Page 132 - Mathematics_Form_Two

Logarithms

Example 6�14
( )
(c) log 0.1 6 6log 0.1
( ) =
Find the value of each of the following
= 6log10 − 1 expressions.
= 6 ( 1)log10− æ 27 ö
= 6 ( 1) 1−  6. (a) log 3 ç 3 è 9 ø ÷ .
FOR ONLINE READING ONLY
6 6
=
= − −

(b) log (9 243)
Mathematics for Secondary Schools Quotient rule and q = log y. (1) (c) log(10 0.001) 27   9  = log 27 log 9 3
6
Therefore, log(0.1) = −
Solution
log x
Let p =
Let

27 

a
a
log 27 log 9
=
log
−
−
(a) log
3 
3 
3
3
3

Expressing equation (1) in exponential
log 3
form gives
log 3 −
=
3
2
3
3
3
q
p
3log 3 2log 3−
=
x = a and y = a . (2)  9  = log 3 − − 3 3 log 3 3 2
=
3log 3 2log 3 3
3
3
=
3(1) 2(1)−
From equation (2), divide x by y to obtain = 3(1) 2(1)
−
x = a  p a = q a pq− = 1 = 1
y (by laws of exponents)  27 
x Therefore, log 3   = 1.
Thus, a pq- . (3)  9 
\ =
y
(a)
Express equation (3) in logarithmic form (b) log (9 243) log 9 log 243 3 = 3 − 3
2
to obtain = log 3 − log 3 5
3
3
x
 = 2log 3 5log 3−
log a  = log a pq− 3 3
y
 a = 2(1) 5(1)−
= ( pq− )log a = − 3
a
 Therefore, log (9 243) 3 = − 3.
x
−
Thus, log a  = pq= log a . pq− (4)
a
y
  10 
(c) log(10 0.001) log
Substitute the expressions for p and q (b)  = 10   0.001  
from equation (1) into (4) to obtain = log 10 log 0.001
−
10
10
1
 x  = log 10 − log 10 − 3
log a   = log x − log y 10 10
−−
y
  a a = 1log 10 ( 3)log 10
10
10
13 4
 x  = +=
Therefore, log a   = log x − log y .
y
  a a Therefore, log(10 0.001) 4. =
126
Student's Book Form Two
11/10/2024 20:12:57
MATHEMATIC F2 v5.indd 126
MATHEMATIC F2 v5.indd 126 11/10/2024 20:12:57