Cambridge Additional Mathematics

140 Logarithms (Chapter 5)

SOLVING LOGARITHMIC EQUATIONS

Logarithmic equations can often be solved using the laws of logarithms. However, we must always check

that our solutions satisfy the original equation, remembering thatlgxis only defined for x> 0.

Example 12 Self Tutor

Solve forx:

a lg(x¡6) + lg 3 = 2 lg 6 b

a lg(x¡6) + lg 3 = 2 lg 6

) lg(x¡6) = lg 6^2 ¡lg 3

) lg(x¡6) = lg

¡ 36

3

¢

) x¡6=12

) x=18

Check: x¡ 6 > 0 ,sox> 6 X

b lgx+lg(x+ 5) = lg 14

) lg(x(x+ 5)) = lg 14

) x(x+5)=14

) x^2 +5x¡14 = 0

) (x+ 7)(x¡2) = 0

) x=¡ 7 or 2

But x> 0 and x+5> 0

) x=2is the only valid solution.

EXERCISE 5D.2

1 Solve forx:

a lg(x¡4) = lg 3 + lg 7 b lg(x+5)¡lg 8 = 2 lg 3

c lg(2x)=1+^12 lg 16 d log 2 x= 3 log 25 ¡ 6

e lgx¡lg(x¡4) = lg 5 f log 5 (x¡2)¡log 5 (x+ 2) = log 53

g log 3 x¡2 = log 3 (x¡1) h lg(x+2)¡1 = lg(x¡3)¡lg 12

2 Solve forx:

a lgx+lg(x+ 1) = lg 30 b log 5 (x+ 9) + log 5 (x+ 2) = log 5 (20x)

c log 7 x= log 78 ¡log 7 (6¡x) d log 6 (x+ 4) + log 6 (x¡1) = 1

e lgx+lg(2x+8)=1 f lg(x+ 2) + lg(x+ 7) = lg(2x+2)

g 2 log 2 x¡log 2 (8¡ 3 x)=1 h log 2 x+ log 2 (2x¡7) = 2

Example 13 Self Tutor

Solve forx: logx3 + logx12 = 2

logx3 + logx12 = 2

) logx(3£12) = logx(x^2 )

) 36 =x^2

) x=6 fsince x> 0 g

3 Solve forx:

a logx 32 ¡logx4=1 b logx45 = 2 + logx 5

c logx54 = 3¡logx 4 d 2 logx 2 ¡3 = logx

¡ 1

16

¢

lgx+lg(x+ 5) = lg 14

The base of a

logarithm must

be positive.

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_05\140CamAdd_05.cdr Wednesday, 29 January 2014 9:15:03 AM BRIAN