24 Higher Engineering Mathematics
1
2
log4 = log x
Hence,
log
√
4 = log x
becomes
log2 = log x
i.e.
2 = x
from which,
i.e. the solution of the equation is: x = 2
Problem 21. Solve the equation:
log
x 2 − 3
− log x = log2.
log
x
2
− 3
− log x = log
x 2 − 3
x
from the second law of logarithms
log
x 2 − 3
x
= log 2
Hence,
x 2 − 3
x
= 2
from which,
x
2
− 3 = 2x
Rearranging gives:
x
2
− 2x − 3 = 0
and
(x − 3)(x + 1) = 0
Factorizing gives:
x = 3 or x = −1
from which,
x = −1 is not a valid solution since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is: x = 3
Now try the following exercise
Exercise 12 Further problems on laws of
logarithms
In Problems 1 to 11, write as the logarithm of a
single number:
1. log 2 + log 3
[log 6]
2. log 3 + log 5
[log 15]
3. log 3 + log 4 − log 6
[log 2]
4. log 7 + log 21 − log49
[log 3]
5. 2 log 2 + log 3
[log 12]
6. 2 log 2 + 3 log5
[log 500]
7. 2 log 5 −
1
2
log 81 + log 36
[log 100]
8.
1
3
log 8 −
1
2
log81 + log 27
[log 6]
9.
1
2
log 4 − 2 log3 + log45
[log 10]
10.
1
4
log 16 + 2 log3 − log 18
[log 1 = 0]
11. 2 log2 + log 5 − log 10
[log 2]
Simplify the expressions given in Problems 12
to 14:
12. log 27 − log9 + log 81
[log 243 or log 3 5 or 5 log3]
13. log 64 + log 32 − log 128
[log16 or log2 4 or 4 log2]
14. log 8 − log4 + log 32
[log64 or log 2 6 or 6 log2]
Evaluate the expressions given in Problems 15
and 16:
15.
1
2 log 16 −
1
3 log 8
log 4
[0.5]
16.
log 9 − log3 +
1
2 log 81
2 log3
[1.5]
Solve the equations given in Problems 17 to 22:
17. log x 4 − log x 3 = log5x − log 2x
[x = 2.5]
18. log 2t 3 − log t = log 16 + logt
[t = 8]
19. 2 logb 2 − 3 logb = log8b − log 4b
[b = 2]
20. log (x + 1) + log(x − 1) = log 3
[x = 2]
21.
1
3
log 27 = log(0.5a)
[a = 6]
22. log
x 2 − 5
− log x = log 4
[x = 5]
3.3 Indicial equations
The laws of logarithms may be used to solve certain
equations involving powers—called indicial equations. For example, to solve, say, 3 x = 27, logarithms to a base of 10 are taken of both sides,
i.e. log 10 3
x
= log 10 27
and x log 10 3 = log 10 27, by the third law of logarithms
1
2
log4 = log x
Hence,
log
√
4 = log x
becomes
log2 = log x
i.e.
2 = x
from which,
i.e. the solution of the equation is: x = 2
Problem 21. Solve the equation:
log
x 2 − 3
− log x = log2.
log
x
2
− 3
− log x = log
x 2 − 3
x
from the second law of logarithms
log
x 2 − 3
x
= log 2
Hence,
x 2 − 3
x
= 2
from which,
x
2
− 3 = 2x
Rearranging gives:
x
2
− 2x − 3 = 0
and
(x − 3)(x + 1) = 0
Factorizing gives:
x = 3 or x = −1
from which,
x = −1 is not a valid solution since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is: x = 3
Now try the following exercise
Exercise 12 Further problems on laws of
logarithms
In Problems 1 to 11, write as the logarithm of a
single number:
1. log 2 + log 3
[log 6]
2. log 3 + log 5
[log 15]
3. log 3 + log 4 − log 6
[log 2]
4. log 7 + log 21 − log49
[log 3]
5. 2 log 2 + log 3
[log 12]
6. 2 log 2 + 3 log5
[log 500]
7. 2 log 5 −
1
2
log 81 + log 36
[log 100]
8.
1
3
log 8 −
1
2
log81 + log 27
[log 6]
9.
1
2
log 4 − 2 log3 + log45
[log 10]
10.
1
4
log 16 + 2 log3 − log 18
[log 1 = 0]
11. 2 log2 + log 5 − log 10
[log 2]
Simplify the expressions given in Problems 12
to 14:
12. log 27 − log9 + log 81
[log 243 or log 3 5 or 5 log3]
13. log 64 + log 32 − log 128
[log16 or log2 4 or 4 log2]
14. log 8 − log4 + log 32
[log64 or log 2 6 or 6 log2]
Evaluate the expressions given in Problems 15
and 16:
15.
1
2 log 16 −
1
3 log 8
log 4
[0.5]
16.
log 9 − log3 +
1
2 log 81
2 log3
[1.5]
Solve the equations given in Problems 17 to 22:
17. log x 4 − log x 3 = log5x − log 2x
[x = 2.5]
18. log 2t 3 − log t = log 16 + logt
[t = 8]
19. 2 logb 2 − 3 logb = log8b − log 4b
[b = 2]
20. log (x + 1) + log(x − 1) = log 3
[x = 2]
21.
1
3
log 27 = log(0.5a)
[a = 6]
22. log
x 2 − 5
− log x = log 4
[x = 5]
3.3 Indicial equations
The laws of logarithms may be used to solve certain
equations involving powers—called indicial equations. For example, to solve, say, 3 x = 27, logarithms to a base of 10 are taken of both sides,
i.e. log 10 3
x
= log 10 27
and x log 10 3 = log 10 27, by the third law of logarithms