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INDICES &
LOGARITHMS
Name
........................................................................................
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
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
2
1
4 4 2
4
1
16 [2]
1
532 [2]
(ii) mnn mn
m
aaa
3
2
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
Indices and Logarithms
ACTIVITY 1:
Find the value for each of the following;
(a) 1
264 64 8
(b) 3
1
8
[2]
(c) 4
1
16
[2]
(d) 5
2
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
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
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
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 a
1 = 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
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
[-1]
(d) log8 2
[3]
(e) log2 23
[3]
(f) loga a4
[4]
Indices and Logarithms
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 2 32= 5
2log 2 5
(b) log 3 27
[3]
(c) log 3 1
[ 0]
(d) log 3 9
[2]
(e) log 8 64
[2]
(f) log 2 8
[3]
2. Find the value of
(a) log 2 6 + log 2 12 – log 2 18
2
2
6(12)log
18
log 4
2
(b) log 3 18 + 2log 3 6 – log 3 72
[2]
LAWS OF
LOGARITHMS
loga xy = loga x + loga y
loga xm = m loga x
y
xalog = loga x - loga y
Indices and Logarithms
(c) 2log 4 2 - 4log 3+ log 4 12
[2]
(d) log 5 45+ 5 5 5log 100 log 10 log 18
[2]
3.. Given that log 2 3 = 1.585 and log 2 5 = 2.322 . Evaluate each of the following.
(a) log 2 15
[3.907]
(b) log 2 75
[6.229]
(c) log 2 20
[4.322]
(d) log 2 1.5
[0.585]
5. Given that log3 2 = 0.6309 and log3 5 = 1.4650. Evaluate each of the following.
(a) log 3 10
[2.0959]
(b) log 3 18
[2.6309]
(c) log 3 45
[3.4650]
(d) log 3 0.3
[-1.0959]
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
[log a + log b]
(b) log 3 2a b
[3log a + 2log b]
(c) log 3
2ab
[3log a +6 log b]
(d) log ab
c
[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
[log log ]a ax y
(b) log a x2y
3
[2log 3log ]a ax y
(c) log a 2x
y
[2log log ]a ax y
(d) log a 2 3a x
y
[2 3log log ]a ax y
Indices and Logarithms
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 3– 2 lg 6
[log 2]
(c) log 2 x + log 2 y 2
[2
2log xy ]
(d) lg 6 + 2 lg 4 – lg 8
[log 12]
(e) lg x +2lg y - 1
[
2
log10
xy
]
(f) 3 lg x – 2
1lg y
4 + 2
[
3
2
100log
x
y
]
(g) 2log a x - 1 + loga y
[2
loga
x y
a
]
(h) log 3 x + 2log 3 y – 1
[
2
3log3
xy
]
(i) log b x + log b y + 1
[ logb xyb ]
(j) log a x + log a y – 1
[ loga
xy
a
]
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 :
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) 3log 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) 4log 16 . log 3 125 =
[8]
(d) log 4 5. log 5 3. log 3 7. log 7 64 =
[3]
a
bb
c
ca
log
loglog
When c = b, then a
bb
b
ba
log
loglog
= ablog
1
If c = b, so loga b =
log 2
log 5
0.3010
0.6990
0.431
3 3
3 3
log 9 log 8
log 2 log 3
2 3 6
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 33
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] [
4
b]
(c) log 4 a
(d) log a 32a
[
2
b]
[5
1b ]
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) 3
9log
x
y
[2r+s] [2+r-s]
4. If mx 2log and ny 2log , express each of the following in terms of m and n.
(a) 4log xy (b) 2logx y
[1
[ ]2
m n ]
[2n
m]
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 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+1
= 4
[1
3 ]
(d) 9x+1
= 3
[1
2 ]
(e) 9x. 3
x-1 = 81
[5
3]
(f) 2x + 3
- 42x
= 0
[1]
(g) 814 x
[ 3 ]
(h) 3
125
1 x
[5 ]
43 3
4
x
x
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) 87 x
[-1.07]
(c) 42x+1
= 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]
lg 2 lg3
lg3 1.58
lg 2
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
Indices and Logarithms
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 = a n.
ACTIVITY 10:
1. Solve the following equations.
(a) lg x = lg 3 + 2 lg 2 –lg 2
(b) 2 lg 3 + lg (2x) = lg (3x + 1)
[1
15]
(c) lg (4x – 3) = 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 + lg x – lg (x – 1) = 2
(b) lg 4 + 2 lg x = 2
[5]
23 2 lg
2
6x
2
25lg 2
1
25 10 ( 1)
25 100 100
75 100
4
3
x
x
x x
x x
x
x
Indices and Logarithms
(c) lg x + lg (2x – 1) = 1
[5
2,2
]
(d) 4log58log xx
[2]
(e) 2log54log3 xx
[2]
(f) 5 5 5log (2 5) 2log 6 log 4x
[7]
(g) log2 x2
= 3 + log2 (x + 6)
[-4,12]
(h) 3log)6(log 24 x
[3]
Indices and Logarithms
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 m and p.
[4 marks] [2p – m -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 of p and r .
[4 marks] [3r – 2p + 1]
8. SPM 2006
Solve the equation 2 3
2
18
4
x
x
. [3 marks] [x =1]
9. SPM 2006
Given that 2 2 2log 2 3log logxy x y , express y in terms of x.
[3 marks] [y=4x]
10. SPM 2006
Solve the equation 3 32 log ( 1) logx x . [3 marks] [1
18
x ]