1. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 46
CHAPTER 5- INDICES AND LOGARITHMS
5.1 INDICES AND LAW OF INDICES
In Form Three, we have learned about indices and law of indices. For the early of this chapter, we will
only revise the concept about it.
Example 1:
Solve the equation 0255 262
=− − xx
Solution:
xx 26
255
2
−
=
)26(2
55
2
xx −
=
xx 4122
−=
01242
=−+ xx
0)2)(6( =−+ xx
6−=x or 2=x
Example 2:
Find the value of y that satisfies the equation 36
216
1
.)6( 2
=y
y
Solution:
2
3
2
6
6
1
.6 =y
y
232
66.6 =− yy
Law of indices
1.
nmnm
aaa +
=×
2.
nm
n
m
a
a
a −
=
3.
mnnm
aa =)( or
nm
a
4.
xxx
abba )(=×
5. 10
=a
6.
mnn
m
aa )(= or )(n mn
m
aa =

2. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 47
2)3(2
66 =−+ yy
232 =− yy
2=− y
2−=y
Example 3:
Express
mmm
22)2(36 32
−+ +−
in the simplest form.
Solution:
mmm
22.22.2.36 32
−+= −
mmm
22.82.
2
36
2
−+=
mmm
22.82.
4
36
−+=
mmm
2.12.82.9 −+=
m
2)189( −+=
m
2)16(=
m
2)2( 4
=
4
2 +
= m
Example 4:
Given the equation
xxxx 2112
3832 ×=× −+
. Show that
3
2
6 =x
.
Solution:
xxxx 23112
3232 ×=× −+
1
2
3
12
3
3
2
2
−
+
= x
x
x
x
12312
32 −−−+
= xxxx
11
32 −−
= xx
xx
3.32.2 11 −−
=
x
x
−
=
2
3
3
2
Factorize 2m

3. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 48
ya x
= xya =log
xx
2.3
3
2
=
x
)23(
3
2
×=
3
2
6 =x
(proof)
EXERCISE 5.1
1. Express 12122
555 −+
+− nnn
in the simplest terms.
2. Solve the following equations.
(a) 322 23
=+x
(b) 02781 321
=− −+ xx
(c) 039 531
=− +− xx
3. Solve the following equations.
(a) 481
=−x
(b) xx 313
255 −
=
4. Given
81
1
273 114
=× ++ pp
, find the value of p.
5.2 LOGARITHMS AND LAW OF LOGARITHMS
1. Logarithms can be used to simplify the operations multiplication and division of large numbers. We
can convert from index form to logarithmic form and vice versa.
2. If yax
= (this is in index form), then xya =log (this is in logarithmic form) where a is known as base
and x is index.
3. The logarithm to the base of 10 is known as the common logarithm. xya =log is read as logarithm
of y to the base a.
For example, 100102
= can be converted into logarithmic form which is 2100log10 = .
(i) 38log2 = is 823
= .
(ii) 11.0log10 −= is equal to 1)
10
1
(log10 −= which is
10
1
10 1
=−
(ii) 6000001.0log10 −=
(iv) =∝0log10

4. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 49
5.2.1 Law of Logarithms
1. )(logloglog mnnm aaa =+
Example:
)1000100(log1000log100log 101010 ×=+
)100000(log 10=
5=
2. )(logloglog
n
m
nm aaa =−
Example:
)
100
10000
(log100log1000log 101010 =−
)100(log 10=
2=
3. mnm a
n
a loglog =
Example:
3
1010 10log1000log =
10log3 10=
13 ×=
3=
5.2.2 Change of base of Logarithms
xba =log
then x
ab =
Take logarithm to base c on both sides,
x
c ab 10loglog =
axbc 10loglog =
a
b
x
c
c
log
log
=
Substitute into ,
a
b
b
c
c
a
log
log
log =
1
21
2

5. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 50
Example 1:
Evaluate
(i) 2log8
(ii) 125log25
Solution:
(i)
8log
2log
2log
2
2
8 =
3
2 2log
1
=
2log3
1
2
=
3
1
=
(ii)
25log
125log
125log
5
5
25 =
2
5
3
5
5log
5log
=
5log2
5log3
5
5
=
2
3
=
Example 2:
Given that xa =2log and ya =3log . Express the following in terms of x and/or y.
(i) a2log
(ii) 3log2
Solution:
(i)
2log
log
log2
a
a a
a =
x
1
=
a
b
b
c
c
a
log
log
log = c is the new base that we can choose

6. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 51
(ii)
2log
3log
3log2
a
a
=
x
y
=
EXERCISE 5.2.1
1. Express the following equations in index form.
(a) 4log3 =p (b) 25log =p (c) x=3log2
2. Express the following equations in logarithm form.
(a) 5
3=y (b) y
510 = (c) r
nm =
5.2.3 Solving problems involving laws of logarithms
Example 1:
Given 7924.03log4 = and 404.17log4 = . Without using the logarithm tables, calculate
(i) 75.1log4
(ii) 48log7
Solution:
(i) )
4
7
(log75.1log 44 =
4log7log 44 −=
1404.1 −=
404.0=
(ii)
7log
48log
48log
4
4
7 =
7log
)34(log
4
2
4 ×
=
7log
3log4log
4
4
2
4 +
=
7log
3log4log2
4
44 +
=
404.1
7924.02 +
=
404.1
7924.2
=

7. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 52
9889.1=
Example 2:
Solve the following equations:
(i) 1)13(log5log 44 =−− xx
(ii)
2
1
8log2log4 =− xx
Solution:
(i) 1)13(log5log 44 =−− xx
1)
13
5
(log4 =
−x
x
1
4
13
5
=
−x
x
)13(45 −= xx
4125 −= xx
47 =x
7
4
=x
(ii)
2
1
8log2log4 =− xx
2
1
8log2log 4
=− xx
2
1
)
8
2
(log
4
=x
2
1
)
8
16
(log =x
2
1
2 x=
222
1
)2()( =x
4=x
Example 3:
Given
2
1
136log9log212log =−+ kkk . Find the value of k.
Solution:
2
1
136log9log212log =−+ kkk

8. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 53
2
1
136log9log12log 2
=−+ kkk
2
3
)
36
8112
(log =
×
k
2
3
)27(log =k
272
3
=k
3
2
3
2
2
3
)27()( =k
23
)27(=k
9=k
EXERCISE 5.2.2
1. Given 3log =px and 5log =qx , find the values of:
(a) pqxlog
(b)
q
p
xlog
2. Given that xp =2log and yq =2log , express the following in terms of x or/and y.
(a)
q
p
2log
(b)
( )
4
log
2
2
pq
3. Solve the equation 12log)93(log 33 =−+ tt .
4. Given w3333 log2log8log4log2 =−+ , find the value of w.
5.3 EQUATIONS INVOLVING INDICES AND LOGARITHMS
5.3.1 Solving equations involving indices
Example 1:
Solve 322 2
=k
Solution:
322 2
=k
162
=k
4±=k

9. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 54
Example 2:
Given that pnm
6373 == . Show that
nm
mn
p
2+
=
Solution:
nm
73 =
mnmm
11
)7()3( =
m
n
73 =
pn
637 =
pn
)37(7 2
×=
ppn 2
377 ×=
Substitute into ,
pm
n
pn 2
)7(77 ×=
)7(77
2
m
np
pn
×=
m
np
p
n
2
77
+
=
m
np
pn
2
+=
m
np
m
mp
n
2
+=
m
npmp
n
2+
=
npmpmn 2+=
mnnmp =+ )2(
nm
mn
p
2+
= (proof)
1
2
1 2

10. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 55
5.3.2 Solving equations involving logarithms
Example 1:
Solve
(i) 13log)35(log 33 −=−− tt
(ii) 59 3log
=x
.
Solution:
(i) 13log)35(log 33 −=−− tt
1)
3
35
(log3 −=
−
t
t
1
3
3
35 −
=
−
t
t
3
1
3
35
=
−
t
t
tt 3915 =−
912 =t
12
9
=t
4
3
=t
(ii) 59 3log
=x
5log9log 3
log
3
3
=x
5log9loglog 333 =×x
5log2log 33 =×x
5loglog2 33 =x
5loglog 3
2
3 =x
52
=x
5±=x
0>x
5=x

11. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 56
Example 2:
Solve the simultaneous equations.
)3(log2log 22 −+= xy
yx
16)64(4 =
Solution:
yx 23
4)4(4 =
yx 231
44 =+
yx 231 =+
124
)3(4loglog
)3(log4loglog
)3(log2log
22
222
22
−=
−=
−+=
−+=
xy
xy
xy
xy
Substitute into ,
5
255
24831
)124(231
=
=
−=+
−=+
x
x
xx
xx
Substitute 5=x into ,
8
1220
12)5(4
=
−=
−=
y
y
y
Example 3:
Show that 5
log
5
log
5
=+
abab ba
.
Solution:
ab
a
ab b
b
b log
5
log
log
5
+=
aba
ab
bb
b
log
5
log
log
5 +÷=
1
2
2 1
2
Change the log to base b

12. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 57
abab
a
bb
b
log
5
log
log
5 +×=
abab
a
bb
b
log
5
log
log5
+=
ab
a
b
b
log
5log5 +
=
ab
ba
b
bb
log
log5log5 +
=
ab
ab
ab
ba
b
b
b
bb
log
)(log5
log
)log(log5
=
+
=
5= (proof)
Example 4:
A liquid cools from its original temperature of C°80 to a temperature CT° in x minutes. Given that
X
T )88.0(80= , find
(a) the temperature of the liquid after 10 minutes
(b) the time taken for the liquid to cool down to C°27
Solution:
(a)
X
T )88.0(80=
,10=x
10
)88.0(80=T
C°= 28.22
(b) X
T )88.0(80=
,27=T
)
80
27
log(88.0log
80log27log88.0log
)88.0log(80log27log
)88.0(80log27log
)88.0(8027
=
−=
+=
=
=
x
x
X
X
X
Add logb b that is actually equal to 1
Factorize

13. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 58
88.0log
4717.0
4717.088.0log
−
=
−=
x
x
5.8=x minutes
EXERCISE 5.3
1. Solve the equation 13
74 −+
= nn
.
2. Given 72 =x
, find the value of x.
3. Given that 3 3log (1 ) 2[log 1]y x− = − . Express y in terms of x.
4. Find the value of x given that 27log9log 33
−= x .
CHAPTER REVIEW EXERCISE
1. Solve the equation
2 1
2
27
9
3
x
x
x
+
+
=
2. Solve the equation ( )2
3 4 21x−
=
3. Solve the equation 4log 2 log 8 1x + =
4. Given that log 3 2 = m and log 3 7 = n. Express each of the following in terms of m and / or n.
(a) log 3 3.5
(b) log 4 7
5. (a) Solve the equation ( )
2 1
3 1
x −
= .
(b) Mr. Ammar Fauzan invested RM 5,000 in a unit trust for his new born son’s education in the
future. The amount of the invested money will become n
RM5,000(1.09) for a period of n
years.
(i) Find the amount of his investment in the fifth year, correct to two decimal points.
(ii) Find the minimum number of years required so that the amount of his investment is more
than RM 20,000.
6. Find the value of x and of y that satisfy both the equations below.
2
2
9
3
27
x
y
= , 4 4 4log 2 log (2 3 ) log ( 5)x y x+ − = +
7. Given that qp
62 = , show that
qp
q
−
=2log3 .
8. Given that 3
log xp n= and nq xlog= , show that 3=pq .
9. Given 3loglog2)3(log2 555 ++=+ xx , show that 09692
=+− xx .

14. Additional Mathematics Module Form 4
Chapter 5- Indices And Logarithms SMK Agama Arau, Perlis
Page | 59
10. (a) Given that 2log3=x , find the value of x
9 . Hence find the value of k
9 given xk += 2 .
(b) Given 5022.02log =x , prove that 5066.28log =xx .
11. (a) Given that px =2log and qx =7log , express 





2
56
log
x
x in terms of p and q.
(b) Solve the equation 2log32log2 xx =+
12. Without using a scientific calculator, find the value of
3
1
3log75.0log5.1log2 +−
13. Solve the equation 9643 22
=× +xx
14. Find the values of x given that 16loglog2 xx =
15. Simplify xx
x
21
32
39
6
−+
+
.