Logarithms 115
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
Hence,
log
x 2 − 3
x
= log 2
from which,
x 2 − 3
x
= 2
Rearranging gives
x 2 − 3 = 2x
and
x 2 − 2x − 3 = 0
Factorizing gives (x − 3)(x + 1) = 0
from which,
x = 3 or x = −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 is x = 3.
Now try the following Practice Exercise
Practice Exercise 60 Laws of logarithms
(answers on page 346)
In problems 1 to 11, write as the logarithm of a
single number.
1. log 2 + log 3
2. log 3 + log5
3. log 3 + log4 − log 6
4. log 7 + log21 − log 49
5. 2 log2 + log 3
6. 2 log2 + 3 log5
7. 2 log5 −
1
2
log 81 + log36
8.
1
3
log 8 −
1
2
log81 + log 27
9.
1
2
log4 − 2 log3 + log45
10.
1
4
log 16 + 2 log3 − log 18
11. 2 log2 + log 5 − log 10
Simplify the expressions given in problems
12 to 14.
12. log 27 − log9 + log 81
13. log 64 + log 32 − log 128
14. log 8 − log4 + log 32
Evaluate the expressions given in problems 15
and 16.
15.
1
2
log16 −
1
3
log 8
log 4
16.
log 9 − log3 +
1
2
log 81
2 log3
Solve the equations given in problems 17 to 22.
17. log x 4 − log x 3 = log5x − log 2x
18. log 2t 3 − log t = log 16 + logt
19. 2 logb 2 − 3 logb = log8b − log 4b
20. log(x + 1) + log(x − 1) = log3
21.
1
3
log 27 = log(0.5a)
22. log(x 2 − 5) − log x = log 4
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, 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
Rearranging gives x =
log 10 27
log 10 3
=
1.43136 ...
0.47712 ...
= 3
which may be readily
checked.
Note,
log 27
log 3
is not equal to log
27
3
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
Hence,
log
x 2 − 3
x
= log 2
from which,
x 2 − 3
x
= 2
Rearranging gives
x 2 − 3 = 2x
and
x 2 − 2x − 3 = 0
Factorizing gives (x − 3)(x + 1) = 0
from which,
x = 3 or x = −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 is x = 3.
Now try the following Practice Exercise
Practice Exercise 60 Laws of logarithms
(answers on page 346)
In problems 1 to 11, write as the logarithm of a
single number.
1. log 2 + log 3
2. log 3 + log5
3. log 3 + log4 − log 6
4. log 7 + log21 − log 49
5. 2 log2 + log 3
6. 2 log2 + 3 log5
7. 2 log5 −
1
2
log 81 + log36
8.
1
3
log 8 −
1
2
log81 + log 27
9.
1
2
log4 − 2 log3 + log45
10.
1
4
log 16 + 2 log3 − log 18
11. 2 log2 + log 5 − log 10
Simplify the expressions given in problems
12 to 14.
12. log 27 − log9 + log 81
13. log 64 + log 32 − log 128
14. log 8 − log4 + log 32
Evaluate the expressions given in problems 15
and 16.
15.
1
2
log16 −
1
3
log 8
log 4
16.
log 9 − log3 +
1
2
log 81
2 log3
Solve the equations given in problems 17 to 22.
17. log x 4 − log x 3 = log5x − log 2x
18. log 2t 3 − log t = log 16 + logt
19. 2 logb 2 − 3 logb = log8b − log 4b
20. log(x + 1) + log(x − 1) = log3
21.
1
3
log 27 = log(0.5a)
22. log(x 2 − 5) − log x = log 4
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, 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
Rearranging gives x =
log 10 27
log 10 3
=
1.43136 ...
0.47712 ...
= 3
which may be readily
checked.
Note,
log 27
log 3
is not equal to log
27
3