22 Higher Engineering Mathematics
In Problems 12 to 18 solve the equations:
12. log 10 x = 4
[10000]
13. lg x = 5
[100000]
14. log 3 x = 2
[ 9 ]
15. log 4 x = −2
1
2
1
32
16. lg x = −2
[ 0 .01]
17. log 8 x = −
4
3
1
16
18. ln x = 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) = log A + log B
The following may be checked by using a calculator:
lg 10 = 1
Also, lg 5 + lg 2 = 0.69897...
+ 0.301029... = 1
Hence,
lg (5 × 2) = lg 10 = lg 5 + lg 2
(ii) To divide two numbers:
log
A
B
= log A − log B
The following may be checked using a calculator:
ln
5
2
= ln 2.5 = 0.91629...
Also,
ln 5 − ln 2 = 1.60943... − 0.69314...
= 0.91629...
Hence,
ln
5
2
= ln 5 − ln 2
(iii) To raise a number to a power:
log A
n
= n log A
The following may be checked using a calculator:
lg 5
2
= lg 25 = 1.39794...
Also, 2 lg 5 = 2 × 0.69897... = 1.39794...
Hence,
lg 5 2 = 2 lg 5
Here are some worked problems to help understanding 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 logarithm of a single number.
log 16 − log 2 = log
16
2
by the second law of logarithms
= log 8
Problem 12. Write 2 log 3 as the logarithm of a
single number.
2 log 3 = 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: log 64 − log 128 + log32.
64 = 2 6 , 128 = 2 7 and 32 = 2 5
Hence, log 64 − log 128 + log32
= log 2
6
− log 2
7
+ log 2
5