1

1.10 INDICES
Index is the power to which number is raised to
and the number is called a base.
For example 2 3 , 3 is index of 2 and 2 is a base.
1.11. The fundamental rules of indices
1.
2.
But
a
5
3
But
5
But 5
a
4
m
5
125 625
5
5
750 78125
a
34
78125 78125
3
4
125 625 5
m
a
m
2
3
n
5
n
7
7
34
a
mn
mn
a a (Addition of indices)
2
But 2
4 2
2
8 4 2
a
2
8 4 2
m
a
n
4
2 2
3
2
1
a
n
a
2
32
1
2
a
mn
3 34
3. 2
2
8
4
2
12
4096 4096
m n mn
a
4.
32
mn
INDICES, SURDS AND LOGARITHMS:
(Subtraction of indices)
a (Multiplication of indices)
1
4
3
4
3
0.015625
1
64
1
64
0.015625
0.015625
m
1
a (Negative index)
m
a
3 2
5.
3
2
64 3 2
64
64
6.
3
16
4096
4
16
m
n n m
a a n a m
; n> 0
(Rational indices)
0
100 = 1
1 a
a
0
1
1
1
4
a
1
a
11
a
(index Zero )
7. 81
4
81 3
8. 9
1
a m a ; m 0
(S imple fractional index)
a
3
2
p
q
m
9
1
3
2
1
2 3
9
0
1
2
1
3
9 27
1 1 1
p q
p
;
p
a
q
q
a
a
(Negative rational indices)
2
2
9. 2
5 2 5 4 25 100
m m m
ab a b
(Distribution of index)
1.12 Simplification of an expression
Example 1. (a) Simplify the following
(a)(i)
2
1
(i)
4 3
t t 2
(ii)
Solution
4 3
t
2a
7c
t
3
4
1
2
b
d
2
3
(ii)
2a
7c
-3
4
b
d
2
3

1
3
1 2
t 4 t
2
3 1
t 4 t
4
t
= t
3
1
4 4
2
a
1
3
2c4b2
7a3d3
Example 2: Simplify the following
Solution
(i)
(ii)
(i)
27
(ii)
8
27
1
3
81
16
3
4
1
1
8 3
3
27
8
81
16
3
2
3
4
4
8
1
27
1
3
3
4
3
4
4
27
8
1
3
1
3
81
16
3
2
3
4
3
4
3
4
4
3
4
4
Example 3: Simplify:(i)
3
2
3
1
3
1
3
3
3
2
3
3
3
2
1
b2
7 d3
c4
1
3
3
1
3
3
27
8
7
4 3
2 5
x
yz x
yz
xz 2
2n1
n1
32
42
(ii)
Solution:
2
n1
2
n
7
4 3
2 5
(i) x
yz x
yz
xz 2
4 3
2
5
x yz x yz
=
7
xz 2
1
2
3
2
5
1 1
8 2 6
x y z
x 2 y 2 z 2
= 7 7
x 2 z 2
5 1 1
8
2
6
2 2 2
=
x
y
x
7
2
z
z
7
2
5 1 1 7 7
8 2
6
=
x 2 y 2 z 2
x 2 z 2
5 7
5
1 7
8
6
=
x 2 2
y 2
z 2 2
=
x
5
2 2 9
y
z
n1
n1
32
42
(ii)
2
n1
Example 4:
Solution:
3 2
=
x x 3
x
x
1
3
x
2 1
3 3
x
1
1
2
2
n
n 1
3
2
2 4
2
2
n n
2
2 2
n
6
2 2
2
2
2 2
=
n n
2
=
2
n
n
Simplify:
=
1
=
x
2
3
x
x
6 2
2 1
x
1
2
1
3
n
4
1
3
2
x x 3
=
x
x 1
x 1
23
=
5
6 x 6
x
3
1
6
x
x
2
3
x
5
5
x 2 xy
Example 5: Simplify
3
xy y
1
2
x
1
1
3
x
x y
n

Solution
x
x x xy
3
xy y
x
x xy
x
x y
=
2
yx
y x y
3
x
x xy x y
xxy
y
3
y
y x y
2
y
xy
x
x
x
y
x
x xy
xy
x xy
x xy
x xy
x xy
2
y
x y
x y
2
2
3
2
xy
x
x y
y
y
x y
4
3
1.13 Solving Equations
Example 1: Find the value of x in the
5
x
6 6 9
equation 6
36
Solution:
6
5
x 5
x
6 6 5
x2
x
6 6
3
2
36
5
6
x
6 6 3 x
6 6
36
Comparing the powers;
3 x 9
x 6
Example 2:Find the value of x and y if
5
x
25
2y
Solution:
5
5
x
x
25
5
x4y
2y
2 2y
1
and
1
0
5
and
0
3
5x
9
9
3
y
5x
3
1
9
9
5x
y
3
5x2y
3
2y
1
9
x
9
2
1
5 5
3 3
Comparing powers with both cases we have;
x 4y
0 …… ………………(i) and
5x
2y
2…..….……………(ii)
2(ii) – (i)
3
10x
4x
4
x 4y
0
9x
4
4
x
9
From equation (i) if we replace x we have
4
4y
0
9
4
4y
9
1
y
9
Exercise 1 (a)
1. Solve the equation:
3
2
i) 64
x1
y , ii 9 3
81
2. Find the values of x and y in the
equations.
2y 3
3 2 x and 2
2y x
64
3. Simplify the following expressions
i)
ii)
iii)
1
t
x
t
x
x
xy
1
t
1
2
t
2 3
x
xy
n1
n1
9
6
2n1
n
3
2
4. Simplify the following
i)
ii)
iii)
x
n1
n1
42
3 2
2
n1
t 2
2t
1
8
4
t
t
2
42
5
2
2
3
x
2 2x
2 2
1.20SURDS:
These are numbers which cannot be evaluated
exactly hence they are irrational e.g.
7 , 61, 20 .etc.In short they are numbers
n

which when converted to decimal cannot be
exact but they are approximated e.g. 7 =
2.6458 (4dpts)
1.21: Definition of rational number and
irrational number.
Rational number is a number which can be q
p
where p and q are integers but q 0 . All
proper fractions, improper fractions, mixed
fractions and integers are rational numbers and
so are terminating or recurring decimal
fractions.
An irrational number is not a rational number.
If an irrational number is written in decimal it
has no repeating pattern e.g. 7,
e , . etc.
Surds are numbers which cannot be evaluated
exactly hence they are irrational numbers but
irrational numbers are not surds like π is an
irrational number but not a Surd. Surd is an
expression containing 1 or more square roots of
prime numbers e.g. 2 3, 7, 3 5 4 2 . etc.
1.22: Some identities in Surds
(1) 2 3 6
t n tn
t t
(2)
n n
2
(3) 7 5 7 5 2 7 5
2
t n t n 2 tn
(4) 5
2
3 5 3
2 5
3
t
2
n t n 2 tn
2
(5) 2
3 4 3
4 3
2 2
a
b a b 2a
b
(6) 7 3 7 3
7 3
a b a b
a b
(7) 3
2 3
2
3
2 2
2
a
b a
b
a b
1.23 Simplification of Surds is a process which
surd and an expression is reduced to its
lowest form.
Example 1Simplify 150 to its simplest form.
Solution 150 25 6
25 6 5 6
Example 2Simplify 8 512 8
simplest form.
3 to its
Solution 3 4
2 256 2 4
2
3 4 2 256 2 4 2
3
2
216
2 2
2
6
216
2 2 2
6
218
2
216
Example 3 Reduce the expression
98 50 8 32 72 to its simplest
form.
Solution
49 2 25 2 4 2 16 2 36 2
7
49
16
2 5
2
2
2 2
25
36
2
2 4
Example 4: Simplify
at denominator.
Solution
3
1
3
x
3
1
3
2
x x
x
x
2
x 3 1
1 1
x 2 3
4
2 6
2
1
2
4
2
x x 3
without surd
x
1
2
x
1
1
3

x
2
3
x
x
=
1
5
6
2
3
x
x
2
3
1
5
6
6
5
1
x
1.24 Rationalization of Surds
Rationalization is a process of removal of surds
from denominator.
Example 1: Simplify
1 1
(a)
3 5 3 5 3
(b)
3
3
Solution:
2
2
(a)
3
5 3
3
5 3 3
5 3
3 5
9
3
3
5 3
5 3 5 9
5
42
3
5 3
4
6 5 2 3 21 5 63
84
63 2 3 15
84
3
4
3
42
5
5 5
28
3 2 3 2
(b)
3 2 3 2
6
5
6
5
5
3 2 2
3 2
5 3
3
5 3
6
5 2
6
Example 2: Simplify
Solution:
3
3
2 2
2 2
3
2 2 33
2 2 3
3
2 2 33
2 2 3
9
2
6 6 6
9 4
2
30 12
6
18 12
6 4
3
30 12
6
Example 3: Simplify
Solution:
=
1
1
2
2
5
5
1
1
2 5 2 5
2 5 2 5
2
5 2
2
6
2 5 2 5
2 3
2 5 3
3
2
3
5
3
10
6 3
Example 4: (a) Express
3
3
3
6
5 2
2
2
5
10 5
6
1
3
5
3 10
a b c, where a , b and c are rational.
32
3
(b)
5 3
a b
c
Solution:
make it in form
1
3 1
3 6
3 10
(a)
6 3 10
6
3 10
6
3
6
in the form

6
=
3 10
6310
6
2
16 3 28
=
108 100
16 3 28
=
8
7
= 2 3
2
310
32
3
(b)
5 3
32
3 5 3
=
5 3 5 3
2
3 3 5 3
=
5 2 15 15 3 2
2
=
3
15 2 3
5 3
2
5 3
6 2 15 3 5 3 3
=
2
3
= 3
15 5 3
2
Example 5: Simplify
1 1 1
(a)
1 x 1
x 1
x
(b)
x
Solution:
(a)
1
x
11
x
1
x 1
x
1
=
1
x
x
1
1
x 1
1
x
3
3
11
x 1
1
x 1
x 1
x
x
x
1
1
x
1
x 1
x 1
=
=
x 1
3 3
x 1
1 x
(b)
x
x
1
x
x 1
x
x
2
2
x
x
x
x 1
x 1
x
x x
x
x x x 1
x
1 1
x x
x x
=
x
x
x
x
1
x
1
x
x
1
1.25 Equations with Surds
They are solved by isolating surds on one side
of the equation, then squaring throughout until
surd has disappeared.
However this method introduces extra solutions
because a negative sign when squared is equal
to a positive when squared i.e.
2 2
x
x x.
Hence solution of the equation with surds
should be checked by substituting the solutions
got in the original equation before giving final
solution.
NB: This part should be learnt after the
quadratic equation is taught.
Example 1Simplify 14 4 6
Solution
Suppose 14 4 6 a 2 b
Squaring both sides of the equation we have
2
14 4 6 a 2 b 2
4 6 a 4b
4 ab
14
By comparing R.H.S and L.H.S.
14 a 4b……………………..(i)
ab 6……………………………(ii)
a 14 4b Substituting for a
14
4b
b
6
7b
2b 2 3
2b
2 7b 3 0

2
7 7 43
2
b
4
7 5 1
= 3 or
4 2
When b 3
a 14
43 2
1
When b
2
1
a 14 4
12
2
When a 2 b 3
2 2 3 2
14 4 6
∴ 14 4 6 2 2 3 does not satisfy
1
When a 12
and b
2
14 4
6
2
12 2
3
2
1
2
14
4 6 2 3 2
Example 2:
equation
2
satisfies
Solve for x in the following
(i) x 1 3
(ii)
Solution:
(i) x 1 3
2 2
x 1 3
x 1 9
x 8
(ii)
2 x 5 1
x
(iii) 3x
5 x 3
(iv) 3 x x 5 3
2 x 5 1
x
2
2x
5 x
1 2
2x
5 x
5 x
x
2
2
4
x 2
1
2
2x
1
When x 2
4 5 1
2
3 1 2 .Satisfies
When x 2
4 5 1
2
1
1 2
Does not satisfy
Hence x 2
(iii) 3x
5 x 3
3 x 5 3 x
2
3x
5 3
x 2
x
2
9x 4 0
2
9 9 41
4
x
2
9 8.0623
2
x 8.5311 or 0.4689
When x 8. 5311
3
8.5311 5 8.5311 3
Does not satisfy.
When x 0. 4689
3
0.4689 5 0.4689 3
3 3Satisfies
Hence x = 0.4689
(iv) 3 x x 5 3
2 2
3 x x 5 3
x
5 9
9x
6 x x 5
10 x 5 6 xx
5 9
6 x x 5 10
x 4
2
6 x x 5 10x
4 2
2
36 x x
5 100 x 80 x 16
9x
2
45x
25x
2
16x
2 65x 4 0
65
x
2
20x
4
65 416
4
32

65 63
x
32
= 4 or 0. 0625
When x 4
3 4 4 5 3
6 9 3
3 =3
Since the LHS = RHS.
Hence it satisfies the equation.
When x 0. 0625
3 0.0625 0.0625 5 3
0.75 – 2.25 ≠ 3Does not satisfy
Hence x 4 .
Example 3 Solve the following equation,
verify your solutions in each case.
(i) 2x
1
x x 3
(ii)
Solution:
x 6 4 x 1
3x
(i) 2x
1
x x 3
2x
1
x 2 x 3 2
x 1
x 2 x 2x
1
x
3x
1 x
3 2 x2x
1
2 3
2
2x
4 2 2 x2x
1
4x
2 16
x 16
4x
2x
1
2
4x 16x
16
8x
4x
4x
2 12x 16
0
12
x
12
12 20
x
8
x 4 or -1
2
2
4
416
8
When x = -1, 1
1
4
Does not satisfy
When x 4
8 1 – 4 = 4 3
3 – 2 = 1 Satisfies
x 4
(ii)
x 6 4 x 1
3x
2
x 6 4 x 1
3x
2
x 6
4
x
2 x
64
x 1
3x
2 x 64
x 10
1
3x
2 x
64
x 3x
9
2 x
64
x 2
3x
9 2
2
4x 64
x 9x
81 54x
2
2
x x 24 6x
9x
54 x 81
4 4
8x
4x
2
96 9x
13x
2 62 x 15
0
2
54x
81
62 62 13
415
x
213
62 68
=
26
6
= or 5
26
3
= or 5
13
When x 5
we will have
2
5 6 4
5 1
15
1 9 16
13
4 it Satisfies
3
When x we will have
13
3
13
6
81
13
9
4
49
13
7
3
13
2
4
13
13 13 13
16 2
13 13
16 2 Does not satisfy
Hence x = -5 only.
1
3
13

Example 4: Solve the following equation
3 x 7
x 16
2x
Solution:
Squaring both sides we have
3 x
2 3
x7
x 7
x 16 2x
2 3
x7
x 10
16
2x
2 3
x 7
x 2x
6
Squaring both sides we have;
2
3
x7
x 36 24x
4
4 x
2x
x
2
2
10
x 12
0
5x
6 0
x 6
or 1
When x 6
is substituted back in the above
equation we have
3 6 7
6 16
12
3 1 2
Hence x = -6 satisfies
When x = 1 is substituted back in the above
equation we have
3 1 7
1 16
21
2 8 18
2 2 2 3 2
2
3 2
1
Dividing through by
2
1 3does not satisfy
Hence the root is x = -6
Example 5 Find the square root of
(i) 5 + 2
Solution:
6 (ii) 18 – 12 2
(i) Let the square root of
a
b
5
2 6 a b
Squaring both sides we have
5 2
6
5 2 be
2
6 a b a b 2 ab
Comparing parts we have
5 a b ………………………(i) and
6 ab…………………………(ii)
From equation b 5 a and substituting b in
(ii)
2
We have 6 a5
a 5a
a
a
2
5a
6 0
a
2 a
3
0
a 2 or 3
When b = 3 then a = 2 and
When a = 3 then b = 2
Hence square root of 5 2 6 is 3 2
(ii) Let square root of 18 12 2 be x y
18
12
2 x y
Squaring both sides
18 12
18 2
2
2 x y x y 2 xy
2
6
2 x y 2 xy`
Comparing parts we have
18 x y ……………………...(i) and
72 xy ………………………...(ii)
From (i)
(ii)
y 18 x and substituting for y in
We have
2
x
2
72 x 18 x 18x
x
18x 72 0
2
18 18 41
72
x
2
18 6
x
2
18 6 18 6
= or
2 2
= 12 or 6
Hence square root of
18 12 2 is 12
6
Exercise 1 (b)
1. Find the square root of:
i) 14 6 5 , ii) 13 2 42

3
2. express (i)
3
form
p q
r
2
2
and ii)
3
3
2
2
3. Express the following without a surd in
the denominator.
2 3
i)
1
3 3 2 3
1
1
ii)
1
x 1
1
x 1
4. Rationalize the denominators and
simplify as far as possible the following.
1
i)
2 5 5 1
3 2 4
ii)
3 2 2
5.Given that to six significant figures.
6 2.44949 , evaluate as accurately as
possible 1
, justify the accuracy
2
3 6
you have given.
6. Solve for x in the following equations.
i)
ii)
7. Rationalize the surd
8. Solve:
i)
ii)
iii)
iv)
2
2
9. solve the following:
i)
ii)
iii)
2
x
6 4
x 1
3x
3x
4 2 x
2
3
1
5
1
3 3
x
2 x
3
1
x 4 4 x
1
4x
2 x
1 7
5x
x
2 3x
4 2
6
x 1
x 5
3x
x
1 3 2x
5 x
2
4x
13 x
1 12
x
10. Simplify the following expressions
2
5
in
0
i.
ii.
iii.
48
28
112
72
243
8
98
11. Solve the equation
x 3 x 2x
1
1
2 1
2
12. Express in the form
5 3 5 3
a b c d , hence state values of a, b, c
and.
1.30LOGARITHM:
Definition: Logarithm is the power or index to
which a number (base) is raised to produce a
given number.
Example 1: Logarithm of 100 to base 10 is 2
and mathematically can be written as
log 10
100 2 .
Which means 100 = 10 2 = 10 10
Example 2: log 2 8 3
8 = 2 3 = 2
22
Example 3: log 5 625 = 4
625 = 5 4 = 5
555
Example 4: log 10
1000 3
1000 = 10 3 = 10 10 10
Example 5: log 10 2 0. 3010
0.3010
2 = 10
1.31Simple Elementary Rules for
Logarithms
1. log ab log a log b
x
a
2. log x
log x a log x b
b
n
3. log a nlog
a
4.
5.
x
log
log x P
log
log a
x
1
log
c
a
c
x
x
P
x
n 1
6. log x m log x m
n
x
x

1
7. log P t x t
log P
8. log x 1
x
9. log 1
0
x
10. log y does not exist
x
1.32Proving the laws of logarithms
Example 1. To prove that
log a b log ab log a log
c
x
b
Solution; If
Given that
log
And given that
x
x
log c a
x
x
a x and log
x
log c b
y
x
a c
y
x
x
y
b c
b y
Then a b c c
Using the standard rules of indices we have
x
y
ab c
And hence
log
c
ab log
c
c
x
y
Using the standard rules of logarithms
log ab x ylog
c
x y
c
c
y
x 1
log
c
ab log ab log a log b
Example 2.Prove that
a
log
c log
c
a log
b
c
c
b .
Solution.If c x a then log a x and if c y b
then
log
c
b y
Using the standard rules of indices we have
x
a c x
y
c
y
b c
a
x
log
c log
c
c
b
x log a , y log
log
c
c
a
log
b
c
y
c
b
a log
Example 3.Prove that
c
x
y1
x y
c
b
n
log a nlog
c
c
c
But
a
Solution Let
Then
log
c
x log c
n
a c
nx
;
Taking log
But
x
log
x
a c
a log
a
c
log c
c
a
n
a
nx
a
nlog
c
c
x
nxlog
c
a
Example 4. Prove that
Solution. Let
t
P x
log P
c
log
x
t log
c
log
x
x
log
t
log
xlog
c
P t
c
c
But t log x
log c P
P
log x
c
P
x
P
Example 5. Prove that
Solution. Let
taking
log
a
a
log
1 t log
1
t
log
log
x
x
a t
c
c nx
c x
log
log x P
log
log a
x
1
log
a
c
c
x
P
x
t
a x
log a on both sides we have
log
x
a
a
x
a
t
x
a
1
log
Example 6: Prove that
Solution: But
n
x
a
x
log m log
Let m = x y
n 1
log x m log
n
1
m n
x
x
m

log
log
then
log
m
x
m
x
1
n
m
x
y log
y
m
x
1
y log
n
1
y
n
But y log
n 1
log
x m log x m
n
1.33 Change of Base in the logarithm
When solving equation involving more than
one base we must first change those bases to
one base only then we can solve the equation
when it is in one base only by using
a log
log b
log
a
c
b
c
1
n
m
x
x
x
1
y
n
Example 1. Solve log 5 x log x 5 2. 5
Solution:
Let log 5 n
x
n
5 x
log 5 5 log
log
n
log
5
5
n 5
5
x
x
1
log
5
x
Then log 5 x log x 5 2. 5 becomes
1 5
log 5 x
log x
2
log x
log
5
5
2
5
1
x
5
2
5
5 log
2
2
log
1 x
x 5
2
log
2 5 x
2 5 log
Let log 5 x y
2y 2 2 5y
x 5
2y
2 5y 2 0
x
x
5
y
2
5 4
2
2
2
2
5
5 9 5 3 5 - 3
y or
4 4 4
1
Since y log 5 x 2 or
2
The log 5 x 2
x 5 2 25
1
or log 5 x
2
1
x 5 2 5
x 25 or 5
Example 2. Solve the equations
(i) 0.3 x1 0.
7 x
x
(ii) 0.1 0.2 5
(iii)
Solution:
2
3 x 1 2
x
Inx
3
(iv) e Ine 8
(i) 0.3 x1 0.
7 x
x
log
0.3 log 0. 7
1 10 x 10
3 7
x 1log
10 xlog
10
10 10
x log 3 log 10 x log
x
2
25 16
4
1
or
2
1 10 10 10 7 log 10 10
x
1 0.4771
1 x0.8451
1
x 1 0.5225 0.
1545x
0.5225
x 0.5225 0.
1545 x
0.5225
0.1545
x 0. 5225 x
0.5225
0.3684 x
0.5225
x 1.4184
0.3684
x
(ii) 0.1 0.2 5
x
1 2
10
10
1
x log 10 5log 10
10
5
2
10

log 101
log 10 10 5log
10 2 log 10 10
0 1 50.3010
1
x
x
x 3.4950
x 3.495
3 x 1 2
x
(iii) 2 3
x 1
log 2 x
2log
3
3 10 10
log
log
0.4771
0.3010
10
3x
1 x
2 x
2
10
3
2
3x
1 x
21.5850
3x
11.585x
3.1701
3x
1.585x
3.1701 1
1.415
x 2.1701
2.1701
x
1.415
x 1.5336
Inx
(iv) e Ine 8
x
But
e Inx x and
Ine x x
x x 8
2x
8
x 4
Example 3. Solve the simultaneous equations
log b a 2log a b 3 and log 9 a log 9 b 3
Solution:
From log a 2log b 3
log
b
b
a
a log b 2 3
a
********************************
Side work;
Let
2 t
b a
log
b
2log
b
b
log
2
t
a
b
2
t log
b t log
2 t log
2
log
b
t
b
a
b
b
a
a
a
*****************************
log
2
a
log
b
3
a
b log
a 2 3log
a
2
log
a 3log
a
2 0
Let
2
b
log
b
a m
m 3m 2 0
m 1 m 2
b
0
m 1 or 2
log a 1
or log
b
1
a b
or
b
a b
2
a 2
a b or a b …………….(i)
From log 9 a log 9 b 3
log 9 ab 3
ab 9 3 729 ……………..(ii)
When a = b then b 2 = 9 3
b
9 3 27
2
Therefore when a = 27 then b = 27 and when a
= -27, b = -27
When a = b 2 then substituting for a in equation
(ii)b 3 = 9 3 , b= 9
When b = 9 then a = 81
Example 4. Given that log 2 x 2log 4 y 4,
Show that xy 16 . Solve simultaneous
equations.
log 10 x y 1
And log 2 x 2log 4 y 4
Solution:
log x 2log 4 y
2
log 2 x log 4 y 4
Let
y
2
log 4
t
y
2
4 2
2
t
2t
2 2t
log 2 y log 2 2
2log
2 y 2t
log 2
t log 2 y
log 2 x log 4 y 4
Becomes log 2 x log 2 y 4
log 2 xy 4
2
2
b
4
2
b

4
xy 2 = 16
From x
y 1
log 10
1
x y 10
x y
10 …………….(i)
xy 16 …………………(ii)
From (i)
x 10 y
Substituting for x in equation (ii)
10
y y
10 y y 2 16
y
y
2
2
16
10
y 16
0
10
y 16
0
10
y
10
10
2
10 6
2
16
or
2
2
10
4 116
2 1
100 64
2
36
4
2
= 8 or 2
When y = 8, x = 2 from x = 10 – y
y = 2, x = 8
Example 5 Solve the equation
3
Solution. x 1
3x
2 22
22
3 x
and 3
x
x2 2
3 3
3 x 2
3 3 x
2 2
2
3x
3
x
2
3
2
2
3 x 1 2
x
3x
2
2
3
log
10 log
x
10
3 2
3x
log 10 2 xlog
10 3 2log 10 3 log 10 2
0.3010
x0.4771
20.4771
0. 3010
3x
0.9030
x 0.4771 x 0.9542 0.3010
3
0.4260
x 0.6532
x 1.5333
3 x 1 2
3
x
3x
1
10 2
x
log 10 3
2
3x
1
log 10 2 x 2 log 10
Alternative 1 2
log
3
3x
1log
10 2 x
2log
10 3 0
3x
1 0.3010 x
20.4771
0
0.9030
x 0.3010 0.4771 x 0.9542 0
0.426
x 0.6532
x 1.5333
Example 6.Find the values of x and y if
x4y
5 5
0
and 3
5x2y
3
x4
y
Solution.From 5 5
sides
0
2
We have x
ylog
5 0log 5
4 5 5
taking log
5
on both
x 4y
0 ………………………….(i)
5x2y
2
And from 3 3 taking log
3
on both
sides
5x
2ylog
3
3 2log
3
3
5x
2y
2…………………..(ii)
From i – ii, x + 4y = 0
10x + 4y = - 4
- 9x 4
- 4
x
9
x
From i, y
4
4
y 4
9
1
9
Exercise 1(c)
1. Solve the simultaneous equations
1
log x y and 1
log 5 x log y 5
2
2. Solve for z in the equation
log 46
z log 2 z
3. Solve for x in equations
(i) log 5 3log 5 x 2
x .
(ii) log 10 x log x 100 3

4. Solve the simultaneous equations:
x
y x
y
log 25 log 5 and x y 8
5. Solve the following simultaneous
equations:
1 3
log 2 x log 2 y 2log 4 6 and
3
128 2
x log 8 y log 8
x 3 4
6. Solve log 4
log x 2. 5
2
5
p
2
p
5
16
2
y
7. Solve log log 3
8
y
8. Solve log log 1
x x x 5
9. If log 3 log 9 log 27 . Find x.
3
10. Solve the simultaneous equations
a
b
a
ab
log b 2log 3 and log 9 3
2
x
11. Solve log4
6log x 1
0
12. Solve the simultaneous equations
xy
x
5
log
25
1 and log 3log
5
1.34 Simplification of logarithmic
expression.
This is done by reducing logarithmic
expression to its simplest form.
Example 1.Use logarithm to evaluate
(i) log 3
18 and (ii) (1.75) 0.67
Solution:
(i) Let log 3 18 n
n
18 3
log 1018
nlog
10
3
log
n
log
10
10
18
3
1.2553
0.4771
= 2.6310
log 3 18 2.6310
0.67
(ii) Let 1
.75 t
4
y
2
2
0.67 log 101.75
log 10 t
0.2430
t
0.67
log 10
10 0. 1628
t or anti - log 0.1628 t =
1.4548
0.67
1.75 1. 4548
Example 2.If log 7 2 0.356 , log 7 3 0. 566
Find the value of
7 25 7
2log
7 log 7
2log 7
15
12 3
Solution
log
log 7
7
7
15
7
15
2
25
log 7
12
2
25
12
2
7
3
2
7
log 7
3
49 25 9
= log 7
225 12 49
1 1 9
= log 7
9 12 1
1
= log 7 log 7 1
log 7 12
12
2
= 0 log 3
7 2
= log 7 3
2log 7 2
= -0.566 – 2(0.356)
= -1.278
Example 3. (i) Express as a single logarithm
1
4
2log
x 3log x 4 log x
2
5
(ii) Express as a single logarithm
5
log n x log nx
1 log n x
2
Solution
1
4
(b) (i) 2log
x 3log x 4 log x
2
5
log
x
1
2
2
log
x
4
3
log
x
4
5

Using the rules of logarithm
2
1 3
4
2
1 5
log
3
x
log x 4 log x 20
4 4 4
5
Therefore
1
4
2log
x
3log x 4 log x
log x 20
2
5
5
(ii) log n x log nx
1 log n x
2
log x 2 log
log
n
n
x
5
5
2
2
log x x 1
n
n
x
1 log x
x 1
log
x
n
x
5
2
n
1
x
1x
2
100 a
Example 4:Express log in terms of
3
b c
log a,log
b and log c
Solution
100 a
log
3
2
1
2
b c
is common logarithm
since no base is given we assume it
2
100 a
log
1
log
3 2
b c
1
= 2 2log 10 a 3log 10 b log 10 c
2
Example 5:
log
3
1
3
Solution
3
log
2
100 a
3
10 log 10 log 10 b log 10
Simplify;
3
2
1
243 log 327
2
2
log a 2
3
8
3
log
3
a
3
c
1
2
1
3
1
log
3
3
log
3
1
243 log
327
2
log
3
2
a
2
8
3
log
3
a
3
1
8
3 3
1
3
5
log 3
log 3 log 32
3 3 3
3log 3 a 1
2log 3 a 2
3 8
3
log 3 5log 3 log 32
3 3log 3 1
=
3 3 3 a
2log
3 a 1
3log
=
3 3 5log 3 3 4log 3 3 3log 3 a 1
2 log a 1
=
=
3 5 4 3log
3
a 1
2 log a 1
3
3 3log 3 a
2 log a 1
3
=
3
3
2
log
log
3
3
a
a
1
1
= 2
3
2
Example 6: Find the value of 512
9
2
Solution.To find value of 512
9
Let 9
2
x
512
Take common log on both sides
2
log 10 x log 10 512
9
log 10 x 0.602060
0.60206
1
x 10 hence 512
9
4
Exercise 1 (d)
1
2
1
4
4 2
1. Simplify 1
log 10
2log
10 x
2
x
2. Simplify
log
3
1
3
3
log
3
243 log
log
3
3a
2
3
55
log
2
3
3
a 1

3. Express as a single logarithm
1
4
2 log 3log 4 log
7
2 7 7
5
4. Express as a single logarithm
5
log x log x 1
log
2 40 40 10
5. Find the value of
9 25
2log 7 log
105
7
12
x
If given that log
log
7
2 0.356 and log 7 3 0.566
1. Simplify log 7 98 log 7 30 log 7 75
1.35 Solve equations involving logarithm
Example 1If log a a 2 2 , find the value of
a
Solution
2
a 2 a
a
2
a
2a
1
a 2
But
a 2 0
Hence
1 cannot be base.
a 2
0
or a 1
Example 2. Solve the equations
1 2
(a) log 8 x log 8
6 3
x
2
x 6
(b) 3 3 0
Solution:
1
(a) log x log 8
6
8
2
3
log x log 81
log 8 6
8
x
log 8
6
x
6
2
3
2
3
2
3
8 2
3
2
3
x
6
2
3
2
2 3 2
2
x 2 6
2
x 24
x
6
2
x 6
(b) 3 3 0
3
3
3
3
x
x
6
6
Let
3
t
6
2
3
3
3
2
x
2
x
6
3
0
6
0
3
x 2
3
x 0
3
x 2
3
x 0
3
x t
2
t
t 0
6
3 t
t 3 6 OR t 0
But
3
x
x
t 3
3
6
Or 3
x
0
Comparing powers
x 6
Or
log
x 6
3 3 log 3 3
x log 3 3 6log 3 3
x 6
Or
log
x 6
10 3 log 10 3
x log 10 3 6log 10 3
6log
10
3
x
log 3
x 6
10
Example 3Find x if log 8
log 2 16 1
Solution
From log 8 log 2 16 1
Let
x
x
x
log 2 16 t
x
2 t 2t
x
x
16
log x 16 2t
log x x 2t
x

1
t log x 16
2
1
log x 8 log x 16 1
2
1
log 8 log 16 2
x x 1
log
log
log
x
x
x
8 log
x
8
1
4
2 1
2 x x
x 2
1
4 1
Example 4 Solve the equation
2 x 5
4.381 8. 032
x
Solution
2 5
To solve equation x
4.381
x 8. 032
2
x 5
log 10 4.381 log 108.
032 x
2x
5log
10 4.381 xlog
10 8. 032
2x
50.641573
x0.904824
1.28314x-3.207865=0.904824x
0.378322
x 3.207865
x 8.4779192
x 8.5
3
10 10
Example 5 log x 8
Solution
3
10 log 10 x 8
8
log 10 x 0.008
3
10
0.008
x 10 = 1.0186
Exercise 1(e)
1. Solve simultaneous equations
9
log 25 xy and log 5 x 10
log 5 y
2
3
2. Given that y log a x and z log x a .
Show that zy=3, hence find the values of
y and z if
log 3log
a
x
log log
a log 27
a
a
2
2
3. Solve the equation log 4x
log x
4. Solve
i)
x
a
25 5
3
2
x 1
ln
log e log 5
ln x
log 10e
10
10
2
ii ) log 5 xlog
5(
) 1
0
x
5. Solve the simultaneous equations
2 1
log 9xy
and log 3 x log
3 y 3
2
log 4 log 2 x log 2 log 4 x
7. Solve for x in the equation
log 14 x log 7 4x
8. Solve the simultaneous equations
log y 2log
y log
30
x
6. Solve for x if
x 1 and 0
2x
1
x
9. Solve equation 5 4 215
10. Solve the following simultaneous
equations
x
y
2 y
i) 2 4 12 and32
22
16
x
y
ii) 2 3 5 and 2
3
23
x
y
x
x 3
y2
x 2 y
2
iii) 2 3 44 and 2 3 221
iv)log 4 x 2 – 6 log x 4 = 1
1.36 Proof of identities involving
logarithm
Example 1 Given that log 4 p 2log 4 q 2
Show that pq = 16.
Hence solve for p and q in simultaneous
equation
log 10 p q 1
And log 2 p 2log 4 q 4
Solution:
log 4 p 2log 16 q 2
*******************************
Side work;
Let log 16 q t

log
log
4
2
q 16
q t log
2
q
4
t
4
2t
log
2
4
4
1
t log 4 q
2
******************************
1
log 4 p 2
log 4 q 2
2
log 4 p log 4 q 2
log 4 pq 2
pq 4 2 16 As required
If y 1
log 10
x and xy = 16 ………..*
p q 10 1 p q 10 .……………...**
From ** q 10 p and substituting for y in
equation * we have
p 10 p
p
16
2
10
p 16
0
10
p
10 36
2
10 6
2
p = 8 or 2
2
10 416
1
21
When p = 2 and q = 8 and when p = 8
and q = 2
Example 2 Prove that
2
2y
y
2log
a x y 2log a x log a 1
2
x x
Solution
Let
2 log
t log
a
a
x log
a
2y
y
1
x x
2
2
x 2xy
y
x log a
2
x
2
2
t
2
2
2
x 2xy
y
t log a x
2
x
From rules of logarithms
t log
Hence
2log
a
a
2
x
2 2xy
y
2
log x
y 2
a
t 2 log
x
y
x
y 2log
a x log a 1
2
x x
2y
Example 3 If
a log b c,
b log c a and c log a b .
Prove that abc 1
Solution
Convert them to common base say base b Let
t
log c
a t a c log a t log c
log
t
log
Let
log
b
log
b
b
a
a
c
b n
b n log
b
a
b
b a
n
b
log
n
log
Then , from abc log
clog
alog
b
a
Substituting back into right side of equation we
have
log b a log b b
abc log b c
log c log a
Hence abc 1
b
b
b
b
b
b
a
Example 4 Given that
x
1
1
log 1 a,
log
x 1
b
8 15 and
1
log
x 1
c,
Show that
24
log
x
1
1
80
a b c
Solution:
From
1 1 1
a b c log x 1
log x 1
log x 1
8 15 24
c
y
2

99 16 25
log x log x log x
88 15 24
Using the rules of logarithms
9 15 24
a b c log x
8 16 25
81 1
log x log x1
80 80
Example 5 Given that
4log
64 N p and 2log 416N
qand
q p 8. Find N .
Solution
Let log
log
4 64
N
be
4 4
64
N t N 64
4 3t
log 4 N log 4 4
4
log 4 N 3t
1 4
t log 4 N log
3
t
4
N
4
3
2
16
N log N 8
log 3
log
16N
N
2
N
4
N 3
N
N
4 4
2
16 N
4
N 3
2
4
4
3
2
4
2
3
2
3
N
4
4
4
4
4
4
8
4
8
4
3
4 6
4 2 4 4096
Example 6 Given that
log y w a and log x w b
t
where w 1
log
Prove that
n x log n y a b
log n x log n y a b
Verify your answer without using a calculator
or tables when
y 4,
n 2, x 8 and w 4096
Solution:
log y w a
(a)
w y
log
n
log
a
log
w alog
n
n
w
y
a b log
a b log
log
n
w
n
n
1
log
n
n
w
y
a
y
y
log
log
1
log
n
n
n
w
x
log
n
log
log
log
log
log
x
n
log
x
w b
w x
w blog
n
n
n
n
w
b
x
w log
y log
1
w
log n y
n
n
n
b
x
w
x
1
log
log
=
n x log n y
log x y
n log n
log n y log n x log n y log n x
log n x log n y
log x log y
n
n
Given y = 4, n = 2, x = 8, and w = 4096
a b
log 4 4096 6 ,log 8 4096 4
6 4 log 2 8 log 4
2
6 4 log 8 log 4
2 3 2
10 3 2
1 1
5 5
Hence it is verified
Example 7
2
Solution
Let log y a
a
y x
log
y
x
y a log
1 a log
a
1
log
y
y
y
x
2
Prove that
x
x
log y
x
1
log
y
x
n
x

Hence log y
Example 8
x
1
log
y
x
Show that
Given that log 3 2 0. 631
log
6
log 3 x
x .
1
log 2
Find without using tables or calculator log 6
4
correct to3 significant figures.
Solution
Let
log 6 x t
t
x 6 2
3
t
log 3 x t log 3 2 log 3 3
log 3 x tlog
3 2 1
log 3 x
t
1
log 2
3
Since t log 6 x
log
Hence
3 x
log 6 x
1
log 2
log
6
log 3 4 2 log 3 2
4
1
log 2 1
log 2
3
2 0.631
0.774 (3sign figures)
1
0.631
Example 9 Show that
1
log a nlog
a m
n
m
Solution
Let log m x
log
log
a
n
m
a
m
1
Hence
n
x
m a
m
n
a
a
nxlog
nx
=
=
log
nx
a
a
m
nlog a
log
a
n
m
a
1
m
n
log
a
3
m
3
n
nlog
a
m
3
Exercise 1 (f)
1. Prove that log ab log a log b .
Hence solve the equation
x 3 x
3
log log 2
4 4
c
2. Given that
log 5 x p and log 30 x q .
Show that
i)
ii)
iii)
log
3log
log
5 log 6
2
x
x log
2
p 6log
4 log
x
4
c
q
. Hence solve:
p q
8 2
x
2
p
3
c
2 7 0
2 2
3. If a b 23ab,
show that
a b
log
10
a log
10b
2log
10
5
4. Prove that log ab log b + log c a.
Hence solve the equation
log
4x
3 3log
4x
3
c
5. Given that 3
2log 2 log y , show
c
x 2
2
that y 8x , hence, find the root of the
equation 2log x log 14x
3
3
2 2
6. Prove that if x log bc , y log ac and
z log c ab then x y z 2 xyz
n
7. If y a bx is satisfied by the values
x 1 2 4
5
y 7 10 15 show that n log 2
3
and deduce the values a and b
x y x y
8. If 2 3 3 4 6, show by taking
logarithm to base two and eliminating
log 2
3 that x y 2x
3y
a
2 2 2
1
9. (a) Show that log a x log c x
log a
(b) Prove that
log y
x
c
1
log
y
x
b

Miscellaneous Exercise 1
1. (a) Rationalizing each fraction as a separate
item
5 5
(i)
5 3 5 3
x
(ii) 1
x
1
x
(b) Express the following surd number with
rational denominators
(i)
(ii)
6
6
3
3
1
2
2
2 2
1
x
x
1
log 3 x
2. (a) Evaluate (i)
log
x
8
(ii) log 3
2log 15 log 0.14 10 log 7
2 6 7 7 98
3. Show that log b
a
1
log
Using the result to prove that
log b
log c log c
a
b
4. Express the following as a single logarithm
1
3
(i) 2log a log 10b 3
log 2 5 4
2
(ii) log 6 8log 16 9log 18
7 3 2 10
5.(a) Simplify the expression
(i) 5
4 20 8
(ii)
3
9
5n
a
b
3n1
2n
2n2
6
2n3
n2
6
2 6. (a) If a log x pq,
b log p rq,
c log r pq
prove that a b c abc 2
(b) Show that log blog
a 1and hence
a
5 64
evaluate log 2log 4 24
7. Solve simultaneous equations
x2
y1
(a) 5 7 3468 and 7 5 76
(b) 6log
3 x 6log 27 y 7 and
log x 4log y 9 .
4 9 3
8. Solve (a)
2
a
b
y x
log 9 x log 9 x log 9 x log 9 x 5
3
4
3
(b) log 4 x log 4 7 .
2
9.a) Express
5
(i) log a x log ax
1 log a x
2
and(ii) 7log
a 2 3log a 12 5log a 3 as a
single logarithm.
1
(b) Prove that log 2 10 and
log 2
evaluate log e
100 correct to 3 significant
figures taking e as 2.718.
log 10 a
10.(a) Prove that log e a ,hence show
log e
that log 718 2 ln 2
2.
(b) (i) Prove that
10
10
a
log log c a log c b
b
2
(ii) Find x if log 8 log 2 1
11. (a) Solve the simultaneous equations
p and 3 2 2 2 16
2 4 12
(b) If
q
Find x
x
p
x
2q
log x log 9 x log 81 x
3
1
3 5
(c) Simplify
2 5
16
28
12. (a) (i) Find log 8
64 4 without using
calculator or tables.
2 3
(ii) Simplify log
q log
p
p
(b) Find the values of t and m such that
(i) log 9 t log 9 m 1and
log 9 t log 9 m log 91.5
log
8
t 8 log
8
t 8
(ii)
3
13. (a) Given that
log
log
10
10
7
3
9
5
q
, solve for
p and q the simultaneous equations
p 3 1
3
q
p 1
7 and 9
(b)(i) Prove that log
x
2q1
63
q log
r
.
p log
r
pq

Hence, solve theequations
log y 3 log 3 y
2
e e .
(ii) log 5 x log x 5 2. 5
14. (a)Rationalize the surd
1 1
3 3 3 5 3
4
2
(b) Simplify 1 log 2log
4 10 x
8x
Express as a single logarithm and simplify
your answer
1 1
log
2 x
x 1
log
2 x 1
x
15.(a) Solve the equation
2
2
2 2 x
4
(b) Find the square root of 70 14 25
4 2
1
16.(a)Simplify (i) d
d
d 3 d 3
n1
n1
32
42
(ii)
2
n1
2
(b) Using logarithm tables evaluate
1
(i) .0713 12
0
5
(ii) 0 . 00237
(iii) log 5 0. 24
17.(a) Rationalize the denominators of
2 2 5
11 3 7
(i)
(ii)
3 2
7 4 11
(b) Express the following in the form
x y z
n
(i) 3 21
2 3
(ii)
(c) Simplify (i)
24 3 6 294
216
(ii) 48 2 12 27
18.(a)Show that (i)
log
c
1
3
d log
1
9
d
1
27
a log
(ii) x a
1
log x
a (iii) log x 1
x
(b) Express as single logarithm
3
(i) 3log
a a log a a 2log
b a
2
(ii) 2log 3 3log 2 3log 6
19.(a) Solve the pair of equations
x
y
x
y
27 9 and 2x y 64 .
(b) Find values of
12
3
2
1
6
2
16
27 18
(c)Express the following in simplest form
as possible
(i) (a+ b ) (c - b ) (ii)(2 + 3) 3 + (2 – 3) 3
2
1
8
1
2
20. Solve 2 x
1 x
4 1
21. Solve the following pair of
simultaneous equations.
(i) log 2 x + 2log 4 y = 4 and x + 12y = 52
(ii) 5 x+2 + 7 y+1 = 3468 and 7 y = 5 x − 76
c
a