LESSON 2
FURTHER MATHEMATICS
SS 1
Topic:
Theory of Logarithms
I:
Relationship between logarithms and indices
If
23 = 8
(Index form)
8
Then Log 2 = 3
(logarithmic form)
In general: If N = ax, then logNa = x
DEFINITION:
The logarithm of any number N to any base a is the power or index x to
which the base must be raised to equal the number N.
TYPES OF LOGARITHMS
(i)
Common logarithms is logarithms to base 10 are called common
(ii)
logarithm Examples log510, log710
Natural logarithms : logarithms to base e are called natural logarithm
e.g log5e or In5
Example 1:
Evaluate the following without using tables or a calculator.
(i) log16
(ii) log1/64
2
0.5
SOLUTIONS
(i)
Let log16
2 = x
Convert to index form
2x = 16
2x = 24
:. x = 4
(ii)
Let log1/64
0.5 = y
Convert to index form
0.5y = 1/64
(5/10)y = 1/64
(1/2)y = 1/26
2-y = 2-6
-y = -6
:. y = 6
Example 2
Express in index form
log1/16
= -4
2
SOLUTION
2-4 = 1/16
Example 3
Express 163/2 = 64 in logarithmic form
SOLUTION
3
log64
16 = /2
II.
Laws of logarithms
(the same base)
Rule 1
log(MN)
= logMa + logNa
a
Example 4
Simplify log510+ log210
SOLUTION
log10 (5 x 2) = log1010= 1
Rule 2
N
log(M/N)
= logM
a
a - log a
Example 5
2
Simplify log10
10 – log 10
SOLUTION
log10/2
= log510
10
Rule 3
n
logaM = n logMa
Example 6
Simplify 2/3 log27
3
SOLUTION
3
2
/3 log33 = 2/3 x 3 log3 3
= 2 log33 = 2 x 1 = 2
Special logarithms
(i)
logaa = 1 e.g log55 = 1
(ii)
log1a = 0 e.g log16 = 0
1/5
(iii) log1/a
a = -1 ; log 5 = -1
(iv)
x
1/5
5
log1/x
a = - log e.g log a = - log
Further worked examples
Example 7
Evaluate without using tables or calculator
log45 – 2 log5/65 + log625/144
5
Solution
log5 {4 ÷ (5/6)2 x 625/144}
log5 {4/1 ÷ 6/5 x 6/5 x 625/144}
log5 (25) = log 552 = 2 log55= 2 x 1 = 2
Example 8
Evaluate without using tables or calculator
log 48 – log 3
log 20 – log 5
SOLUTION
(i)
log 48 – log 3 = log48/3 = log16
(ii)
log20 – log5 = log20/5 = log4
2
log16 ÷ log4 = log16/log4 = log4/log4
= 2 log4/log4 = 2
Example 9
(iii)
If log2 = 0.3010, log3 = 0.4771 and log7 = 0.8451, evaluate without using
tables or calculator.
(i) log21
(ii) log0.8
SOLUTIONS
(i)
log21 = log(3x7)
log3 + log7
= 0.4771 + 0.8451 = 1.3222
(ii)
0.8 = 8/10
log8 – log10
3 log2 – log10 = 3 x 0.3010 – 1 = 0.903 – 1 = -0.097
CHANGING THE BASE OF A LOGARITHM
logb
= logbc/logac
a
logca = 1/logac
Example 10
Evaluate log 364 x log 8243
SOLUTION
(i)
(ii)
Change of base log 364 = log 648 /log 38
5
log 64
log 64 x log 3
2 x 5 log 3
/
/log8 3 =810
8 /log 3 x log 243 =
8
8
log
3 8=
8
8
ASSIGNMENT
(1)
Evaluate the following without using tables or calculator
(a) log 3729 (b) log ½0.125
(c) 2 log 6 + ½ log 625 – ½ log 81
(2)
(3)
Show that log√7 + log√20 – log√14 = ½
Simplify log√27/log√81
(4)
(5)
Find p in terms of q if log 3p + 3 log q3 = 3
Simplify 3 log 69 + log 612 + log 664 – log 672