Cambridge International Mathematics

Consider two positive numbersxandy. We can write them both with basea: x=ap and y=aq, for

somepandq.

) p= logax and q= logay ...... (*)

Using exponent laws, we notice that: xy=apaq=ap+q

x

y

=

ap

aq

=ap¡q

xn=(ap)n=anp

) loga(xy)=p+q= logax+ logay ffrom *g

loga

μ

x

y

¶

=p¡q= logax¡logay

loga(xn)=np=nlogax

loga(xy) = logax+ logay

loga

μ

x

y

¶

= logax¡logay

loga(xn)=nlogax

Example 5 Self Tutor

If log 3 5=p and log 3 8=q, write in terms ofpandq:

a log 340 b log 325 c log 3

¡ 64

125

¢

a log 340

= log 3 (5£8)

= log 3 5 + log 38

=p+q

b log 325

= log 352

= 2 log 35

=2p

c log 3

¡ 64

125

¢

= log 3

μ

82

53

¶

= log 382 ¡log 353

= 2 log 38 ¡3 log 35

=2q¡ 3 p

C RULES FOR LOGARITHMS [3.10]

Logarithms (Chapter 31) 629

Example 4 Self Tutor

Simplify: a log 27 ¡^12 log 2 3 + log 25 b 3 ¡log 25

a log 27 ¡^12 log 2 3 + log 25

= log 2 7 + log 25 ¡log 23

(^12)
= log 2 (7£5)¡log 2
p
3
= log 2
³
p^35
3
́
b 3 ¡log 25
= log 223 ¡log 25
= log 2
¡ 8
5
¢
= log 2 (1:6)
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Y:\HAESE\IGCSE01\IG01_31\629IGCSE01_31.CDR Tuesday, 18 November 2008 11:13:47 AM PETER