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# Indices

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7.1Indices

In this Math tutorial, you will learn:

• What are indices? What is the other name used for indices?
• How indices are related to powers?
• What are the properties of indices?
• What is the reciprocal of a number?
• How is the reciprocal involved in raising a number in a negative power?
• How to calculate the powers of negative numbers?
• What differs when raising a negative number to an odd power to raising the corresponding positive number at the same power?

## Introduction

This tutorial is dedicated to indices (or exponents, as they are randomly known). We have included indices our tutorials in various occasions, but this is the right moment to have them explained thoroughly. In this tutorial a number of properties of indices (also known as the "indices laws") are explained by illustrating each property with examples. Moreover, negative indices and negative numbers raised in a given power are (among others) new situations discussed in this tutorial.

## Powers, Indices, and Exponents

In previous tutorials, we have provided a general description of powers and exponents. Let's recall what we have said about powers in tutorial 1.3.

If we have to multiply the same factor several times by itself, we use a shorter notation known as "power" to represent such a recurring multiplication by the same number. In other words, the power of a number says how many times to use that number in a multiplication.

For example, instead of writing 5 × 5 × 5, we can write 53 which reads "5 at power 3".

In general,

The recurring factor is called the base, the number that shows how many times this factor appears in a recurring multiplication is called the exponent and the result of this operation is called power. Therefore, the popular terminology used to express this operation is not 100% correct. Thus, in the expression 53 = 125, we should have said "the power of 5 with exponent 3 is 125" instead of "5 raised in power 3 gives 125". However, we are now familiar with the second way of expression and nobody complains about that.

Thus, the schematic representation of power is as below:

Another name used for the exponent is index (indices in plural). Thus, for example, in the operation

34 = 81

the base is 3, the exponent (or index) is 4, and the power is 81.

### Example 1

Write the following expressions more shortly and calculate the result if possible.

1. 2 × 2 × 2 × 2 × 3 × 3
2. m × m × m × n × n × n × n × n

### Solution 1

1. We have four '2's and '3's in this expression. Thus,
2 × 2 × 2 × 2 × 3 × 3
= 24 × 32
= 16 × 9
= 144
2. We have three 'm's and five 'n's in this expression. Thus,
m × m × m × n × n × n × n × n
= m3 × n5

## Properties of Indices

Now, we will explain some properties of indices by illustrating them with numbers as proof.

Property 1: When multiplying powers with the same base, we add the indices.

In symbols,

am × an = am + n

For example,

23 × 24 = 23 + 4
= 27
= 128

Indeed,

23 = 8 and 24 = 16

Thus,

23 × 24 = 8 × 16
= 128

Property 2: When dividing powers with the same base, we subtract the indices.

In symbols,

am ÷ an = am ÷ n

For example,

25 ÷ 23 = 25 - 3
= 22
= 4

Indeed,

25 = 32 and 23 = 8

Thus,

25 × 23 = 32 ÷ 8
= 4

Property 3: When raising a power into another power, we multiply the indices without changing the base.

In symbols,

(am )n = m × n

For example,

(34 )2 = 34 × 2 = 38 = 6,561

Indeed,

34 = 81

and

812 = 6,561

Property 4: If two numbers of different bases are raised at the same power, the bases multiply without changing the index.

In symbols,

ac× bc= (a × b)c

For example,

32 × 42 = (3 × 4)2
= 122
= 144

Indeed,

32 = 9 and 42 = 16

Thus,

32 × 42 = 9 × 16
= 144

Property 5: Any number raised to the first power gives the number itself.

In symbols,

a1 = a

For example,

51 = 5

Indeed, we can write using the second property of indices

51 = 53-2
= 53 ÷ 52
= 125 ÷ 25
= 5

Property 6: Any number raised at power zero gives 1.

In symbols,

a0 = 1

For example,

70 = 1

Indeed, we can write using the second property of indices

70 = 72 - 2
= 72 ÷ 72
= 49 ÷ 49
= 1

### Example 2

Calculate the following expressions using the properties of indices.

1. 121 437/121 435
2. 35 × 37/38

### Solution 2

1. From the second property of indices, we have
121 437/121 435 = 121437 - 1435
= 122
= 144
Obviously, it would have been impossible to calculate the result of this expression through the standard way, i.e. calculating the value of numerator and denominator separately and then, dividing them.
2. First, we do the operations in the numerator using the first property of indices. Then, we divide the resulting numerator by the denominator, i.e.
35 × 37/38 = 35 + 7/38
= 31 2/38
= 312 - 8
= 34
= 81
Indeed, if we calculated the powers one by one we would obtain
35 × 37/38 = 243 × 2,187/6,561
= 531,441/6,561
= 81

## Negative Indices. The Meaning of Reciprocal

We have briefly mentioned in the previous tutorial the concept of negative powers when dealing with powers of algebraic fractions. In that case, we explained that if an algebraic fraction is raised at a negative power, the fraction is inverted upside down. We call the new number obtained (whether it may be a fraction or integer) the reciprocal of the original number.

By definition, the reciprocal represents a number, expression, or function so related to another that their product is unity (one, therefore). In other words, the reciprocal of a quantity represents the quantity obtained by dividing the number one by a given quantity.

For example the reciprocal of 3 is 1/3, the reciprocal of 2x is 1/2x, etc.

There is a property involving negative indices that says:

If a number or expression is raised to a negative power, it is equal to the reciprocal raised at the corresponding positive power.

In symbols,

a-b = (1/a)b

In fractions, the reciprocal is obtained by swapping the positions of numerator and denominator. For example, the reciprocal of a/b is b/a, the reciprocal of 3/5 is 5/3 and so on.

### Example 3

Calculate the value of the following expressions.

1. (32 ∙ 43/25 )-4
2. (24/52 )-3

### Solution 3

1. We have
(32 ∙ 43/25 )-4 = (25/32 ∙ 43)4
We can express 43 as (22)3 = 22 × 3 = 26. Thus, we obtain
(25/32 ∙ 43)4 = ( 25/32 ∙ 26)4
= (25 - 6/32 )4
= (2-1/32 )4
= (1/21 ∙ 32)4
= (1/2 ∙ 9)4
= (1/18)4
= 14/184
= 1/104,976
2. We have
(24/52 )-3 = (52/24)3
= 52 × 3/24 × 3
= 56/212
= 15,625/4,096

## Powers of Negative Numbers

In this paragraph, we discuss the sign of the result obtained by raising a number at a given power. Thus, we don't have concerns if a positive number (i.e. if the base is positive) is raised at a given power because the result is always positive regardless of the sign of the index. Thus, if the index is positive, it is obvious that we always obtain a positive number as a result because in such situations we multiply a positive number several times (determined by the value of index) by itself.

However, if a negative number is raised at a given power, we must be very careful about the sign of the final result because from the table of sign rules provided in tutorial 6.2, it is clear that:

If a negative number is raised at an even power (the index is an even number), the result is positive and when a negative number is raised at an odd power (the index is an odd number) the result is negative.

For example,

(-2)3 = -8

because

(-2)3 = (-2) ∙ (-2) ∙ (-2)
= [(-2) ∙ (-2)] ∙ (-2)
= ( + 4) ∙ (-2)
= -8

On the other hand,

(-2)4 = + 16

because

(-2)4 = (-2) ∙ (-2) ∙ (-2) ∙ (-2)
= [(-2) ∙ (-2)] ∙ [(-2) ∙ (-2)]
= ( + 4) ∙ ( + 4)
= + 16

### Example 4

Calculate the value of the following expressions:

1. (- 3)4/(-2)3
2. (- 1)167 · (- 2)3 · (- 5)2

### Solution 4

1. We have
(-3)4/(-2)3 = 81/-8 = -81/8
2. We have
(-1)167 ∙ (-2)3 ∙ (-5)2
= (-1) ∙ (-8) ∙ ( + 2)
= [(-1) ∙ (-8)] ∙ ( + 2)
= ( + 8) ∙ ( + 2)
= + 16

### Putting All Together

Now, let's see a couple of examples where all the above properties of indices can be applied.

### Example 5

Simplify the value of the following algebraic expressions.

1. 32 + x + 3x/2x ∙ 5x
2. (24 ∙ 4-2/35 ∙ 9-3)-2

### Solution 5

1. We have:
32 + x + 3x/2x ∙ 5x = 32 ∙ 3x + 3x/(2 ∙ 5)x
= 3x (32 + 1)/(2 ∙ 5)x
= 3x ∙ 10/10x
= 3x ∙ 101/10x
= 3x ∙ 101 - x
2. We have
(24 ∙ 4-2/35 ∙ 9-3)-2 = [24 ∙ (22 )-2/35 ∙ (32 )-3 ]-2
= [24 ∙ 22 ∙ (-2)/35 ∙ 32 ∙ (-3)]-2
= [24 ∙ 2-4/35 ∙ 3-6]-2
= [24 + (-4)/35 + (-6) ]-2
= [ 20/3-1 ]-2
= [3-1/20 ]2
= (1/1 ∙ 3)2
= (1/3)2
= 12/32
= 1/9

## Whats next?

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