Mathematics notes, chapter 1

Indices

Textbook Mathematics Form 3 (DLP), pages 1–29

What this chapter covers

Represent repeated multiplication in index form and describe its meaning; rewrite a number in index form and vice versa (1.1)
Relate multiplication, division and power of numbers in index form to repeated multiplication and state the laws of indices (1.2)
Verify that a⁰ = 1 and a⁻ⁿ = 1/aⁿ, and state the relationship between fractional indices, roots and powers (1.2)
Perform operations involving the laws of indices (1.2)
Solve problems involving the laws of indices, including equations with the unknown in the index (1.2)

1.1Index notation

Index notation is a short way to write repeated multiplication. In aⁿ, a is the base and n is the index (also called the exponent or power). It tells us how many times the base is multiplied by itself: aⁿ = a × a × … × a (n factors). We read 4³ as "4 to the power of 3".

Repeated multiplication	Index form	Base	Index
5 × 5 × 5 × 5 × 5 × 5	5⁶	5	6
(−2) × (−2) × (−2)	(−2)³	−2	3
0.3 × 0.3 × 0.3 × 0.3	(0.3)⁴	0.3	4
25 × 25 × 25	(25)³	25	3

A negative or fractional base must go inside brackets. (−2)⁴ = 16 but −2⁴ = −16, because only the 2 is raised to the power in the second. Also 2⁴ means 2 × 2 × 2 × 2 = 16, not 2 × 4 = 8.

Writing a number in index form

A number can be written in index form with a suitable base. Divide repeatedly by the base until you reach 1 (the number of divisions is the index), or multiply the base repeatedly until you reach the number.

Example 2 (textbook p. 4): 64 with different bases

64 ÷ 2 ÷ 2 ÷ 2 ÷ 2 ÷ 2 ÷ 2 = 1 (6 divisions), so 64 = 2⁶.
4 × 4 × 4 = 64, so 64 = 4³.
8 × 8 = 64, so 64 = 8².

Finding the value of a number in index form

Multiply out the repeated multiplication, or use a scientific calculator with the ^ key. Put a negative or fractional base in brackets on the calculator.

Example 5 (textbook p. 5): values

(−7)³ = (−7) × (−7) × (−7) = −343
(23)⁴ = 2⁴3⁴ = 1681
(−0.5)⁶ = 0.015 625 (an even index gives a positive answer)

Negative base: an even index gives a positive value and an odd index gives a negative value.

Write 625 in index form with base 5, and (−3)⁴ as a number.

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625 = 5 × 5 × 5 × 5 = 5⁴
(−3)⁴ = 81

1.2Law of indices

The laws for the same base

Writing out the repeated multiplication shows each law. For example, 2³ × 2⁴ = (2 × 2 × 2) × (2 × 2 × 2 × 2) = 2⁷, and 4⁵ ÷ 4² = (4 × 4 × 4 × 4 × 4) ÷ (4 × 4) = 4³.

Law	Rule	Example
Multiplication	a^m × aⁿ = a^{m + n}	7² × 7³ = 7⁵
Division	a^m ÷ aⁿ = a^{m − n}	5⁴ ÷ 5² = 5²
Power of a power	(a^m)ⁿ = a^mn	(3⁴)² = 3⁸
Power of a product	(a^mbⁿ)^q = a^mqb^nq	(5m⁴n³)² = 25m⁸n⁶
Power of a quotient	(a^mbⁿ)^q = a^mqb^nq	(2x³3y)⁴ = 16x¹²81y⁴

The laws of multiplication and division work only when the bases are the same. 2³ × 3² cannot be written as 6⁵. Also, a^m × aⁿ adds the indices, but (a^m)ⁿ multiplies them.

Different bases and coefficients

Group the terms with the same base first, then add the indices. Operate on the coefficients (the numbers in front) separately.

Example 7(d) and 8(d) (textbook pp. 8-10)

Simplify (−m²) × 2n⁵ × 3m³ × 14n².
Coefficients: (−1) × 2 × 3 × 14 = −32
Bases: m²⁺³ × n⁵⁺² = m⁵n⁷
Answer: −32m⁵n⁷

Simplify 25x²y³ ÷ 5xy.
= (25 ÷ 5) × x²⁻¹ × y³⁻¹ = 5xy²

Simplify (a) 3a⁴ × 5a² (b) 36p⁷q³ ÷ 9p²q (c) (2m³n)⁴

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(a) 15a⁶
(b) 4p⁵q²
(c) 16m¹²n⁴

Zero index and negative index

Since a^m ÷ a^m = 1 and also a^{m − m} = a⁰, we get a⁰ = 1. In the same way 2³ ÷ 2⁵ = 2 × 2 × 22 × 2 × 2 × 2 × 2 = 12² and also 2^{3 − 5} = 2⁻².

Zero index: a⁰ = 1, for a ≠ 0.
Negative index: a⁻ⁿ = 1aⁿ and 1a⁻ⁿ = aⁿ, for a ≠ 0. A negative index means "reciprocal", not "negative number".

Term	In positive index form
x⁻⁴	1x⁴
2m⁻³	2m³ (only m has the negative index)
18⁻⁵	8⁵
(23)⁻²	(32)² = 94

2⁻³ is 18, not −8 and not −6. And 3⁻¹ × 3⁻¹ = 3⁻² = 19.

Write in positive index form and find the value: (a) 5⁻³ (b) (−3)⁰ (c) (12)⁻⁴

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(a) 15³ = 1125
(b) 1
(c) 2⁴ = 16

Fractional indices

The square root undoes squaring, so a^1/2 × a^1/2 = a¹ shows that a^1/2 = √a. In general the root is the reciprocal index.

Fractional index: a^1/n = ⁿ√a, for a ≠ 0.
a^m/n = (a^m)^1/n = (a^1/n)^m = ⁿ√(a^m) = (ⁿ√a)^m

Take the root first, then the power. It keeps the numbers small: 16^3/4 = (⁴√16)³ = 2³ = 8.

Example 12 and 14 (textbook pp. 17-19)

³√(−27) = (−27)^1/3 = −3
9^5/2 = (√9)⁵ = 3⁵ = 243
8^−2/3 = 18^2/3 = 1(³√8)² = 14

Calculate (a) 64^1/3 (b) 81^3/4 (c) 125^−2/3

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(a) 4
(b) (⁴√81)³ = 3³ = 27
(c) 1(³√125)² = 125

Using the laws of indices (1.2.8 and 1.2.9)

Simplifying expressions

Write roots as fractional indices.
Apply the power laws to every bracket.
Group the same bases and add or subtract the indices.
Give the answer in positive indices, unless told otherwise.

Simplify (4xy²)² × x⁵y ÷ (2x³y)⁵ (like textbook Example 11)

Top: 16x²y⁴ × x⁵y = 16x⁷y⁵
Bottom: 32x¹⁵y⁵
= 1632 x⁷⁻¹⁵y⁵⁻⁵ = 12x⁻⁸y⁰ = 12x⁸

Numbers with different bases

Change each number into its prime factors so the bases match.

Different bases, no calculator (idea of textbook Example 17, p. 22)

Calculate 8^2/3 × 9^−1/2 × 6².
8^2/3 = (2³)^2/3 = 2² = 4
9^−1/2 = 13
6² = 36
4 × 13 × 36 = 48

Equations with the unknown in the index

Make the bases the same, add or subtract the indices, then equate the indices: if a^m = aⁿ then m = n.

Example 18 and 19 (textbook pp. 22-23)

Solve 3^x × 9^x+5 ÷ 3⁴ = 1.
3^x × 3^2(x+5) ÷ 3⁴ = 3⁰
3^{x + 2x + 10 − 4} = 3⁰ so 3x + 6 = 0 and x = −2.

Solve 3^x² × 3^2x = 3¹⁵.
x² + 2x = 15, x² + 2x − 15 = 0, (x − 3)(x + 5) = 0
x = 3 or x = −5

Check by substituting your answer back into the original equation. Both sides must give the same value.

Simultaneous equations in indices (textbook Example 20, p. 23)

Solve 25^m × 5ⁿ = 5⁸ and 2^m × 12ⁿ = 2.
5^{2m + n} = 5⁸ gives 2m + n = 8 (1)
2^{m − n} = 2¹ gives m − n = 1 (2)
(1) + (2): 3m = 9, m = 3, then n = 2.

Solve 2^x × 4^x+1 = 2¹¹.

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2^{x + 2(x+1)} = 2¹¹
3x + 2 = 11
x = 3

A car costs RM55 000 and its value after n years is RM55 000 × (89)ⁿ (textbook p. 26). Find the value after 2 years.

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55 000 × 6481 = RM43 456.79, about RM43 457

Summary

Remember	Key fact
Index notation	aⁿ = a × a × … × a (n factors); a is the base, n is the index
Same base	a^m × aⁿ = a^m+n; a^m ÷ aⁿ = a^m−n; (a^m)ⁿ = a^mn
Zero index	a⁰ = 1 (a ≠ 0)
Negative index	a⁻ⁿ = 1aⁿ
Fractional index	a^1/n = ⁿ√a; a^m/n = (ⁿ√a)^m
Equation	Same base, then equate the indices