Indices & Standard Form Notes

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    Chapter 1 Indices & Standard Form

    Section 1.1 Simplifying

    ( ) .

    : 2+ 3 + 42 5 + 10

    = 2+ 42+ 3 5 + 10

    = 52 2 + 10

    .: 22 3

    = 2 2 3

    = 2

    3

    2

    = 63

    Consolidation

    1) 2 + 5 + 3 4

    2) 2+ 4 6 + 22+ 12

    3) 4 5 +

    4) 2 5

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    5) 32 23

    68 E 5A 6

    70 E 5B 1 2

    Section 1.2 Index Laws

    5 .

    The Index Laws

    Consolidation: Simplify the following:-

    1) 4 3=

    2) 62= 3) (3)2= 4) 24 57= 5) (22)3=

    =

    +

    (

    )=

    0

    = 1

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    6) 18231=

    7) 524 23=

    8) 2434323=

    9) =

    73 E 5C 1 8

    Section 1.3 Zero and Negative Powers

    , .

    :

    32

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    Consolidation: Simplify the following:-

    1) 31

    2)

    3)

    Find the value of n for each of the following:-

    1)

    2)

    3)

    426 E 26A 1 10

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    Section 1.4 Standard Form

    .

    10 = 1000, 4 10 = 4000 . 4000 4 10 .

    .

    . H,

    ( , 3), .

    1 10, 10( ).

    81 900 000 000 000 = 8.19 1013

    .

    .

    0.000 001 2 = 1.2 106

    I 106

    6

    1.2

    Use of Calculator

    , : ( 1 10).

    .

    10 .

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    Consolidation: Write the following as ordinary numbers:-

    1) 4.2 104 2) 3.544 105 3) 2 103 4) 1.2 101 5) 7.5 103 6) 3 100

    Consolidation: Write the following numbers in standard form:

    1) 6 000 2) 5 3) 0.4 4) 0.000 259 5) 0.001 97 6) 375 500

    ( 103) (. 105)

    = 16.1 103 105

    = 16.1

    103+(5)

    = 16.1 102

    = 1.61 101 102

    = 1.61 101+(2)

    = 1.61

    101

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    (3 104) (2 108)

    =

    Consolidation: Evaluate the following, giving your answers in standard form:

    1) (6 109) (5 103) 2) (4 108)(2 103)

    3)

    (3.2

    10

    10

    )

    (6.5

    10

    6

    )

    4) (2.46 1010)(2.5 103)

    429 E 26B 1 12

    431 E 26C 1 12

    Section 1.5 Fractional Indices

    What is square root?

    , ,

    .

    : 4 4 = 16, 16 4.

    : 36 = 6 ( 6 6 = 36)

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    What is cube root?

    ,

    .

    : 3 3 3 = 27, 27 3.

    Proof of fractional indices

    +

    .

    I

    .

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    4= 4 = 2

    125

    27 = 27 = 3

    What about more complicated fraction powers?

    43/2

    ? (3)

    (1/2), .

    .

    A ( /) :

    () ,

    (/)

    , / (/) :

    , / (/) :

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    A :

    A :

    D

    43/2= 43(1/2)= (43) = (444) = (64) =

    43/2

    = 4(1/2)3

    = (4)3= (2)

    3=

    E .

    274/3

    = 274(1/3)

    = (274) = (531441) =

    274/3

    = 27(1/3)4

    = ( 27)4= (3)

    4=

    I 2 !

    ,

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    C: E :

    1)2)3)

    4)5)

    . 434 E 26D 1 6

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    Section 1.6 Using indices to solve powers

    .

    . .

    5= 5

    3

    ()

    :

    = 3

    101

    = 104

    ()

    :

    1 = 4

    1 = 4 +

    1 4 =

    3 =

    3= 9

    .

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    : 9 = 32

    :

    3= 3

    2

    = 2

    22 4

    =1/8

    C1/8 2

    ?

    1/8= 2

    3

    22 4

    = 23

    2 4 = 3

    2 = 4 3

    2 = 1

    =

    42 + 4

    = 82

    4 = 22 8 = 2

    3

    (22)2+4= (23)2

    4 + 8 = 6

    8 = 2

    = 4

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    Consolidation

    :

    1) 43 1= 16

    2) 3410= 1/9

    3) 9 4= 272 +

    H