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Index Notation
A way of representing numbers (constants) and variables (e.g.x x and y) that have been
multiplied by themselves a number of times.
Examples:
ξ₯΅
οΈ
ξ‘ ξ΄½
οΉ
ξ‘ ξ₯΄ξ΅
ο»
ξ‘
σ°
ξ₯Άξ΅
οΆ
ξ΅
οΈ
σ°
ο»
ξ‘ ξ΅
οΏοΉ
οΆ
Term:
A single number (constant) or variable
E.g.,
In the expression 4x β 7 both 4x and β7 are terms.
Coefficients:
The number which the variable is being multiplied by
E.g.,
In 2x
3
the coefficient is 2.
Integer:
A whole number
E.g.,
1, 7 and 1003.
Index (also called exponent or powers):
The index number is the number of times you multiply a number/variable by itself. It is normally
E.g.,
The index number in 5
4
is 4.
Note: the plural of index is indices
Note: you will see index number as a superscript
Base Number:
The number/unknown that is being multiplied by itself an amount of times
E.g.,
The base number in 5
4
is 5 and in 2x
3
the base number is x.
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Square Numbers:
A number/variable that is βsquaredβ is multiplied by itself
E.g.
4 Γ 4 can be written as 4
2
and is spoken as β4 squaredβ or β4 to the power of 2β.
A square number is found when we multiply an integer (whole number) by itself.
E.g.
1Γ1=1Therefore 1 is a square number2Γ2=4Therefore 4 is a square
number3Γ3=9Therefore 9 is a square number10Γ10=100Therefore 100 is a square
number1Γ1=1Therefore 1 is a square number2Γ2=4Therefore 4 is a square
number3Γ3=9Therefore 9 is a square number10Γ10=100Therefore 100 is a square
number
Cube Number:
A number/variable that is βcubedβ is multiplied by itself three times.
E.g.
4 Γ 4 Γ 4 can be written as 4
3
and is spoken as β4 cubedβ.
A cube number is found when we multiply an integer (whole number) by itself three times.
E.g.
1Γ1Γ1=1Therefore 1 is a cube number2Γ2Γ2=8Therefore 8 is a cube
number3Γ3Γ3=27 Therefore 27 is a cube number10Γ10Γ10=1000Therefore 1000 is
a cube number
How to simplify expressions involving index notation
1. Identify whether the base numbers for each term are the same
In higher tier questions you may need to manipulate the base numbers first
2. Identify the operation/s being undertaken between the terms
3. Follow the rules of index notation to simplifying the expression
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How to use the laws of indices
We can use the laws of indices to simplify expressions.
Law 1: multiplying indices
When multiplying indices with the same base, add the powers.
2 Multiply any coefficients.
Example:
Simplify each of the following, leaving your answer in index notation where appropriate.
1. a. ξ₯΄
ο·
ξ΅ξξ₯΄
ο»
This can be written as
ξ₯΄ξξ΅ξξ₯΄ξξ΅ξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξξ΅ξξ₯΄ξ
You will notice that we multiplied 2 ten times.
Therefore, ξ₯΄
ο·
ξ΅ξξ₯΄
ο»
ο ξ₯΄
ο΅ο΄
Check:
ξ₯΄
ο·
ο ξ₯Ίξξ΄½ξ΅ξ΅ξξ₯΄
ο»
ο ξ₯³ξ₯΄ξ₯Ί
Multiply them and you will get ξ₯Ίξξ΅ξξ₯³ξ₯΄ξ₯Ί ο ξ₯³ξ₯²ξ₯΄ξ₯Ά
ξ₯΄
ο΅ο΄
ο ξ₯³ξ₯²ξ₯΄ξ₯Ά
If both terms have the same base, we can simply write it as ξ₯΄
ο·
ξ΅ξξ₯΄
ο»
ο ξ₯΄
ο·οΎο»
ο ξ₯΄
ο΅ο΄
b. σ°οξ₯Άσ°
οΊ
ξ΅ξσ°οξ₯Άσ°
οΉ
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This can be written as
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξξ΅ξ
σ°
οξ₯Ά
σ°
ξ΅ξ
σ°
οξ₯Ά
σ°
ξ΅ξ
σ°
οξ₯Ά
σ°
ξ΅ξ
σ°
οξ₯Ά
σ°
ξ΅ξ
σ°
οξ₯Ά
σ°
You will notice that we multiplied (-4) eleven times.
Therefore, σ°οξ₯Άσ°
οΊ
ξ΅ξ
σ°
οξ₯Ά
σ°
οΉ
ο ξ₯Ά
ο΅ο΅
Check:
σ°οξ₯Άσ°
οΊ
ο ξ₯Άξ₯²ξ₯»ξ₯Έξξ΄½ξ΅ξ΅ξσ°οξ₯Άσ°
οΉ
ο οξ₯³ξ₯²ξ₯΄ξ₯Ά
Multiply them and you will get
ξ₯Άξ₯²ξ₯»ξ₯Έξξ΅ ο ξ₯³ξ₯²ξ₯΄ξ₯Ά ο οξ₯Άξ‘ξ₯³ξ₯»ξ₯Άξ‘ξ₯΅ξ₯²ξ₯Ά
σ°οξ₯Άσ°
ο΅ο΅
ο οξ₯Άξ‘ξ₯³ξ₯»ξ₯Άξ‘ξ₯΅ξ₯²ξ₯Ά
If both terms have the same base, we can simply write it as
σ°οξ₯Άσ°
οΊ
ξ΅ξσ°οξ₯Άσ°
οΉ
ο σ°οξ₯Άσ°
οΊοΎοΉ
ο σ°οξ₯Άσ°
ο΅ο΅
c. ξ΅
οΌ
ξ΅ξξ΅
ο·
This can be written as
ξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξξξΊξξ
You will notice that we multiplied x eleven times.
Therefore, ξΊ
ο‘
ξξξΊ
ο
ο ξ΅
ο΅ο΅
If both terms have the same base, we can simply write it as
ξ΅
οΌ
ξ΅ξξ΅
ο·
ο ξ΅
οΌοΎο·
ο ξ΅
ο΅ο΅
d.
σ°
ξ₯΅ξ΅
οΆ
σ°
ξ΅ξσ°ξ₯Ίξ΅
ο»
σ°
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This can be written as
σ°
ξ₯΅ξ΅
οΆ
σ°
ξξξξξξξξξξξξξξξξξξξ₯΅ξξξξ΅ξξξξ΅
σ°
ξ₯Ίξ΅
ο»
σ°
ξξξξξξξξξξξξξξξξξξ₯Ίξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξ
We can write is as
ξ₯΅ξξξξ₯Ίξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅
You will multiply the coefficients, 3 and 8 and add the indices 2 and 7
Therefore ξ₯΅ξξξξ₯Ίξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ξξξξ΅ ο ξ₯΄ξ₯Άξξξξ΅
ο½
or ξ₯΄ξ₯Άξ΅
ο½
If both terms have the different base, we can simply write it as
σ°ξ₯΅ξ΅
οΆ
σ°ξξ΅ξσ°ξ₯Ίξ΅
ο»
σ° ο
σ°
ξ₯΅ξξ΅ξξ₯Ί
σ°
ξ΅ξξ΅
οΆοΎο»
ο ξ₯΄ξ₯Ά
ο½
Law 2: dividing indices
When dividing indices with the same base, subtract the powers.
1 Subtract the indices.
2 Divide any coefficients of the base number or letter.
Example:
Simplify each of the following, leaving your answer in index notation where appropriate
2. a. ξ₯·
οΌ
ο ξ₯·
οΉ
Identify the indices and subtract: 8 and 5, 8 β 5 = 3
We can write it as
οΉξο«ξοΉξο«ξοΉξο«ξοΉξξο«ξοΉξο«ξοΉξο«ξοΉξο«ξοΉ
οΉξο«ξοΉξο«ξοΉξο«ξοΉξο«ξοΉ
Dividing them will give us
οΉξο«ξοΉξο«ξοΉξο«ξοΉξξο«ξοΉξο«ξοΉξο«ξοΉξο«ξοΉ
οΉξο«ξοΉξο«ξοΉξο«ξοΉξο«ξοΉ
ο ξξξ₯·ξξ΅ξξ₯·ξξ΅ξξ₯·ξξ΅ξ΅ξξ₯·
ο·
Therefore, ξ₯·
οΌ
ο ξ₯·
οΉ
ο ξ₯·
ο·
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If both terms have the same base, we can simply write it as
ξ₯·
οΌ
ο ξ₯·
οΉ
ο ξ₯·
οΌοΏοΉ
ο ξ₯·
ο·
b. σ°οξ₯Ήσ°
ο΅ο΅
ο σ°οξ₯Ήσ°
οΈ
Identify the indices and subtract: 11 and 4, 11 β 4 = 7
We can write it as
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°
σ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°
Dividing them will give us
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξ
σ°
οΏο»
σ°
ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°
σ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°ξο«ξσ°οΏο»σ°
ξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξο ξξξ
σ°
οξ₯Ή
σ°
ξξ΅
σ°
οξ₯Ή
σ°
ξ΅
σ°
οξ₯Ή
σ°
ξ΅ξ
σ°
οξ₯Ή
σ°
ξ΅ξ
σ°
οξ₯Ή
σ°
ξξ΅ξσ°οξ₯Ήσ°ξξ΅ξσ°οξ₯Ήσ°ξξ΅ξ΅ξσ°οξ₯Ήσ°
ο»
Therefore, σ°οξ₯Ήσ°
ο΅ο΅
ο σ°οξ₯Ήσ°
οΈ
ο σ°οξ₯Ήσ°
ο»
If both terms have the same base, we can simply write it as
σ°οξ₯Ήσ°
ο΅ο΅
ο σ°οξ₯Ήσ°
οΈ
ο σ°οξ₯Ήσ°
ο΅ο΅οΏοΈ
ο σ°οξ₯Ήσ°
ο»
c. ξ₯Έξ΅
ο»
ο ξ΅
ο·
Identify the indices and subtract: 7 and 3, 7 β 3 = 4
We can write it as
οΊξξο«ξξ΅ξο«ξξ΅ξο«ξξ΅ξξο«ξξ΅ξο«ξξ΅ξο«ξξ΅ξο«ξξ΅ξ
ξ΅ξο«ξξ΅ξο«ξξ΅ξ
Dividing them will give us ξ
ξ₯Έξξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξ
ξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξ
If there is no coefficient, then that means that the coefficient is 1.
ξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξο ξξξξ₯Έξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξξ΅ξ΅ξξ₯Έξ΅
οΈ
Therefore, ξ₯Έξ΅
ο»
ο ξ΅
ο·
ο ξ₯Έξ΅
οΈ
If both terms have the different base, we can write it as
ξ₯Έξ΅
ο»
ο ξ΅
ο·
ο
σ°
ξ₯Έ ο ξ₯³
σ°
ξ΅ξσ°ξ΅σ°
ο»οΏο·
ο ξ₯Έξξξξ΅
οΈξ
ο ξ₯Έξ΅
οΈ
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d.
σ°
οξ₯³ξ₯·ξ΅
ο½
σ°
ο ξ₯·ξ΅
οΈ
Identify the indices and subtract: 9 and 4, 9 β 4 = 5
We can write it as
οΏο΅οΉξξο«ξο¬ξο«ξο¬ξο«ξο¬ξξο«ξο¬ξο«ξο¬ξο«ξο¬ξο«ξο¬ξο«ξο¬ξο«ξο¬ξ
οΉξο«ξο¬ξο«ξο¬ξο«ξο¬ξο«ξο¬ξ
Dividing them will give us ξ
οξ₯³ξ₯·ξξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξ
ξ₯·ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξ
ξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξξο ξξ οξ₯΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξ΅ξξξ΅ξ΅ξοξ₯΅ξ΅
οΉ
Therefore, σ°οξ₯³ξ₯·ξ΅
ο½
σ° ο ξ₯·ξ΅
οΈ
ο οξ₯΅ξ΅
οΉ
If both terms have the different base, we can write it as
σ°οξ₯³ξ₯·ξ΅
ο½
σ° ο ξ₯·ξ΅
οΈ
ο
σ°
οξ₯³ξ₯· ο ξ₯·
σ°
ξ΅ξσ°ξ΅σ°
ο½οΏοΈ
ο οξ₯΅ξξξξ΅
οΉξ
ο οξ₯΅ξ΅
οΉ

### Unformatted Attachment Preview

Index Notation A way of representing numbers (constants) and variables (e.g.x x and y) that have been multiplied by themselves a number of times. Examples: 34 , π5 , 2π₯ 7 , (4π¦ 2 π₯ 4 )7 , π§ β5 2 Term: A single number (constant) or variable E.g., In the expression 4x β 7 both 4x and β7 are terms. Coefficients: The number which the variable is being multiplied by E.g., In 2x3 the coefficient is 2. Integer: A whole number E.g., 1, 7 and 1003. Index (also called exponent or powers): The index number is the number of times you multiply a number/variable by itself. It is normally shown about the base number E.g., The index number in 54 is 4. Note: the plural of index is indices Note: you will see index number as a superscript Base Number: The number/unknown that is being multiplied by itself an amount of times E.g., The base number in 54 is 5 and in 2x3 the base number is x. Square Numbers: A number/variable that is βsquaredβ is multiplied by itself E.g. 4 Γ 4 can be written as 42 and is spoken as β4 squaredβ or β4 to the power of 2β. A square number is found when we multiply an integer (whole number) by itself. E.g. 1Γ1=1Therefore 1 is a square number2Γ2=4Therefore 4 is a square number3Γ3=9Therefore 9 is a square number10Γ10=100Therefore 100 is a square number1Γ1=1Therefore 1 is a square number2Γ2=4Therefore 4 is a square number3Γ3=9Therefore 9 is a square number10Γ10=100Therefore 100 is a square number Cube Number: A number/variable that is βcubedβ is multiplied by itself three times. E.g. 4 Γ 4 Γ 4 can be written as 43 and is spoken as β4 cubedβ. A cube number is found when we multiply an integer (whole number) by itself three times. E.g. 1Γ1Γ1=1Therefore 1 is a cube number2Γ2Γ2=8Therefore 8 is a cube number3Γ3Γ3=27 Therefore 27 is a cube number10Γ10Γ10=1000Therefore 1000 is a cube number How to simplify expressions involving index notation 1. Identify whether the base numbers for each term are the same In higher tier questions you may need to manipulate the base numbers first 2. Identify the operation/s being undertaken between the terms 3. Follow the rules of index notation to simplifying the expression How to use the laws of indices We can use the laws of indices to simplify expressions. Law 1: multiplying indices When multiplying indices with the same base, add the powers. 1 Add the powers. 2 Multiply any coefficients. Example: Simplify each of the following, leaving your answer in index notation where appropriate. 1. a. 23 π₯ 27 This can be written as 2 π₯ 2 π₯2 π₯ 2 π₯ 2 π₯ 2 π₯ 2 π₯ 2 π₯ 2 π₯ 2 You will notice that we multiplied 2 ten times. Therefore, 23 π₯ 27 = 210 Check: 23 = 8 πππ 27 = 128 Multiply them and you will get 8 π₯ 128 = 1024 210 = 1024 If both terms have the same base, we can simply write it as 23 π₯ 27 = 23+7 = 210 b. (β4)6 π₯ (β4)5 This can be written as (β4) π₯ (β4) π₯ (β4) π₯ (β4) π₯ (β4) π₯ (β4) π₯ (β4)π₯ (β4)π₯ (β4)π₯ (β4)π₯ (β4) You will notice that we multiplied (-4) eleven times. Therefore, (β4)6 π₯ (β4)5 = 411 Check: (β4)6 = 4096 πππ (β4)5 = β1024 Multiply them and you will get 4096 π₯ β 1024 = β4,194,304 (β4)11 = β4,194,304 If both terms have the same base, we can simply write it as (β4)6 π₯ (β4)5 = (β4)6+5 = (β4)11 c. π₯ 8 π₯ π₯ 3 This can be written as πxπxπxπxπxπxπxπxπxπxπ You will notice that we multiplied x eleven times. Therefore, ππ x ππ = π₯11 If both terms have the same base, we can simply write it as π₯ 8 π₯ π₯ 3 = π₯ 8+3 = π₯ 11 d. (3π¦ 2 )π₯ (8π¦ 7 ) This can be written as (3π¦ 2 ) is equivalent to 3 x π¦ x π¦ (8π¦ 7 )is equivalent to 8 x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ We can write is as 3x8xπ¦xπ¦xπ¦xπ¦xπ¦xπ¦xπ¦xπ¦xπ¦ You will multiply the coefficients, 3 and 8 and add the indices 2 and 7 Therefore 3 x 8 x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ x π¦ = 24 x π¦ 9 or 24π¦ 9 If both terms have the different base, we can simply write it as (3π¦ 2 ) π₯ (8π¦ 7 ) = (3 π₯ 8)π₯ π¦ 2+7 = 249 Law 2: dividing indices When dividing indices with the same base, subtract the powers. 1 Subtract the indices. 2 Divide any coefficients of the base number or letter. Example: Simplify each of the following, leaving your answer in index notation where appropriate 2. a. 58 Γ· 55 Identify the indices and subtract: 8 and 5, 8 β 5 = 3 We can write it as 5π₯5π₯5π₯5 π₯5π₯5π₯5π₯5 5π₯5π₯5π₯5π₯5 Dividing them will give us Therefore, 58 Γ· 55 = 53 5π₯5π₯5π₯5 π₯5π₯5π₯5π₯5 5π₯5π₯5π₯5π₯5 = 5 π₯ 5 π₯ 5 ππ 53 If both terms have the same base, we can simply write it as 58 Γ· 55 = 58β5 = 53 b. (β7)11 Γ· (β7)4 Identify the indices and subtract: 11 and 4, 11 β 4 = 7 We can write it as (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) (β7) π₯ (β7) π₯ (β7) π₯ (β7) Dividing them will give us (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) π₯ (β7) (β7) π₯ (β7) π₯ (β7) π₯ (β7) = (β7) π₯(β7)π₯(β7)π₯ (β7)π₯ (β7) π₯ (β7) π₯ (β7) ππ (β7)7 Therefore, (β7)11 Γ· (β7)4 = (β7)7 If both terms have the same base, we can simply write it as (β7)11 Γ· (β7)4 = (β7)11β4 = (β7)7 c. 6π₯ 7 Γ· π₯ 3 Identify the indices and subtract: 7 and 3, 7 β 3 = 4 6 π₯π₯π₯π₯π₯π₯ π₯π₯π₯π₯π₯π₯π₯π₯ We can write it as π₯π₯ π₯π₯π₯ Dividing them will give us 6 π₯ π₯π₯π₯π₯π₯ π₯π₯π₯π₯π₯π₯π₯π₯ π₯π₯ π₯π₯π₯ If there is no coefficient, then that means that the coefficient is 1. = 6 π₯ π₯ π₯ π₯ π₯ π₯ π₯ π₯ ππ 6π₯ 4 Therefore, 6π₯ 7 Γ· π₯ 3 = 6π₯ 4 If both terms have the different base, we can write it as 6π₯ 7 Γ· π₯ 3 = (6 Γ· 1)π₯ (π₯)7β3 = 6 x π₯ 4 = 6π₯ 4 d. (β15π¦ 9 ) Γ· 5π¦ 4 Identify the indices and subtract: 9 and 4, 9 β 4 = 5 We can write it as β15 π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ Dividing them will give us 5π₯π¦π₯π¦π₯π¦π₯π¦ β15 π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ 5π₯π¦π₯π¦π₯π¦π₯π¦ = β3 π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ π₯ π¦ ππ β3π¦ 5 Therefore, (β15π¦ 9 ) Γ· 5π¦ 4 = β3π¦ 5 If both terms have the different base, we can write it as (β15π¦ 9 ) Γ· 5π¦ 4 = (β15 Γ· 5)π₯ (π¦)9β4 = β3 x π¦ 5 = β3π¦ 5 Name: Description: ...
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