Standard form and Indices
Subject :
Mathematics
Topic :
Class :
SS 1
Term :
First Term
Week :
Week 5
Instructional Materials :
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 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:
Standard form and Indices
Content:
 Revision of standard form
 Introduce indices and examples
 Laws of indices
 Application of indices, simple indicial equation
SUBTOPIC 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
Express the following numbers in stand ard form:
1
(a) 6000
(b) 7000
(e) 80 000
(d) 52
(e) 46
(D3
(g) 520
(h) 6400
(1) 2
() 4000
(k) 152
(1) 64 000 000
Here are the numbers expressed in standard form:
(a) 6000 = 6 x 10^3
(b) 7000 = 7 x 10^3
(c) 80,000 = 8 x 10^4
(d) 52 = 5.2 x 10^1
(e) 46 = 4.6 x 10^1
(f) 520 = 5.2 x 10^2
(g) 6400 = 6.4 x 10^3
(h) 2 = 2 x 10^0 (Note: Any nonzero number to the power of 0 is 1)
(i) 4000 = 4 x 10^3
(j) 152 = 1.52 x 10^2
(k) 64,000,000 = 6.4 x 10^7
EVALUATION:
Express the following in standard form
 (a) 86000 (b) 4730 (c) 307 (d) 1903000
 (a) 0.075 (b) 0.00059 (c) 0.22 (d) 0.0000036
SUBTOPIC 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 2^{3} in index form, 81 can be

expressed as 3^{4} 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 3^{2}_{.} In this case, 3 is the base and 2 is the index or power

.When 625 is expressed in the index form, we have 5^{4} Here, 5 is the base and 4 is the index or power.
In general, index numbers are written in the form x^{m} 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.
SUBTOPIC 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) x^{1/n} =
(7) x ^{m/n} = (x ^{m}) ^{1/n}
=
SUBTOPIC 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} 5^{2} x 5^{7} {ii} 3^{2} x 3^{4} x 3^{3} x 3
{iii} 2^{3}a^{2} b x 2a^{5}b^{3} {iv} x^{3/4} X x^{5/8}
Solution:
(i) 5^{2} x 5^{7} = 5^{2+7}
= 5^{9}
(ii) 3^{2} x 3^{4} x 3^{3} x 3 = 3^{2+4+3+1}
= 310 (Since 3 = 31)
 2^{3}a^{2}bx 2a^{5}b^{3}= 2^{3} x 2^{1} xa^{2} xa^{5} xb^{1} xb^{3}
= 23+1 a2+5 b1+3
= 2^{4}a^{3}b^{4}
= 16a^{3}b^{4}
(iv) x3/4 X x5/8 = x(6+5)/8
= x11/8
LAW 2: x ^{m} ÷ x ^{n} = x ^{mn}
Example 2:
simplify the following (i) 3^{7} ÷ 3^{4}
 21a^{4} b^{3} 7ab^{2}

 9a^{5} b33a^{2}b^{2}
Solution^{:}
(i)3^{7} ÷3^{4} =3^{7} ^{–} ^{4}
=3^{3}
= 27
 21a^{4}b^{3}
7ab2 = 3a4 – 1b3 – 2
= 3a^{3}b
(iii) 9a^{5}b^{3} ÷ 3a^{2}b^{2} = 3a^{5} ^{–} ^{2} b^{3} ^{–} ^{(2)}
= 3a3 b3 + 2
= 3a^{3}b^{5}
EVALUATION
Simplify the following
(i) 73 x 76
 25 x 23 x 22
 34a2b x 3a4b2
 a1/2 x a1/4
(v) 5^{4}¸ 5^{7}
 15a^{3}b^{5} 5ab^{2}
 8a^{6}b^{2}¸ 4ab^{2}
LAW 3: x ^{–n} = ^{1}/x^{n}
Example 3:
Simplify the following
(i) 2^{4} (ii) 6a^{2}b^{3} 3a3 b5
(iii) (^{27}/_{8} )^{2/3}
(iv) 24x^{4}y^{6} (v) _1_ 8x9y3 32
Solution:
(i) 2^{4} = _1_ 24
= _1_
16
 6a^{2}b^{3} = 6b^{3}b^{5} 3a^{3}b^{5} 3a^{3}a^{2}
= 2b3+5
a3+2
= 2b^{8}
a5
(iii) 27 ^{2/3} 8 ^{2/3}
8 = 27
=
2/3
3 3×2/3
= 2^{2}
32
= 4
9
(iv) 24x^{4}y^{6} = 3x^{4} . x^{–} ^{9}. y^{6} . y^{–} ^{3}
8x^{9}y^{3}
= 3×49 y63
= 3x^{5}y^{3}
= 3y^{3}
x5
(v) _1_
32 = 32
= 9
LAW 4: x^{0} = 1
Example 4:
Simplify the following
(i) (20ab^{7} x 15a^{6}bc^{5})^{0}
(ii) (17x^{4}y^{2})^{0} + 1
 3^{2} x 3^{3} x 3
Solution:
(i) (20ab^{7} x 15a^{6}bc^{5})^{0} = 1 (ii) (17x^{4}y^{2})^{0} + 1 = 1 + 1
= 2
(iii) 3^{2} x 3^{3} x 3 = 3^{2} x 3^{3} x 3^{1}
= 323+1
= 3^{0}
= 1
LAW 5:
(xm)n = xm.n
Example 5:
Simplify the following (i) (ab^{2})^{3} x (2a^{4}b)^{2}
(ii) 5a^{3}b^{2} x (2ab)^{2}
(iii) (3x^{2})^{3} ÷ 9x^{3}
Solution:
(i) (ab^{2})^{3} x (2a^{4}b)^{2}
= a3 b2x3 X 22 X a4x2 b2
= a^{3}b^{6} X 4a^{8}b^{2}
= 4a3+8 b6+2
= 4a^{11}b^{8}
 5a3b2 x (2ab)2
= 5a3b2 x 22a2b2
= 5 x 2^{2}a^{3} x a^{2} x b^{2} x b^{2}
= 5 x 1 a32 b22
22
= 5 ab^{0}
4
= 5a 4
(iii) (3x^{2})^{3} ÷ 9x^{3}
= 33x2x3 ÷ 9x3
= 27x^{6} ÷ 9x^{3}
= 3×6 – (3)
= 3×6 + 3
= 3x^{9}
EVALUATION
Simplify the following
(i) 3^{5} (ii) 2a^{3}b^{2} 3a2b4
 15a^{3}b^{5} (iv) 30x^{3}y^{5} 5ab^{2} 6x^{7}y^{2}
(v) 3^{4} 2
(vi) (300ab^{2})^{0}
 (27xy4)0 + 1
 2^{3} x 2^{4} x 2
 (x^{3}y)^{2} x (2xy^{2})^{5}
 2(ab^{2})^{3} x a^{2}b
 3a^{2}b x (2ab)^{3}
LAW 6: x^{1/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
10^{3}
= 33 x 1/3
103 x 1/3
= _3_
10
LAW 7: x^{m/n} = (x^{m})^{1/n}
=
Simplify the following.
 3 a^{8} x 2a^{6} x 4a
 ^{3}
8y6
Solution:
(i)
= 2x4x a8 +62 1/3
= 23a12 1/3
(ii) ^{3} 8y^{6} = 8y^{6} ^{1/3}
= 8a121/3
= 23×1/3a12x1/3
= 2a^{4}
= 23y6 1/3
= 23×1/3xy6×1/3
= 2y^{2}
= 2/ y2
EVALUATION
Simplify the following
(i) _{3} 8^{2}
Ö 27
(ii) 4 51/163
Ö

(iii) 3^{3}/ ^{2/3}
(iv) (v)
(vi)
(vii) (^{1}/_{81})^{1/4}
(viii) (ix) 0.09^{1/2}
SUBTOPIC 5: simple indicial equation
We shall consider the application of the laws of indices in solving index equations
Example
(i) 2x^{2} = 50
(ii) 3^{x1} = 81
(iii) 8^{x} = 64
Solution:
(i) 2x^{2} = 50
x^{2} = 50
2
x^{2} = 25
x^{2} = 5^{2}
∴ x = 5
(ii) 3^{x} ^{–} ^{1} = 81 3x – 1 = 34 x – 1 = 4
x = 4 + 1
∴ x = 5
 8^{x} = 64
23x = 26
3x = 6
x = ^{6}/_{3}
∴ x = 2
EVALUATION
Solve the following index equation
 3^{2x} = 27
 9^{x} = 81
 5a^{3} = 40
 a1/2 = 5
 4^{x} = 32
 x ^{2} = 4
 4^{x} ^{+} ^{1} = 64
 4^{x} x 8 = 64
 3^{2x1} = 27
 27 x 3^{x} = 81
 2x^{1/3} = 16
 3x^{2} = 27
GENERAL EVALUATION
 Given that3x91+x=27x, find x
 Given that32x= 279x, find x.
 Given that4a3 = 405a, find a.
 Given that16v = 4096 – 6v4, find v.
 Given that4x = 32 – 3×2, find x.
 Given that12y = 144 – 4y
SSCE, June 1994.
 If 4x = 2^{½} 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 4^{x} = 32, find the value of 1985[WAEC]
 Solve the equation 125^{x+3} = 5, find
1981[WAEC]
 Given that 8 x 4^{x2} = 64, find
 Given that 125^{2x+1} = 625 x 25^{x}, find
(8) If 3^{m} x 27^{(2m1)} = 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?
READING ASSIGNMENT
New General Mathematics for SSS 1, pages 227; exercises 20a, 20b Mathematical Association of Nigeria (MAN) pages 1424
 12y = 144
 16v = 4096
 6n4 = 1296
 2g = 8
 3i2 = 9
 5h1/2 = 25
 k0.5 = 1
 6m4 = 1296
 10p3= 1000
 8q2= 64
 r0.5 = 1/√r
 s1/3= ∛s
 27t3= 19683
 u1/4= ∜u
 3v2= 9
 w0.75=1/√w
 x2/3= ∛x
 5y4=3125
 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