Cambridge International Mathematics

We have already seen how to solve equations such as 2 x=5using technology. We now consider an

algebraic method.

By definition, the exact solution is x= log 25 , but we need to know how to evaluate this number.

We therefore consider taking the logarithm of both sides of the original equation:

log(2x) = log 5

) xlog 2 = log 5 flogarithm lawg

) x=

log 5

log 2

We conclude that log 2 5=

log 5

log 2

.

In general: the solution to ax=b where a> 0 , b> 0 is x= logab=

logb

loga

:

Example 11 Self Tutor

Use logarithms to solve forx, giving answers correct to 3 significant figures:

a 2 x=30 b (1:02)x=2: 79 c 3 x=0: 05

a 2 x=30

) x=

log 30

log 2

) x¼ 4 : 91

b (1:02)x=2: 79

) x=

log(2:79)

log(1:02)

) x¼ 51 : 8

c 3 x=0: 05

) x=

log(0:05)

log 3

) x¼¡ 2 : 73

Example 12 Self Tutor

Show that log 2 11 =

log 11

log 2

. Hence find log 211.

Let log 2 11 =x

) 2 x=11

) log(2x) = log 11

) xlog 2 = log 11

) x=

log 11

log 2

EXPONENTIAL AND LOGARITHMIC

EQUATIONS [3.10]

E

) log 2 11 =

log 11

log 2

¼ 3 : 46

634 Logarithms (Chapter 31)

IGCSE01

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Y:\HAESE\IGCSE01\IG01_31\634IGCSE01_31.CDR Monday, 27 October 2008 3:02:12 PM PETER