DEPARTMENT OF MATHIMATICS AND

PHYSICAL SCIENCES

SMA 1108: ALGEBRA

Study Guide
BY

DR. KANYI
2020

Page 1 of 8
INDICES
An index is the number of times a number is multiplied
by itself. E.g. 2  2  2  2 . 3 is the index and 2 is the base.
3

LAWS OF INDICES
Given that x  a3 and y  a2 then,
i. in general given a
xy  a 3  a 2  a  a  a  a  a  a 5 m
 a n  a mn
x a3 aaa
ii. 
y a2

aa
 a . in general given a  a  a m n mn

iii.    a a  a
x2  a3
2 3 3 6

In general a   a  a  n m nm m n

e.g. 3  3  2x x 2

CONSEQUENCES OF LAWS OF INDICES

a) a  a  a  a
n n nn 0

n
Also aa  1 n

Therefore a  1 this implies that any number raised to

0

the power of zero is equal to 1.

b) a 0  a n  a 0 n  a  n
a0 1
Also a n
 n
a
1
This implies that a n
 a n

1
e.g. 2 3
 2 3
1
c) a a n

1
e.g. 3
a a 3

 a
3
2


Page 2 of 8
 EXERCISE
2

 27  3
 
i. Find the value of  8 

ii. Solve 3  123   27  0

2x x

EXAMPLE

1  x  2  1 x1  x  2
1 1

Simplify 2
1 x
SOLUTION
1  x  2  x
1

21  x  2
1

1 x
21  x  2
1
Multiply numerator and denominator by

Student to work out

 2n
Simplify 3 n 1
 9 n  27  3 

EXERCISE
a) Simplify:
x1  x  2  1  x  2
1 
1 1

i. 2
x2
1  x 3  1 x1  x  3
1 2

ii. 3
1  x  3
2

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 1 1
1 
 x 2 1  x  2  x 2 1  x  2
1 
1 1

iii. 2
x
2

1 1
xy  x 3  2 y 4
iv.
x y 
1
10 9 12

LOGARITHMS
Logarithm of a number in a given base is the number
of times the base can be multiplied by itself. E.g.
2  32 index notation
5

In logarithmic form log 32  5 2

23  8 log 2 8  3

22  4

21  2

a1  a

101  10

10 2  100

10 3  1000

Logarithms under the base of 10 are known us

common logarithms
NAPERIAN LOGARITHMS/ NATURAL
LOGARITHMS.

Page 4 of 8
 Naperian Logarithms are logarithms under the base
of e . e is a mathematical constant. e  2.718281828...
Logarithm under base e is denoted by ln
i.e. log 4  ln 4 .
e

NB.
i. ln e  1
ii. e  y
ln y

EXAMPLE:
Solve for x :

i. ln x  9

e x5  2
ii.
LAWS OF LOGARITHMS
a) log AB  log A  log B
a a a

b) log BA  log A  log B

a a a

c) log a A p  p log a A

EXAMPLE
Prove that log AB  log A  log B
a a a

Proof
Let log A  m and log B  n .  A  a and
a a
m
B  an

AB  a m  a m

AB  a m  n

Take logarithm under base a on both side

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log a AB  log a a m  n

Or log AB  (m  n) log a
a a

Or log AB  m  n
a

But log A  m and log B  n

a a

Therefore log AB  log A  log a a a B hence proved.

CHANGE OF THE BASE

EXAMPLE
Evaluate:
a) log 14 2

SOLUTION
Let log 14  x 2

 2 x  14

Take log 10 on both side

log 10 2 x  log 10 14

Or x log 10 2  log 10 14
log 10 14
x
log 10 2

log a m
In general given log b m 
log a b
1
Also log b m 
log m b
1
log 3 x 
e.g. log x 3

Page 6 of 8
 EXAMPLE
i. Simplify:
log 2 8  16

SOLUTION
log 2 16  8  log 2 8  log 2 16

 log 2 2 3  log 2 2 4

 3 log 2 2  4 log 2 2

 3 4  7
log 125
ii. log 25

SOLUTION
log 5 3 3 log 5 3
 
log 5 2 2 log 5 2
a 2b 3
iii. Write log in terms of log a , log b and log c
100 c
iv. Solve the equation log x  log 2 x  3  1 5 5

v. Evaluate log 2 7

vi. Solve for x in 3  2 x

vii. If x  log 5 and y  log 5 , find y in terms x.

9 3

EXAMPLE
1 a
i. If log 10 2  a , show that log 8 5 
3a
SOLUTION
10
log 10
log 8 5 
log 10 5
 2  log 10 10  log 10 2
log 10 8 log 10 2 3 3 log 10 2

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 let log 10 2  a
1 a
log 8 5 
3a
viii. Solve for x if
a) log x  log x  6
3 9
2

b) log x  log 16
2 x

c) 4  62   16  0
x x

d)  23   161
 
1
e) log 2 1  log 4
2
 log 9 x

f) log 8x  log 8x  8
2
3
x
3

g) log x  4 log 3  3  0
3 x

h) ln x  5
i) log 10x  log x  3  2 (ans x=2,5
j) log (3x  1)  4
2 (ans x=5)
k) log (4x  3)  log ( x  6) (ans x=3)
7 7

l) 10  19 3 x 1
(ans x=0.760)
x 3
1
m) 4  
x
(ans x=1)
2

n) e x 5
 1.56

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