Higher Engineering Mathematics, Sixth Edition

Logarithms 21

Problem 1. Evaluate log 3 9.

Letx=log 3 9then3x= 9 from the definition

of a logarithm,

i.e. 3 x= 32 from which,x= 2

Hence, log 39 = 2

Problem 2. Evaluate log 10 10.

Letx=log 10 10 then 10x= 10 from the

definition of a logarithm,

i.e. 10 x= 101 from which,x= 1

Hence, log 1010 = 1 (which may be checked

by a calculator)

Problem 3. Evaluate log 16 8.

Letx=log 16 8then16x=8 from the definition

of a logarithm,

i.e.( 24 )x= 23 i.e. 2^4 x= 23 from the laws of indices,

from which, 4 x=3andx=

3

4

Hence, log 168 =

3

4

Problem 4. Evaluate lg 0.001.

Letx=lg 0. 001 =log 100. 001 then 10x= 0. 001

i.e. 10 x= 10 −^3 from which,x=− 3

Hence, lg 0. 001 =− 3 (which may be checked

by a calculator)

Problem 5. Evaluate lne.

Letx=lne=logeethenex=e

i.e. ex=e^1
from which,x= 1

Hence, lne= 1 (which may be checked

by a calculator)

Problem 6. Evaluate log 3

1

81

.

Letx=log 3

1

81

then 3x=

1

81

=

1

34

= 3 −^4

from which,x=− 4

Hence, log 3

1

81

=− 4

Problem 7. Solve the equation: lgx=3.

If lgx=3 then log 10 x= 3

and x= 103 i.e. x= 1000

Problem 8. Solve the equation: log 2 x=5.

If log 2 x=5thenx= 25 = 32

Problem 9. Solve the equation: log 5 x=−2.

If log 5 x=−2thenx= 5 −^2 =

1

52

=

1

25

Now try the following exercise

Exercise 11 Further problemson lawsof

logarithms

In Problems 1 to 11, evaluate the given

expressions:

1. log 10 10000 [4] 2. log 2 16 [4]


2. log 5125 [3] 4. log 218 [−3]


3. log 82

[

1

3

]

6. log 7 343 [3]


7. lg100 [2] 8. lg 0.01 [−2]


8. log 48

[

1

1

2

]

10. log 273

[

1

3

]

11. lne^2 [2]