Indices Log

8/12/2019 Indices Log

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INDICES&

LOGARITHMS

Name

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


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Indices and Logarithms

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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 .....

nfactors

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 mn

m

aaa 32

27 2 23( 27) 3 9

2

38 [4]

3

416 [8]

Notes: Zero Index : 0,10 awherea

Examples :

00 0 0 1

5 1, 2.2 1, ( 3) 1, 12


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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)

2

14

[16]

(i)

11

32

[32]

(j)

21

3

[9]

EXERCISE 11. 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 )

[ 125

]

(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]


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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 3

32 125

(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


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2. Prove that

(a) 11 4344 nnn isdivisible 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 ofp.

[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


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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) 4 = 644

log 64 3

(b) 3 =81

(c) 2-3 = 18

(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 2

49 7

(b) log3 27 = 3

(c) log9 3 =1

2

(d) log10 100 = 2

(e) log51

16= - 4

Notes! Since a1 = a then loga a = 1 Since a0= 1 then loga 1 = 0

xaN xNa log

LogaNis read as logarithm ofN

to the base a


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3. Find the value of x .

a) log2x= 112 2x

b) log10x= -3

[0.001]

c) log3x= 4

[81]

d) logax= 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 log10N =x, then , antilogN= 10x. [lg N = log10N]

ACTIVITY 41. Use a calculator to evaluate each of the following;

(a) log10 16 = 1.2041 (b) log10 0.025 =

[-1.6021]

(c) log102

3

=

[-0.1761]

(d) log1052 =

[1.3979]

(e) antilog 0.1383 =

[1.3750]

(f) antilog (- 0.729) =

[0.1866]

(g) antilog 1.1383 =

[13.7450]

(h) 10- =

[0.01]

2. Find the value of the following logarithms.

(a) log416= 2

4

log 4 2 (b) log327

[3]

(c) log21

2

[-1]

(d) log82

[3]

(e) log22

[3]

(f) logaa

[4]


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2.3 Finding logarithm of a numbers by using the Laws of Logarithms

ACTIVITY 5:

1. Evaluate each of the following without using calculator.

(a) log 232= 5

2log 2 5 (b) log327

[3]

(c) log 31

[0]

(d) log 39

[2]

(e) log 864

[2]

(f) log28

[3]

2. Find the value of

(a) log 26 + log 212log 218

2

2

6(12)log

18

log 4

2

(b) log 318 + 2log 36log 372

[2]

LAWS OF

LOGARITHMS

logaxy = logax + logay

logaxm= m logax

y

xa

log = logax - logay


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(c) 2log42 - 4log 3+ log4 12

[2]

(d) log 545+ 5 5 5log 100 log 10 log 18

[2]

3.. Given that log 23 = 1.585 and log 25 = 2.322 . Evaluate each of the following.

(a) log 215

[3.907]

(b) log 275

[6.229]

(c) log 220

[4.322]

(d) log21.5

[0.585]

5. Given that log32 =0.6309 and log35 = 1.4650. Evaluate each of the following.

(a) log 310

[2.0959]

(b) log318

[2.6309]

(c) log 345

[3.4650]

(d) log30.3

[-1.0959]


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2.4 Simplifying logarithmic expressions to the simplest form .

ACTIVITY 6

1. Express each of the following in terms of log a, log band/or log c.

(a) log ab

[log a+ log b]

(b) log 3 2

a b

[3log a+ 2log b]

(c) log 3

2ab

[3log a+6log b]

(d) logab

c

[log a+ log b-logc]

2. Express each of the following in term of log axand logay.

(a) log axy

[log log ]a a

x y

(b) log ax y

[2log 3log ]a a

x y

(c) loga

2

xy

[2log log ]a a

x y

(d) log a

2 3

a xy

[2 3log log ]a a

x y


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3. Write each of the following expressions as single logarithm:

(a) lg 3 + lg 25 = lg 3 + lg 5

= lg 15

(b) 3 lg 2 + 2 lg 32 lg 6

[log 2]

(c) log2x+ log2y

[ 2

2log xy ]

(d) lg 6 + 2 lg 4lg 8

[log 12]

(e) lgx+2lgy- 1

[

2

log10

xy

]

(f) 3 lgx2

1lgy4+ 2

[

3

2

100log

x

y

]

(g) 2logax - 1 + loga y

[2

logax y

a

]

(h) log3x+ 2log3y1

[

2

3log3

xy

]

(i) logbx+ logby+ 1

[ logb xyb ]

(j)log ax + logay1

[ logaxy

a

]


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3. CHANGE OF BASE OF LOGARITHMS_________ __________________________________________________________________

3.1 Finding logarithm of a number by changing the base of the logarithm to asuitable base

The base of logarithms can be changed to other base by using a formula :

ACTIVITY 7:1. Find the value of each of the following. Give your answer correct to four

significant figures.

(a)

5log 2

(b) 8log3

[1.893]

(c)3

log 4

[1.262]

(d) 5.0log 2

[-1.00]

2. Find the value of each of the following without using calculator..

(a) log 2 9. 3log 8 (b) log 3 7. log 7 2. log 2 3 =

[1]

(c)4

log 16 . log 3 125 =

[8]

(d) log 4 5. log 5 3. log 37. log 7 64 =

[3]

a

bb

c

c

alog

loglog

When c= b, thena

bb

b

b

alog

loglog

=a

blog

1

log2

log5

0.3010

0.6990

0.431

3 3

3 3

log 9 log 8

log 2 log 3

2 3 6


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3.2 Solving Problems involving the change of base and laws of logarithms.

ACTIVITY 81. Given that log 23 = 1.585 and log 2 5 = 2.322.Find the value of each of the

following.

(a) log3 15 (b) log 33

5

[2.4650] [-0.4650]

2. Given that log2 a = b. Without using the calculator, express the following in termsof b:

(a) loga 16 (b) log 16a

[4

b] [

4

b]

(c) log 4 a (d) log a32a

[2

b] [

51

b ]

3. If log 3x= rand log 3y=s, express each of the following in terms of rands.

(a) log 3x2y (b)

3

9log

x

y

[2r+s] [2+r-s]

4. If mx2log and ny2log , express each of the following in terms of mand n.

(a)4

log xy (b) 2logx

y

[ 1[ ]2

m n ] [ 2nm]


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4. EQUATION INVOLVING INDICES AND LOGARITHMS

4.1 Solving equations involving indices

METHOD:1. Comparison of indices or base

(i) If the base are the same , when yx aa , then x = y(ii) If the index are the same , when xx ba , then a = b

2. Using common logarithm

ACTIVITY 9:1. Solve the following equations:

(a) 3x= 81 (b) 16x= 8

[3

4]

(c) 8x+ = 4

[1

3

]

(d) 9x+ = 3

[1

2

]

(e) 9x. 3x- = 81

[5

3]

(f) 2x+ - 4 x = 0

[1]

(g) 814 x

[ 3 ]

(h) 3

125

1 x

[5 ]

43 3

4

x

x


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2. By using common logarithm(log10), solve the following equations and give your

answer correct to two decimal places.

(a) 2x= 3 (b) 87 x

[-1.07]

(c) 4 x+ = 7

[0.2]

(d) xx 3.2= 18

[1.61]

(e) 46 25 x

[2.11]

(f) 493.2 xxx

[21.68]

3. By using replacement method, solve each of the following equations:

(a) 122 34 xx (b) 433 21 xx

[x= -1]

(c) 13 3 9x x

[x= 2]

(d) 3 22 2 12x x

[x= 0]

lg2 lg3lg3

1.58lg2

x

x

4 32 2 2 2 1

2

16 8 1

8 1

1

8

x x

xlet y

y y

y

y

3

2

12

8

2 2

3

x

x

x

Substitute y

x


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4.2 Solving Equations involving logarithms

METHOD:

1. For two logarithms of the same base, if nm aa loglog , then m = n.

2. Convert to index form, if nma log , then m = an.

ACTIVITY 10:1. Solve the following equations.

(a) lgx = lg 3 + 2 lg 2lg 2 (b) 2 lg 3 + lg (2x) = lg (3x+ 1)

[1

15]

(c) lg (4x3) = lg (x+ 1 ) + lg 3

[ 6 ]

(d) lg (10x+ 5)lg (x+ 4 ) = lg 2

[3

8]

2. Solve the following equations :

(a) lg 25 + lgxlg (x1) = 2 (b) lg 4 + 2 lgx= 2

[5]

23 2

lg2

6x

2

25lg 2

1

25 10 ( 1)

25 100 100

75 100

4

3

x

x

x x

x x

x

x


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SPM QUESTIONS

1. SPM 2003

Given that 3loglog 42 VT , express T in terms of V .[4 marks] [T= 8 V ]

2. SPM 2003

Solve the equation xx 74 12 . [4 marks] [x= 1.677]

3. SPM 2004

Solve the equation 684 432 xx . [3 marks] [x= 3]

4. SPM 2004

Given that m2log5 and p7log5 , express 9.4log5 in terms of mandp.

[4 marks] [2pm-1]

5. SPM 2005

Solve the equation 1)12(log4log 33 xx . [3 marks] [2

3x ]

6. SPM 2005

Solve the equation 122 34 xx . [3 marks] [x= - 3]

7. SPM 2005

Given that pm 2log and rm 3log , express

4

27log

mm

in terms ofpand r.

[4 marks] [3r2p+ 1]

8. SPM 2006

Solve the equation 2 32

18

4

x

x

. [3 marks] [x =1]

9. SPM 2006

Given that 2 2 2log 2 3log logxy x y , expressyin terms ofx.

[3 marks] [y=4x]

10. SPM 2006

Solve the equation3 3

2 log ( 1) logx x . [3 marks] [ 1

18

x ]

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