Page 131 - Mathematics_Form_Two

Logarithms

(b) log 30 log 3 log 10= 10 + 10 Example 6�12
10
log 3 l= 10 + Evaluate log 9 .
2

0.4771 l= + 3
Solution
1.4771=
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2
log 9 = 2log 9
3
3
(c) log 45 log (3 3 5)= 10 ×× = 2log 3 2
10
3
log 3 log 3 log 5= 10 + 10 + 10 = 2(2)log 3
3
= 2log 3 log 5+ 10 = 4(1) Mathematics for Secondary Schools
10
2 0.4771 0.6990= × + = 4
2
1.6532= Therefore, log 9 = 4.
3
Power rule Example 6�13

n
Power rule states that log m = n log m . Find the values of each of the following:
a
a
Let p = log m . (i)
a
( )
(a) log 64 5
It is derived as follows: 4
Express equation (i) in exponential form to (b) log (100 ) 25
get
( )
p
a = m (ii) (c) log 0.1 6
Raise both sides of equation (ii) to the power Solution
n to get,
( )
( ) =
a pn = m n (iii) (a) log 64 5 5log 64
4
4
Apply logarithm to base a on both sides of = 5log 4 3
4
equation (iii) to obtain, = 
5 3log 4
log a ( ) loga pn = a m (iv) = 15 4
n
5
Simplify equation (iv) to get pn = log m Therefore,log (64) = 15.
.
n
4
a
But p = log m .
a
25
(b) log (100 ) = 25log100
It follows that, logn a m = log m n . = 25log10 2
a
Therefore, log m = n log m . = 25 2log10
n
a
a
= 25 2 1
From this rule, it follows that = 50
1 Therefore, log(100) =
25
log a x = − log x 50.
a
125
Student's Book Form Two

11/10/2024 20:12:56
MATHEMATIC F2 v5.indd 125 11/10/2024 20:12:56
MATHEMATIC F2 v5.indd 125