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### Transcript of Indices & Standard Form Notes

• 8/13/2019 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