# Standard form and Indices

Subject :

### Mathematics

Topic :

Class :

SS 1

Term :

First Term

Week :

Week 5

Instructional Materials :

• Wall charts
• Online Resources
• Pictures
• Related Audio Visual
• Mathematics Textbooks
• A chart showing modular arithmetic
• Samples of Duty shift
• Menstrual chart

Reference Materials

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• Education Curriculum

Previous Knowledge :

The pupils have previous knowledge of

### Conversion from Base Ten to other Bases and Conversion from one base to another

Behavioural Objectives: At the end of the lesson, the students should be able to

• solve problems on standard forms
• use the standard notation of indices appropriately
• identify induces as shorthand notation
• solve mental sums involving the laws in indices

### Content:

WEEK 5:

DATE……………………….

# TOPIC:

### Content:

• Revision of standard form
• Introduce indices and examples
• Laws of indices
• Application of indices, simple indicial equation

SUB-TOPIC    1: Revision    of standard    form

Example 1; Express the following in standard form

(a) 5.37 (b) 53.7 (c) 537 (d) 35.65 (e) 7500 (f) 1403420

Solution:

(a) 5.37 = 5.37 x 1

= 5.37 x

(b) 53.7 = 5.37 x

(c) 537 = 5.37 x 100

= 5.37 x 10 x 10

= 5.37 x

(d) 35.65 = 3.565 x 10

= 3.565 x

(e) 7500 = 7.5 x 1000

= 7.5 x

(f) 1403420 = 1.403420 x 1000000

= 1.403420 x

Example 2; Express the following in standard form

(a) 0.037 (b) 0.00065 (c) 0.0058 (d) 0.61

Solution:

(a) 3.7 x 10^{-2}

(b) 6.5 x 10^{-4}

(c) 5.8 x 10^{-3}

(d) 6.1 x 10^{-1}

## Method 1:

(a) 0.037 = 3.7 x 0.01

= 3.7 x

(b) 0.00065 = 6.5 x 0.0001

= 6.5 x

(c) 0.0058 = 5.8 x 0.001

= 5.8 x

(d) 0.61 = 6.1 x 0.1

= 6.1 x

## EVALUATION:

Express the following in standard form

1. (a) 86000    (b) 4730     (c) 307    (d) 1903000
2. (a) 0.075     (b) 0.00059     (c) 0.22    (d) 0.0000036

### SUB-TOPIC    2: Introduction of indices and examples

The law of indices, or exponentiation law, is a mathematical rule that defines how to calculate with numbers that are raised to a power. In mathematics, an index (or power) indicates how many times a number is multiplied by itself. For example, the expression “23” simply means “2 multiplied by two, three times”. The number 3 is called the index or power.

The law of indices states that when two numbers with the same base are raised to a power, the powers can be added. So, for example,

25 = 2 × 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 2

25 + 24 = 27

### This law can be extended to more than two numbers with the same base. For example,

23 = 2 × 2 × 2

22 = 2 × 2

21 = 2

23 + 22 + 21 = 27

The law of indices can also be used to simplify expressions that involve powers of the same base. For example, the expression “23 24” can be simplified using the law of indices:

23 × 24 = 2(3+4) = 27

This law is also known as the law of exponents or the exponentiation law.

Laws Of Indices

The following are the laws governing mathematical operations involving index numbers. These laws are true for all values of m, n and x ≠ 0.

• x m× x n =         x m+ n
• x m ÷ x n =         x m / x n = x m –n
• x –n =         1/ x n

x 0 = 1

(xy) n = x n y n

(x/y) n = x n / y n

x –m = 1 / x m

The zero power rule: Anything to the power of zero is equal to one

The negative exponent rule: Anything to the power of a negative is equal to one over that number

The product rule: When two powers with the same base are multiplied, add their exponents

The power of a power rule (or exponential growth): When a power is raised to a power, multiply the exponents

The quotient of two powers with the same base: When a base is both divided and raised to a power, subtract

Indices is the plural of the word index. An index is the power of a given number. Numbers are sometimes expressed in index form e.g. 8 can be expressed as 23 in index form, 81 can be

 5

expressed as 34 in index form, 1/125 can be expressed as 1/ 3 or 5-3 in index form etc.

A number when expressed in index form must have a base and a power, e.g. when 9 is expressed in index form, we have 32. In this case, 3 is the base and 2 is the index or power

 .

.When 625 is expressed in the index form, we have 54 Here, 5 is the base and 4 is the index or power.

In general, index numbers are written in the form xm where x is the base and m is the index. It should be noted that mathematical operations (ie +, _ ,× and ÷) involving these types of numbers does not follow the conventional way. There are laws that govern its mathematical operations.        Hence, this topic indices deals with mathematical operations involving index numbers.

SUB-TOPIC    3: Laws Of Indices

The following are the laws governing mathematical operations involving index numbers. These laws are true for all values of m, n and x ≠ 0.

• x m× x n =          x m+ n
• x m ÷ x n =          x m / x n = x m –n
• x –n =          1/ x n

(4) x o               =          1

(5) (x m)n          =          x m.n

(6) x1/n             =

(7) x m/n            =          (x m) 1/n

=

SUB-TOPIC    4: Applications        Of The Laws

The following are some examples of how to apply the laws of indices stated above.

LAW 1: x m X x n = x m + n

Example 1: Simplify the following

{i} 52 x 57                    {ii} 32 x 34 x 33 x 3

{iii} 23a-2 b x 2a5b3       {iv} x3/4 X x5/8

### Solution:

(i) 52 x 57 = 52+7

= 59

(ii) 32 x 34 x 33 x 3 = 32+4+3+1

= 310 (Since 3 = 31)

• 23a-2bx 2a5b3= 23 x 21 xa-2 xa5 xb1 xb3

= 23+1 a-2+5 b1+3

= 24a3b4

= 16a3b4

(iv) x3/4 X x5/8    = x(6+5)/8

= x11/8

LAW 2: x m ÷ x n = x m-n

### Example 2:

simplify the following (i) 37 ÷ 34

• 21a4 b3 7ab2
 ÷
• 9a5 b33a2b-2

### Solution:

(i)37 ÷34           =37 4

=33

= 27

• 21a4b3

7ab2                   = 3a4 – 1b3 – 2

= 3a3b

(iii) 9a5b3 ÷ 3a2b-2 = 3a5 2 b3 (-2)

= 3a3 b3 + 2

= 3a3b5

## EVALUATION

Simplify the following

(i) 73 x 76

• 25 x 23 x 22
• 34a-2b x 3a4b2
• a1/2 x a1/4

(v) 54¸ 57

• 15a3b5 5ab2
• 8a6b2¸ 4ab-2

LAW 3: x –n = 1/xn

### Example 3:

Simplify the following

(i) 2-4                (ii)        6a-2b3 3a3 b-5

(iii) (27/8 )-2/3

(iv) 24x4y6        (v) _1_ 8x9y3   3-2

### Solution:

(i)         2-4 = _1_ 24

= _1_

16

• 6a-2b3 = 6b3b5 3a3b-5 3a3a2

= 2b3+5

a3+2

= 2b8

a5

(iii)     27 -2/3       8 2/3

8     =     27

=

2/3

3 3×2/3

= 22

32

= 4

9

(iv)     24x4y6 = 3x4 . x 9. y6 . y 3

8x9y3

= 3×4-9 y6-3

= 3x-5y3

= 3y3

x5

(v)        _1_

3-2              = 32

= 9

LAW 4:            x0 = 1

### Example 4:

Simplify the following

(i)         (20ab7 x 15a6bc5)0

(ii)        (17x4y2)0 + 1

• 32 x 3-3 x 3

### Solution:

(i)         (20ab7 x 15a6bc5)0 = 1 (ii)    (17x4y2)0 + 1    = 1 + 1

= 2

(iii)       32 x 3-3 x 3       = 32 x 3-3 x 31

= 32-3+1

= 30

= 1

## LAW 5:

(xm)n        =          xm.n

### Example 5:

Simplify the following (i)      (ab2)3 x (2a4b)2

(ii)        5a3b2 x (2ab)-2

(iii)       (3x2)3 ÷ 9x-3

## Solution:

(i)         (ab2)3 x (2a4b)2

= a3 b2x3 X 22 X a4x2 b2

= a3b6 X 4a8b2

= 4a3+8 b6+2

= 4a11b8

• 5a3b2 x (2ab)-2

=          5a3b2 x 2-2a-2b-2

=          5 x 2-2a3 x a-2 x b2 x b-2

=          5 x 1 a3-2 b2-2

22

=          5 ab0

4

=     5a 4

(iii)       (3x2)3 ÷ 9x-3

=          33x2x3 ÷ 9x-3

=           27x6 ÷ 9x-3

=          3×6 – (-3)

=          3×6 + 3

=          3x9

## EVALUATION

Simplify the following

(i) 3-5          (ii) 2a-3b2 3a2b-4

• 15a3b5 (iv) 30x3y5 5ab2 6x7y2

(v)      3-4 2

(vi) (300ab2)0

• (27xy4)0 + 1
• 23 x 2-4 x 2
• (x3y)2 x (2xy2)5
• 2(ab2)3 x a2b
• 3a2b x (2ab)-3

LAW 6: x1/n    =

### Example 6

Simplify the following. (i.)

(ii.)

(iii.)      (0.027)1/3

### Solution:

(i.)

= (33)1/3

= 33 x 1/3

(ii).

=   3

1/3

=

=

= 4 2 x ½

5 2 x ½

= 4

5

(iii)       (0.027)1/3 =        0.027 1/3

1

=        _27_ 1/3

1000

=        33 1/3

103

=         33 x 1/3

103 x 1/3

=           _3_

10

LAW 7:    xm/n = (xm)1/n

=

Simplify the following.

• 3 a8 x 2a6 x 4a

• 3

8y-6

## Solution:

(i)

= 2x4x a8 +6-2     1/3

=       23a12 1/3

(ii)       3 8y-6      = 8y-6 1/3

=      8a121/3

= 23×1/3a12x1/3

= 2a4

=    23y-6 1/3

= 23×1/3xy-6×1/3

= 2y-2

= 2/ y2

## EVALUATION

Simplify the following

(i) 3 82

Ö 27

(ii) 4 51/16-3

Ö

 8

(iii)     33/ -2/3

(iv)          (v)

(vi)

(vii)     (1/81)-1/4

(viii)   (ix) 0.091/2

SUB-TOPIC    5: simple   indicial    equation

We shall consider the application of the laws of indices in solving index equations

### Example

(i)          2x2      = 50

(ii)           3x-1   = 81

(iii)           8x     = 64

Solution:

(i)         2x2       =          50

x2         =          50

2

x2         =          25

x2         =          52

∴   x          =          5

(ii)  3x 1 = 81 3x – 1 = 34 x – 1 = 4

x    =     4 + 1

∴ x     =     5

• 8x =          64

23x               =          26

3x         =          6

x          =          6/3

∴ x         =          2

## EVALUATION

Solve the following index equation

1.  32x = 27
2. 9x = 81
3. 5a3 = 40
4. a-1/2 = 5
5. 4x = 32
6. x -2 = 4
7. 4x + 1 = 64
8. 4x x 8 = 64
9. 32x-1 = 27
10. 27 x 3x = 81
11. 2x1/3 = 16
12. 3x2 = 27

## GENERAL EVALUATION

• Given that3x91+x=27-x, find x
• Given that32x= 27-9x, find x.
• Given that4a3 = 40-5a, find a.
• Given that16v = 4096 – 6v4, find v.
• Given that4x = 32 – 3×2, find x.
• Given that12y = 144 – 4y

### SSCE, June 1994.

• If 4x = 2½ x 8, find x [WAEC]
•  If 8x/2 = 23/8 x 43/4, find x [WAEC]
• If 2(a + b) = a(a + 4b), find the value of (ab)
• If 3(2x – 5) = 4(x + 1), find the value of x
• In ∆ABC, if a:b:c= 2:3:4 and B= 60°, find A and C
• In trapezium PQRS, if PS|| QR, PQ= 9cm, QR= 12cm and SR= 8cm, find the area of ∆PSR.
• Given that 4x = 32, find the value of 1985[WAEC]
• Solve the equation 125x+3 = 5, find

### 1981[WAEC]

• Given that 8 x 4x-2 = 64, find
• Given that 1252x+1 = 625 x 25-x, find

(8) If 3m x 27(2m-1) = 81, find m.

### [WAEC]

• Find the value of x in 8-1 x 2(2x+1) = 64.      [WAEC].
• Find the value of x, given that 7 x 49(x+2) =

The law of indices is a mathematical rule that governs the behavior of numbers when they are raised to a power. It states that when a number is raised to a power, the resulting value is equal to the product of the number and all of the preceding integers up to and including the exponent. In other words, the law of indices says that a number raised to a power is equal to the number multiplied by itself as many times as the exponent dictates.

For example, according to the law of indices, 8 raised to the 3rd power is equal to 8x8x8, or 512. This is because 8 multiplied by itself 3 times equals 512. Similarly, 9 raised to the 4th power is equal to 9x9x9x9, or 6561.

The law of indices can be used to simplify expressions that involve powers. For instance, if you wanted to calculate the value of 8 raised to the 5th power, you could first use the law of indices to rewrite the expression as 8x8x8x8x8. This is much easier to calculate than 8 raised to the 5th power, which would be equal to 32768.

Here are five examples of the law of indices in action:

1. 3 raised to the 2nd power is equal to 3×3, or 9.

2. 4 raised to the 3rd power is equal to 4x4x4, or 64.

3. 5 raised to the 4th power is equal to 5x5x5x5, or 625.

4. 6 raised to the 5th power is equal to 6x6x6x6x6, or 7776.

5. 7 raised to the 6th power is equal to 7x7x7x7x7x7, or 117649.

### Presentation

The topic is presented step by step

Step 1:

The subject teacher revises the previous topics

Step 2.

He or she introduces the new topic.

Step 3:

The subject teacher allows the pupils to give their own examples and he corrects them when the needs arise

### EVALUATION:

1. If 2 raised to the 3rd power equals 8, what does 2 raised to the 4th power equal?

2. If 3 raised to the 4th power equals 81, what does 3 raised to the 5th power equal?

3. If 4 raised to the 5th power equals 1024, what does 4 raised to the 6th power equal?

4. If 5 raised to the 6th power equals 15625, what does 5 raised to the 7th power equal?

5. If 6 raised to the 7th power equals 279936, what does 6 raised to the 8th power equal?

New General Mathematics for SSS 1, pages 227; exercises 20a, 20b Mathematical Association of Nigeria (MAN) pages 14-24

1. 12y = 144
2. 16v = 4096
3. 6n4 = 1296
4. -2g = -8
5. 3i2 = 9
6. -5h1/2 = -25
7. k0.5 = 1
8. -6m4 = -1296
9. 10p3= 1000
10. -8q2= -64
11. r-0.5 = 1/√r
12. s1/3= ∛s
13. -27t3= -19683
14. u1/4= ∜u
15. -3v2= -9
16. w-0.75=-1/√w
17. x2/3= ∛x
18. -5y4=-3125
19. z1/6= 1/(√z)³

Conclusion:

The subject teacher wraps up or concludes the lesson by giving out a short note to summarize the topic that he or she has just taught.

The subject teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she makes the necessary corrections when and where the needs arise.

### REFERENCE TEXTS:

• New General Mathematics for senior secondary schools 1 by M.F Macrae et al; pearson education limited
• New school mathematics for senior secondary school et al; Africana publishers limited