Higher Engineering Mathematics, Sixth Edition

22 Higher Engineering Mathematics

In Problems 12 to 18 solve the equations:

12. log 10 x= 4 [10000]


13. lgx= 5 [100000]


14. log 3 x=2[9]


15. log 4 x=− 2

1

2

[

1

32

]

16. lgx=−2[ 0 .01]


17. log 8 x=−

4

3

[

1

16

]

18. lnx=3[e^3 ]

3.2 Laws of logarithms

There are three laws of logarithms, which apply to any

base:

(i) To multiply two numbers:

log(A×B)=logA+logB

The following may be checked by using a calcu-

lator:

lg10= 1

Also, lg 5+lg 2= 0. 69897 ...

+ 0. 301029 ...= 1

Hence, lg( 5 × 2 )=lg10=lg 5+lg 2

(ii) To divide two numbers:

log

(

A

B

)

=logA−logB

Thefollowing may bechecked using acalculator:

ln

(

5

2

)

=ln2. 5 = 0. 91629 ...

Also, ln5−ln2= 1. 60943 ...− 0. 69314 ...

= 0. 91629 ...

Hence, ln

(

5

2

)

=ln5−ln2

(iii) To raise a number to a power:

logAn=nlogA

Thefollowing may bechecked using acalculator:

lg 5^2 =lg 25= 1. 39794 ...

Also, 2lg 5= 2 × 0. 69897 ...= 1. 39794 ...

Hence, lg 5^2 =2lg5

Here are some worked problems to help understand-

ing of the laws of logarithms.

Problem 10. Write log 4+log 7 as the logarithm

of a single number.

log 4+log 7=log( 7 × 4 )

by the first law of logarithms

=log 28

Problem 11. Write log 16−log 2 as the logari-

thm of a single number.

log16−log 2=log

(

16

2

)

by the second law of logarithms

=log 8

Problem 12. Write 2log 3 as the logarithm of a

single number.

2log3=log 3^2 by the third law of logarithms

=log 9

Problem 13. Write

1

2

log 25 as the logarithm of a

single number.

1

2

log 25=log 25

1

(^2) by the third law of logarithms
=log
√
25 =log 5
Problem 14. Simplify: log64−log128+log32.
64 = 26 , 128 = 27 and 32= 25
Hence, log64−log128+log32
=log2^6 −log2^7 +log2^5