Page 133 - Mathematics_Form_Two

Logarithms

Roots law (c) 1 5
n 5 =
The law states that log a m x = m log a , 1
n
x
It is proved as follows: = 5
1
Let p = log a m x n . From the law of = 5 log10 5

1
m
exponents, x = x m n = × 5log10
n
It follows that p = log x n m 5
1
a
n = × 5 1
= log x 5
m a
1
=
n = 1 Mathematics for Secondary Schools
n
Therefore, log m x = log x .
m
a
a
5
Therefore, log 100000 1.=
Example 6�15 1READING ONLY
1
Find the values of each of the following: (d) log 1 = log 3 log 1  1 1 2 = log 3   1   2
(a) log 9 729 (b) log 0.000001 3 27 3   27   27 1  27 
3
1 = 1 log  1  = log 3   1  
(c) log 100000 (d) log 3 27 2 3   27   2  27 
100000
5
3 1
1

FOR ONLINE
1
Solution 1  = log 3  3
3
1 = 2 log 3  2 
3
(a) log 9 729 = 9 log (729) log (729) 2 1  1 1 
2
9 729 =
log
1
9
=
 
3

= 1 log 9 3 1 log 9 3 =  3log 3  2 3log 3 
2
3

9 =
2 2 9 1
1 3
1 =  log 3 − 1 − 1 =  3log 3 − 1
3log 3
=
1
=  3log 9 3log 9 2 2 3 3 2 3
9 = 
2 2 9 1
3 ( 1)log 3
3 = 1 3 ( 1) =  −
=  − log 3×−
= = 3 3 ( 1)log 3 3
3 2
3
2 2 2 2 1
1 3
3 ( 1) 1
3 = −× 1 =  − 
3 ( 1) 1
729 =
log
 Therefore, log 9 2 729 = 3 . =  −  2
2 2

9
= − −
(b) log 0.000001 log(0.000001)= 2 1 3 1 3 = 3 3 = − 3
3
3
2
log 0.000001 log(0.000001)=
1 1 2 2
= log(10 ) log(10 )= − 6 3 - 6 3 Therefore, log 1 = − 3 .
1
3

=
log

−
= log10 log10= − 6 3 - 6 3 3 27 2 3 27 2
= log10 log10= − 2 - 2
= − 2log102log10= - Example 6�16
2 1
= − 2 1× = -´ Determine the value of x given that
Therefore, log 0.000001\ 3 2. = - 2. log x = log 2 log- 2 3.
∴
3
log 0.000001 = −
2
2
127
Student's Book Form Two
MATHEMATIC F2 v5.indd 127 11/10/2024 20:12:58
MATHEMATIC F2 v5.indd 127
11/10/2024 20:12:58