Basic Engineering Mathematics, Fifth Edition

Logarithms 115

Problem 21. Solve the equation

log

(

x^2 − 3

)

−logx=log2

log

(

x^2 − 3

)

−logx=log

(

x^2 − 3

x

)

from the second

law of logarithms

Hence, log

(

x^2 − 3

x

)

=log2

from which,

x^2 − 3

x

= 2

Rearranging gives x^2 − 3 = 2 x

and x^2 − 2 x− 3 = 0

Factorizing gives (x− 3 )(x+ 1 )= 0

from which, x=3orx=− 1

x=−1 is not a valid solution since the logarithm of a
negative number has no real root.

Hence, the solution of the equation isx= 3.

Now try the following Practice Exercise

PracticeExercise 60 Lawsof logarithms

(answers on page 346)

In problems 1 to 11, write as the logarithm of a

single number.

1. log2+log3 2. log3+log5


2. log3+log4−log6


3. log7+log21−log49


4. 2log2+log3 6. 2log2+3log5


5. 2log5−
   1
   2

log81+log36

8.

1

3

log8−

1

2

log81+log27

9.

1

2

log4−2log3+log45

10.

1

4

log16+2log3−log18

11. 2log2+log5−log10

Simplify the expressions given in problems

12 to 14.

12. log27−log9+log81


13. log64+log32−log128


14. log8−log4+log32

Evaluate the expressions given in problems 15

and 16.

15.

1

2

log16−

1

3

log8

log4

16.

log9−log3+

1

2

log81

2log3

Solve the equations given in problems 17 to 22.

17. logx^4 −logx^3 =log5x−log2x


18. log2t^3 −logt=log16+logt


19. 2logb^2 −3logb=log8b−log4b


20. log(x+ 1 )+log(x− 1 )=log3

21.

1

3

log27=log( 0. 5 a)

22. log(x^2 − 5 )−logx=log4

15.3 Indicial equations

The laws of logarithms may be used to solve

certain equations involving powers, called indicial

equations.

For example, to solve, say, 3x=27, logarithmsto a base

of 10 are taken of both sides,

i.e. log 103 x=log 1027

and xlog 103 =log 1027

by the third law of logarithms

Rearranging gives x=

log 1027

log 103

=

1. 43136 ...

0. 47712 ...

= 3 which may be readily

( checked.

Note,

log27

log3

isnotequal to log

27

3

)