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Indices and Logarithms

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 35  3 3 3 33
an  a  a  a ..... a (4)2 
n factors (0.2)3 

a = base(non zero number)
n = index(positive integer)

 1 3 
 5 

(ii) negative indices 21 = 1
an  1 2
an
32 

42 

(b) Fractional indices 1 1

(i) a n  n a 42  4  2
n = positive integer
a 0 1 [2]

 m 16 4 

(ii) a n  n a m  n a m 1 [2]

325 

2

27 3  ( 3 27)2  32  9

2 [4]
[8]
83 

3

164 

Notes : Zero Index : a0  1, where a  0

Examples : 50  1, 2.20  1, (3)0  1,  1 0 1
 2 

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ACTIVITY 1: 1 Indices and Logarithms

Find the value for each of the following; (b) 8 3 [2]
[4]
1
[1 ]
(a) 642  64  8 2

1 2 [16]
[9]
(c) 16 4 (d) 32 5

2 [2]

(e) 273 1

(f) 25 2

[9]

(g)  1 13 (h)  1 2
 8   4 

[2]

(i)  1 1 (j)  1 2
 32   3 

[32]

EXERCISE 1 1 (c)  1 1
 25 
1. Evaluate the following: (b) (16) 4

3

(a) 42

1 [8] [2] [25]

(d)  32 5 (e) (51)2 1 1
 81 
2 (f) 4 2 []

[] 1 2

3 []

25

2. Write in index form 1 (c)  1 1
(a) 1 (b) q3  
 p 
p
[q-3] [p1]
[p-1]

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Indices and Logarithms

LAWS OF INDICES

am  an  amn am  an  amn (a m )n  a mn

(ab)m  ambm  a n  an
b bn

ACTIVITY 2: (b) 52n  53n

1. Simplify each of the following:
(a) a 2  a5  a25

 a7

(c) x6 (d) 4n  2n  23n [55n ]
x2 [1]
[x4 ]
1
1
 (e) p6q4 2
 (f) a4b12 4

1 [ p3q2 ] [ab3 ]
(h) 16 2n4 [28n ]
 (g) 81p8q2 2
[9 p4q1 ]
(i) 32  63  2
11

(j) 325  1253

(k) 2n1  4  2n1 [35  24 ] [2]
a2n  a3n 5
[a5n6 ]
(l) a2  a4

[4]

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Indices and Logarithms

2. Prove that (b) 5n  5n1  5n2 is divisible by 31 for
(a) 4n1  4n  3 4n1 is divisible by 17 all positive integers of n.

for all positive integers of n .

4n 41  4n  3 4n
41

 4n (4 1 3)
4

 17 4n
4

 17(4n1 )

17(4n-1) is a multiple of 17 and hence, [31(5n) is a multiple of 31 and hence, 31(5n) is
17(4n-1) is divisible by 17. divisible by 31].

(c) 3n  3n1  3n2 is divisible by 13 for (d) 2p2  2p3 is divisible by 12 for all

all positive integers of n. positive integers of p.

[13(3n) is a multiple of 13 and hence, 13(3n) is [12(2p) is a multiple of 12 and hence, 12(2p) is
divisible by 13]. divisible by 12].

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Indices and Logarithms

2. LOGARITHMS AND THE LAW OF LOGARITHMS.
_____________________________________________________________________
2.1 Express equation in index form to logarithm form and vice versa

Definition of logarithm Loga N is read as ‘logarithm of N
If a is a positive number and a  1 , then to the base a’

N  ax  log a N  x N = Number
a = base
` x = index

(INDEX FORM) (LOGARITHM FORM)

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

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 log4 64  3

(b) 34 = 81

(c) 2-3 = 1
8

(d)10-2 = 0.01

(e) 1 = 31
3

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

LOGARITHM FORM INDEX FORM

(a) log7 49 = 2 49  72

(b) log3 27 = 3

(c) log9 3 = 1
2

(d) log10 100 = 2

(e) log5 1 =-4
16

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Indices and Logarithms

3. Find the value of x . b) log10 x = -3 [0.001]
a) log2 x = 1 [1]
d) loga x = 0
x  21  2
[81]
c) log3 x = 4

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 =

(c) log10  2  = (d) log10 52 = [-1.6021]
 3  [1.3979]
[-0.1761]
(e) antilog 0.1383 =
(f) antilog (- 0.729) =

[1.3750] [0.1866]

(g) antilog 1.1383 = (h) 10- 2 =

[13.7450] [0.01]
[3]
2. Find the value of the following logarithms. [3]
[4]
(a) log4 16= log4 42  2 (b) log3 27

(c) log2 1 (d) log8 2
 2 
[-1]
(e) log2 23
(f) loga a4

[3]

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Indices and Logarithms

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

LAWS OF
LOGARITHMS

loga xy = loga x + loga y
loga xm = m loga x

log a  x  = loga x - loga y
y

ACTIVITY 5:

1. Evaluate each of the following without using calculator.

(a) log 2 32= log2 25  5 (b) log 3 27

(c) log 3 1 (d) log 3 9 [3]
(e) log 8 64 [2]
[ 0]

(f) log 2 8

2. Find the value of [2] [3]
[2]
(a) log 2 6 + log 2 12 – log 2 18 (b) log 3 18 + 2log 3 6 – log 3 72

 log 2  6(12) 
 18 

 log2 4

2

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(c) 2log 4 2 - log4 3 + log 4 12 Indices and Logarithms

(d) log 5 45+ log5 100  log5 10  log5 18

[2] [2]

3.. Given that log 2 3 = 1.585 and log 2 5 = 2.322 . Evaluate each of the following.

(a) log 2 15 (b) log 2 75

[3.907] [6.229]

(c) log 2 20 (d) log 2 1.5

[4.322] [0.585]

5. Given that log3 2 = 0.6309 and log3 5 = 1.4650. Evaluate each of the following.

(a) log 3 10 (b) log 3 18

[2.0959] [2.6309]

(c) log 3 45 (d) log 3 0.3

[3.4650] [-1.0959]
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Indices and Logarithms

2.4 Simplifying logarithmic expressions to the simplest form .

ACTIVITY 6

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

(a) log ab  (b) log a3b2

[log a + log b] [3log a + 2log b]

 (c) log ab2 3 (d) log  ab 
 c 

[3log a +6 log b]

[log a + log b-logc]

2. Express each of the following in term of log a x and log a y .

(a) log a xy (b) log a x2y3

[loga x  loga y] [2loga x  3loga y]
a2 x3
x2
(c) log a y (d) log a y

[2loga x  loga y] [2  3loga x  loga y]

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Indices and Logarithms

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

(a) lg 3 + lg 25 = lg 3 + lg 5 (b) 3 lg 2 + 2 lg 3– 2 lg 6

= lg 15

[log 2]

(c) log 2 x + log 2 y 2 (d) lg 6 + 2 lg 4 – lg 8

[ log2 xy2 ] [log 12]

(e) lg x +2lg y - 1 (f) 3 lg x – 1 lg y4 + 2
2

 xy2  100x3 
[ log  ]
[ log  10 ]  y2 
 

(g) 2log a x - 1 + loga y (h) log 3 x + 2log 3 y – 1

 x2 y   xy2 
[ log3  ]
[ loga  a ]  3 
 

(i) log b x + log b y + 1 (j)log a x + log a y – 1

[ logb xyb ] [ loga  xy  ]
 a 

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Indices and Logarithms

3. CHANGE OF BASE OF LOGARITHMS
_________ __________________________________________________________________
3.1 Finding logarithm of a number by changing the base of the logarithm to a

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

log a b  log c b
log c a

When c = b, then log a b  log b b
log b a

=1
log b a

ACTIVITY 7:
1. Find the value of each of the fIofllcow= ibn,gs.oGliovgeaybo=ur answer correct to four

significant figures.

(a)  log 2 (b) log3 8
log 5
log5 2

 0.3010
0.6990

 0.431

[1.893]

(c) log3 4 (d) log 2 0.5

[1.262] [-1.00]

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

(a) log 2 9. log3 8  log3 9 log3 8 (b) log 3 7. log 7 2. log 2 3 =
log3 2 log3 3

2 36

[1]

(c) log4 16 . log 3 125 = (d) log 4 5. log 5 3. log 3 7. log 7 64 =

[8] [3]
12
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Indices and Logarithms

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

ACTIVITY 8
1. Given that log 2 3 = 1.585 and log 2 5 = 2.322.Find the value of each of the

following.

(a) log3 15 (b) log 3 3
5

[2.4650] [-0.4650]

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

(a) loga 16 (b) log 16 a

4 b

[] []

b 4
(d) log a 32a
(c) log 4 a

b [ 5 1]

[]
2b

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

(a) log 3 x2y (b) log3 9x
y

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

4. If log2 x  m and log2 y  n , express each of the following in terms of m and n.

(a) log4 xy (b) logx y2

[ 1 [m  n] ] 2n
2
[]

m

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Indices and Logarithms

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 a x  a y , then x = y
(ii) If the index are the same , when a x  bx , then a = b

2. Using common logarithm

ACTIVITY 9: (b) 16x = 8 3
1. Solve the following equations: (d) 9x+1 = 3
(a) 3x = 81 []

3x  34 4
x4

(c) 8x+1 = 4

[ 1 ] [1 ]
3 2

(e) 9x. 3x-1 = 81 (f) 2x + 3 - 42x = 0

5 [1]

(g) x4  81 []

3
(h) 1  x 3
125

[3] [5]

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Indices and Logarithms

2. By using common logarithm(log10), solve the following equations and give your
answer correct to two decimal places.

(a) 2x = 3 (b) 7x  8 [-1.07]
(d) 2x.3x = 18
x lg 2  lg 3
x  lg 3  1.58
lg 2

(c) 42x+1 = 7

[0.2] [1.61]

(e) 652x  4 (f) 2x.3x  9x4

[2.11] [21.68]
[x = -1]
3. By using replacement method, solve each of the following equations:

(a) 2x4  2x3  1 (b) 3x1  3x2  4

2x 24  2x 23  1 Substitute y  2x ,
let y  2x 1  2x
16 y  8y  1 8
8y 1 23  2x
y1 x  3

8 (d) 2x3  2x2  12

(c) 3x1  3x  9

[x = 2] [x = 0]

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Indices and Logarithms

4.2 Solving Equations involving logarithms

METHOD:
1. For two logarithms of the same base, if loga m  loga n , then m = n .
2. Convert to index form, if log a m  n , then m = a n.

ACTIVITY 10: (b) 2 lg 3 + lg (2x) = lg (3x + 1)

1. Solve the following equations.
(a) lg x = lg 3 + 2 lg 2 –lg 2

 lg  3 22 
 2 
 

x6

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

[]

15
(d) lg (10x + 5) – lg ( x+ 4 ) = lg 2

[6 ] 3

2. Solve the following equations : (b) lg 4 + 2 lg x = 2 []
(a) lg 25 + lg x – lg (x – 1) = 2
8
lg  25x   2
 x 1 [5]
16
25x  102 (x 1)

25x  100x 100

75x  100

x4
3

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(c) lg x + lg (2x – 1) = 1 Indices and Logarithms

(d) log x 8  5  log x 4

[ 2, 5 ] [2]
2

(e) 3  log x 4  5log x 2 (f) log5(2x  5)  2log5 6  log5 4

[2] [7]

(g) log2 x2 = 3 + log2 (x + 6) (h) log4 (x  6)  log2 3

[-4,12] [3]
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Indices and Logarithms

SPM QUESTIONS

1. SPM 2003 [T = 8 V ]
Given that log2 T  log4 V  3, express T in terms of V .[4 marks]

2. SPM 2003 [4 marks] [x = 1.677]
Solve the equation 42x1  7 x .

3. SPM 2004 [3 marks] [x = 3]
Solve the equation 324x  48x6 .

4. SPM 2004

Given that log5 2  m and log5 7  p , express log5 4.9 in terms of m and p.
[4 marks] [2p – m -1]

5. SPM 2005 [3 marks] [x 3]
Solve the equation log3 4x  log3 (2x 1)  1. 2

6. SPM 2005 [3 marks] [x = - 3]
Solve the equation 2x4  2x3  1.

7. SPM 2005

Given that log m 2 p and log m 3  r , express log m  27m  in terms of p and r .
 4 

[4 marks] [3r – 2p + 1]

8. SPM 2006 1 [3 marks] [x =1]
Solve the equation 82x3  .

4x2

9. SPM 2006

Given that log2 xy  2  3log2 x  log2 y , express y in terms of x. [y=4x]
[3 marks]

10. SPM 2006

Solve the equation 2  log3(x 1)  log3 x . [3 marks] [ x 11 ]
8

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