114 Basic Engineering Mathematics
= log
4 × 3
25
by the first and second
laws of logarithms
= log
12
25
= log 0.48
Problem 16. Write (a) log 30 (b) log 450 in terms
of log 2, log 3 and log 5 to any base
(a) log30 = log(2 × 15) = log(2 × 3 × 5)
= log 2 + log 3 + log 5
by the first law
of logarithms
(b) log450 = log(2 × 225) = log(2 × 3 × 75)
= log(2 × 3 × 3 × 25)
= log(2 × 3 2 × 5 2 )
= log 2 + log 3 2 + log 5 2
by the first law
of logarithms
i.e.
log450 = log 2 + 2 log3 + 2 log5
by the third law of logarithms
Problem 17. Write log
8 ×
4
√
5
81
in terms of
log 2, log 3 and log 5 to any base
log
8 ×
4
√
5
81
= log 8 + log
4
√
5 − log 81
by the first and second laws
of logarithms
= log 2 3 + log 5
1
4 − log 3 4
by the laws of indices
i.e.
log
8 ×
4
√
5
81
= 3 log2 +
1
4
log 5 − 4 log3
by the third law of logarithms
Problem 18. Evaluate
log 25 − log125 +
1
2
log 625
3 log5
log 25 − log125 +
1
2
log 625
3 log5
=
log5
2
− log 5
3
+
1
2
log 5
4
3 log5
=
2 log5 − 3 log5 +
4
2
log 5
3 log5
=
1 log5
3 log5
=
1
3
Problem 19. Solve the equation
log(x − 1) + log(x + 8) = 2 log(x + 2)
LHS = log(x − 1) + log(x + 8) = log(x − 1)(x + 8)
from the first
law of logarithms
= log(x 2 + 7x − 8)
RHS = 2 log(x + 2) = log(x + 2) 2
from the first
law of logarithms
= log(x 2 + 4x + 4)
Hence,
log(x 2 + 7x − 8) = log(x 2 + 4x + 4)
from which,
x 2 + 7x − 8 = x 2 + 4x + 4
i.e.
7x − 8 = 4x + 4
i.e.
3x = 12
and
x = 4
Problem 20. Solve the equation
1
2
log 4 = log x
1
2
log 4 = log 4
1
2 from the third law of
logarithms
= log
√
4 from the laws of indices
Hence,
1
2
log 4 = log x
becomes
log
√
4 = log x
i.e.
log 2 = log x
from which,
2 = x
i.e. the solution of the equation is x = 2.