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### Transcript of INDICES AND LOGARITHMS - Penditamuda's Blog and Logarithms zefry@sas.edu.my 2 CHAPTER 5 : INDICES...

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INDICES &

LOGARITHMS

Name

........................................................................................

• Indices and Logarithms

zefry@sas.edu.my 2

CHAPTER 5 : INDICES AND LOGARITHMS

1.1 Finding the value of number given in the form of :-

Type of indices In General Examples

(a) Integer indices (i) positive indices

aaaaan .....

n factors

a = base(non zero number)

n = index(positive integer)

53 3 3 3 3 3

2( 4)

3)2.0(

31

5

(ii) negative indices

n

n

aa

1

12 =1

2

23

2

4

(b) Fractional indices (i) nn aa

1

n = positive integer

a 0

21

4 4 2

41

16 [2]

1

532 [2]

(ii) mnn mnm

aaa

32

27 2 23( 27) 3 9

2

38 [4]

3

416 [8]

Notes : Zero Index : 0,10 awherea

Examples :

00 0 0 15 1, 2.2 1, ( 3) 1, 1

2

• Indices and Logarithms

zefry@sas.edu.my 3

ACTIVITY 1:

Find the value for each of the following;

(a) 1

264 64 8

(b) 31

8

[2]

(c) 41

16

[2]

(d) 52

32

[4]

(e) 2

327

[9]

(f) 1

225

[1

2]

(g)

1

31

8

[2]

(h)

21

4

[16]

(i)

11

32

[32]

(j)

21

3

[9]

EXERCISE 1

1. Evaluate the following:

(a) 3

24

[8]

(b) (16)1

4

[2]

(c)

11

25

[25]

(d)

1

532

81

[2

3]

(e) 1 2(5 )

[1

25]

(f) 1

24

[1

2]

2. Write in index form

(a) 1

p

[p-1

]

(b) 3

1

q

[q-3

]

(c)

1

1

p

[p1]

• Indices and Logarithms

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ACTIVITY 2:

1. Simplify each of the following:

(a) 52 aa

(b) 2 35 5n n

(c) 6

2

x

x

(d) 34 2 2n n n

(e) 1

6 4 2p q

(f) 1

4 12 4a b

(g) 1

8 2 281p q

(h) 416 2n

(i) 2 33 6 2

(j) 1 1

5 332 125

(k) 1 12 4 2n n

(l) 2 3

2 4

n na a

a a

LAWS OF INDICES

nmnm aaa nmnm aaa mnnm aa )(

mmm baab )( n

nn

b

a

b

a

2 5

7

a

a

5[5 ]n

[1] 4[ ]x

3 2[ ]p q 3[ ]ab

4 1[9 ]p q 8[2 ]n

5 4[3 2 ] 2[ ]5

[4] 5 6[ ]na

• Indices and Logarithms

zefry@sas.edu.my 5

2. Prove that

(a) 11 4344 nnn is divisible by 17

for all positive integers of n .

17(4n-1

) is a multiple of 17 and hence,

17(4n-1

) is divisible by 17.

(b) 21 555 nnn is divisible by 31 for

all positive integers of n.

[31(5n) is a multiple of 31 and hence, 31(5

n) is

divisible by 31].

(c) 21 333 nnn is divisible by 13 for

all positive integers of n.

[13(3n) is a multiple of 13 and hence, 13(3

n) is

divisible by 13].

(d) 2 32 2p p is divisible by 12 for all

positive integers of p.

[12(2p) is a multiple of 12 and hence, 12(2

p) is

divisible by 12].

1

1

1

44 4 4 3

4

34 (4 1 )

4

174

4

17(4 )

nn n

n

n

n

• Indices and Logarithms

zefry@sas.edu.my 6

2. LOGARITHMS AND THE LAW OF LOGARITHMS.

_____________________________________________________________________

2.1 Express equation in index form to logarithm form and vice versa

Definition of logarithm

If a is a positive number and a 1 , then

`

(INDEX FORM) (LOGARITHM FORM)

N = Number

a = base

x = index

We can use this relation to convert from index form to logarithm form or vice versa.

ACTIVITY 3:

1. Convert each of the following from index form to logarithm form:

INDEX FORM LOGARITHM FORM

(a) 43

= 64

4log 64 3

(b) 34

= 81

(c) 2-3

= 1

8

(d)10-2

= 0.01

(e) 1

3 = 13

2. Convert each of the following from logarithm form to index form:

LOGARITHM FORM INDEX FORM

(a) log7 49 = 2

249 7

(b) log3 27 = 3

(c) log9 3 = 1

2

(d) log10 100 = 2

(e) log5 1

16 = - 4

Notes! Since a1 = a then loga a = 1

Since a0 = 1 then loga 1 = 0

xaN xNa log

Loga N is read as logarithm of N

to the base a

• Indices and Logarithms

zefry@sas.edu.my 7

3. Find the value of x .

a) log2 x = 1

12 2x

b) log10 x = -3

[0.001]

c) log3 x = 4

[81]

d) loga x = 0

[1]

2.2 Finding logarithm of a number

Logarithm to the base of 10 is known as the common logarithm. The value of

common logarithms can be easily obtained from a scientific calculator .

In common logarithm, if log10 N = x , then , antilog N = 10x. [lg N = log10N]

ACTIVITY 4

1. Use a calculator to evaluate each of the following;

(a) log10 16 = 1.2041

(b) log10 0.025 =

[-1.6021]

(c) log10 2

3

=

[-0.1761]

(d) log10 52 =

[1.3979]

(e) antilog 0.1383 =

[1.3750]

(f) antilog (- 0.729) =

[0.1866]

(g) antilog 1.1383 =

[13.7450]

(h) 10- 2

=

[0.01]

2. Find the value of the following logarithms.

(a) log4 16=2

4log 4 2 (b) log3 27

[3]

(c) log2 1

2