Cambridge Additional Mathematics

136 Logarithms (Chapter 5)

More generally, in any basecwhere c 6 =1, c> 0 , we have theselaws of logarithms:

IfAandBare both

positive then:

² logcA+ logcB= logc(AB)

² logcA¡logcB= logc

μ

A

B

¶

² nlogcA= logc(An)

Proof:

² logc(AB)

= logc(clogcA£clogcB)

= logc(clogcA+ logcB)

= logcA+ logcB

² logc

³

A

B

́

= logc

μ

clogcA

clogcB

¶

= logc(clogcA¡logcB)

= logcA¡logcB

² logc(An)

= logc((clogcA)n)

= logc(cnlogcA)

=nlogcA

Example 7 Self Tutor

Use the laws of logarithms to write the following as a single logarithm or as an integer:

a lg 5 + lg 3 b log 324 ¡log 38 c log 25 ¡ 1

a lg 5 + lg 3

=lg(5£3)

=lg15

b log 324 ¡log 38

= log 3

¡ 24

8

¢

= log 33

=1

c log 25 ¡ 1

= log 25 ¡log 221

= log 2

¡ 5

2

¢

Example 8 Self Tutor

Simplify by writing as a single logarithm or as a rational number:

a 2lg7¡3lg2 b 2lg3+3 c

lg 8

lg 4

a 2lg7¡3lg2

=lg(7^2 )¡lg(2^3 )

=lg49¡lg 8

=lg

¡ 49

8

¢

b 2lg3+3

=lg(3^2 ) + lg(10^3 )

= lg 9 + lg 1000

= lg(9000)

c

lg 8

lg 4

=

lg 2^3

lg 2^2

=

3lg2

2lg2

=^32

cyan magenta yellow black

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 4037 Cambridge
Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_05\136CamAdd_05.cdr Friday, 20 December 2013 1:03:10 PM BRIAN