INTRODUCTION TO INDICES
FIRST TERM
LEARNING NOTES
CLASS: JSS 2 (BASIC 8)
SCHEME OF WORK WITH LESSON NOTES
Subject:
MATHEMATICS
Term:
FIRST TERM
Week:
WEEK 2
Class:
JSS 2 (BASIC 8)
Previous lesson:
The pupils have previous knowledge of
WHOLE NUMBERS NOTATION AND NUMERATION OF NUMBERS
that was taught as a topic during the last lesson.
Topic :
INTRODUCTION TO INDICES
Behavioural objectives:
At the end of the lesson, the pupils should be able to
Instructional Materials:
 Wall charts
 Pictures
 Related Online Video
 Flash Cards
Methods of Teaching:
 Class Discussion
 Group Discussion
 Asking Questions
 Explanation
 Role Modelling
 Role Delegation
Reference Materials:
 Scheme of Work
 Online Information
 Textbooks
 Workbooks
 9 Year Basic Education Curriculum
 Workbooks
Content
INDICES
10 × 10 × 10 = 10^{3} is in index form, where 3 is the index or power of 10^{3}.
Similarly, p^{5} is short for p × p × p × p × p. 5 is the index of p in the expression p^{5}. We often say this as ‘p to the power of 5’. The plural of index is indices.
The following are the laws of indices:
Multiplication: X^{a} × X^{b} = X^{a + b}
Example: Multiply the following
(a) m^{3 }× m^{5}
(b) 10^{2} × 10^{7}
(c) 5y^{5} × 3y^{3}
Solutions:
(a) m^{3 }× m^{5}
By expansion,
m^{3} × m^{5 }= m × m × m × m × m × m × m × m = m^{8}
By adding index,
m^{3} x m^{5} = m^{3 + 5} = m^{8}.
(b) 10^{2} × 10^{7}
= 10^{2 + 7} = 10^{9}
(c) 5y^{5} × 3y^{3}
= (5 × 3) × y^{5 + 3}
= 15 × y^{8}
= 15y^{8}
Division: X^{a}÷ X^{b} = X^{a – b}
Example: Solve the following
(a) a^{7} ÷ a^{3 }
(b) 10a^{8} ÷ 5a^{6}
(c) 18x^{5} ÷ 9x^{4}
Solution
(a) a^{7} ÷ a^{3}
by expansion, we have
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By subtracting index,
a^{7} ÷ a^{3} = a^{7 – 3} = a^{4}
(b) 10a^{8} ÷ 5a^{6}
= (10 ÷ 5) × a^{8 – 6}
= 2 × a^{2}
= 2a^{2}
(c) 18x^{5} ÷ 9x^{4}
= (18 × (x^{5 – 4})
= 2 × x^{1}
= 2x
Zero and negative power: Any number to the power of zero is 1 and any number having a negative power becomes a fraction. X^{0 }= 1, X^{a} = ^{1}/_{x}^{a}
Example: Simplify the following
(a) 10^{2}
(b) x^{5} × x^{2}
(c) r^{7}× r^{7 }
(d) 2a^{1} × 3a^{2}
Solution
(a) 10^{2 }= ^{1}/_{10}
(b) X^{5} × X^{2}
= X^{5 + (2)}
= X^{5 – 2}
= X^{3}
= ^{1}/x^{3}
(c) r^{7}r^{7}
= r^{77} = r^{0}
= 1.
(d) 2a^{1} × 3a^{2}
= (2 × 3) × a^{1 + 2}
= 6a^{1}
= 6a.
CLASS ACTIVITY: simplify the following
(i) 2e^{4} × 5e^{10}
(ii) 51m^{9} ÷ 3m
(iii) (3.6 × 10^{7}) ÷ (1.2 × 10^{3})
(iv) (2a)^{1 }× 3a^{2}
(v) (^{1}/_{3})^{2}
Presentation
The topic is presented step by step
Step 1:
The class teacher revises the previous topics
Step 2.
He introduces the new topic
Step 3:
The class teacher allows the pupils to give their own examples and he corrects them when the needs arise
Evaluation
 Express each of the following in standard form

 (a) 7540058
 (b) 720 000 000
 (c) 9 400 000 000
 Express the following decimals in standard form

 (a) 05872
 (b) 0.00489
 (c) 0.000 005
 Write the following in ordinary form

 (a)342 × 10^{3}
 (b) 9.58 × 10^{4}
 The number 0.000 000 000 000 448 2 in standard form is
 Simplify the following
(a) 28z^{12} ÷ 4z^{10}
(b) 5 x 10^{6 }× 2 × 10^{4}
(c) y^{8} ÷ (^{1}/_{y})^{5}
Conclusion
The class 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 class 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.