Cambridge Additional Mathematics

If ,

then.

2=

= log

x a

xaw

148 Logarithms (Chapter 5)

Example 24 Self Tutor

Evaluate log 29 by:

a changing to base 10 b changing to basee.

a log 2 9=log^109

log 102

¼ 3 : 17 b log 2 9=ln 9

ln 2

¼ 3 : 17

The rule can also be used to solve equations involving logarithms with different bases.

Example 25 Self Tutor

Solve forx: log 2 x= log 815

log 2 x= log 815

) log 2 x=

log 215

log 28

fwriting RHS with base 2 g

) log 2 x=log^215

3

) log 2 x= log 215

1

3

) x=^3

p

15

EXERCISE 5G

1 Use the rule logba=

log 10 a

log 10 b

to evaluate, correct to 3 significant figures:

a log 312 b log 1

2

1250 c log 3 (0:067) d log 0 : 4 (0:006 984)

2 Use the rule logba=

lna

lnb

to solve, correct to 3 significant figures:

a 2 x=0: 051 b 4 x= 213: 8 c 32 x+1=4: 069

3 Write:

a log 926 in the form alog 3 b, where a,b 2 Q

b log 211 in the form alog 4 b, where a,b 2 Z

c^6

log 725

in the form alog 5 b, where a,b 2 Z.

4 Solve forx:

a log 3 x= log 2750 b log 2 x= log 413 c log 25 x= log 57

d log 3

p

x+ log 9 x= log 35 e log 8 x^2 ¡log 23

p

x=1 f log 4 x^3 + log 2

p

x=8

5aShow that logab=

1

logba

.

b Solve forx:

i log 3 x= 4 logx 3 ii log 2 x¡4 = 5 logx 2 iii 2 log 4 x+ 3 logx4=7

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_05\148CamAdd_05.cdr Tuesday, 21 January 2014 2:50:02 PM BRIAN