = *b*, (*a* – *b*) = 0 and we have *N ^{0}*

^{.}Therefore for Division we subtract the indices.

** Example 2** 5

^{3n}¸ 5

^{n}

*Solution*** **5^{3n}¸ 5^{n} = 5^{3n-n} = 5^{2n}

3 **Powers** (N^{a})^{b}

By definition (N^{a})^{b} = (*N ^{a}* x

*N*x

^{a}*N*x …. to

^{a}*b*places)

{[(*N* x *N* x *N* x …. to *a* places ) x (*N* x *N* x *N* x to *a* places) x (*N* x *N* x *N* x …. to *a*

places) x *N* x *N* x *N* x …. to *a* places) x ….] to *b* places}

= [*N* x *N* x *N* x …. to (*a* x b*)* places]

= *N ^{ab}* by our definition.

For powers we multiply the indices.

Example 3

Simplify (2^{3})^{2}

(2^{3})^{2 }= 2^{3} x 2^{3} by definition

= 2 x 2 x 2 x 2 x 2 x 2

= 2^{6} which is 2^{(3×2)}

**Fractional Indices (or Roots) N**^{1/a}

Again we define the notation 2^{1/2} as Ö2, the square root of 2; 2^{1/3 }= ^{3}Ö2 i.e. the cube root of 2; 254^{1/4 }= ^{4}Ö254 – the fourth root of 245 and so on and in general.

Example 4

(1728)^{1/3 }= ^{3}Ö1728

= ^{3}Ö(2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3)

= 2 x 2 x 3

= 12

Hence for roots we divide the indices while for powers we multiply the indices.

We can denote *N ^{a/b}* +

^{b}*Ö*

*N*i.e.

^{a}*bth*root of

*N*.

^{a}Since ^{b}*Ö**N ^{a }*=

*Ö(*

^{b}*N*x

*N*x

*N*x …. to

*a*place)

= * ^{b}*Ö

*N*x

*Ö*

^{b}*N*x

*Ö*

^{b}*N*x …. to

*a*places

= *N ^{1}*

^{/b }x

*N*

^{1}^{/b }x

*N*

^{1}^{/b }x to

*a*places

= (*N ^{1}*

^{/b})

*and from above*

^{a} = *N ^{a}*

^{/b}

In fraction indices the numerator is regarded as power while the denominator is regarded as the root.

Example 5

Simplify ^{3}Ö8^{2}

^{3}Ö8^{ }= ^{3}Ö8 x 8

= ^{3}Ö(2 x 2 x 2) x (2 x 2 x 2)

= 2 x 2 = 4

**Meaning of N**^{0}i.e. Zero index or Zero power

*a ^{0}* = 1, 8

^{0}= 1 , i.e. any number N raised to power zero = 1, the multiplicative identity.

**Negative Power/Indices**

This states that 4^{-4} = 1/4^{4} = 1/4 x 4 x 4 x 4 = 1/256

Note: Ö represent Square root

**EXERCISES**

Lets see how much you’ve learnt, attach the following answers to the comment below

- Solve 3
^{x+y}x 3^{2x-y}A. 3^{2x }B. 3^{3xy }C. 3^{3x}D. 3^{2xy} - Solve (16/7)
^{-3 }A. (7/16)^{3}B. (7/16)^{-2-1 }C. (7/16)^{2 }D. (16/7)^{3} - Simplify 3
^{n+1 }x 3^{n-1}/9^{-1 }A. 3^{-4n }B. 3^{6n}C. 3^{4n}D. - Solve 81
^{1/4}A. 2 B. 4 C. 8 D. 3 - Solve, if 4
^{x}= 1/64, what is x? A. 3 B. 6 C. -3 D -4