Logarithms 25
Rearranging gives
x =
log 10 27
log 10 3
=
1.43136 ...
0.4771 ...
= 3
which may be readily checked
Note,
log8
log2
is not equal to lg
8
2
Problem 22. Solve the equation 2
x
= 3, correct to
4 significant figures.
Taking logarithms to base 10 of both sides of 2 x = 3
gives:
log 10 2 x = log 10 3
i.e.
x log 10 2 = log 10 3
Rearranging gives:
x =
log 10 3
log 10 2
=
0.47712125 ...
0.30102999 ...
= 1.585, correct to 4 significant figures
Problem 23. Solve the equation 2 x+1 = 3 2x−5
correct to 2 decimal places.
Taking logarithms to base 10 of both sides gives:
log 10 2 x+1 = log 10 3 2x−5
i.e.
(x + 1) log 10 2 = (2x − 5) log 10 3
x log 10 2 + log 10 2 = 2x log 10 3 − 5 log 10 3
x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771)
i.e. 0.3010x + 0.3010 = 0.9542x − 2.3855
Hence
2.3855 + 0.3010 = 0.9542x − 0.3010x
2.6865 = 0.6532x
from which x =
2.6865
0.6532
= 4.11, correct to
2 decimal places
Problem 24. Solve the equation x 3.2 = 41.15,
correct to 4 significant figures.
Taking logarithms to base 10 of both sides gives:
log 10 x 3.2 = log 10 41.15
3.2 log 10 x = log 10 41.15
Hence log 10 x =
log 10 41.15
3.2
= 0.50449
Thus x = antilog 0.50449 =10 0.50449 = 3.195 correct to
4 significant figures.
Now try the following exercise
Exercise 13 Indicial equations
Solve the following indicial equations for x, each
correct to 4 significant figures:
1. 3 x = 6.4
[1.690]
2. 2
x
= 9
[3.170]
3. 2 x−1 = 3 2x−1
[0.2696]
4. x 1.5 = 14.91
[6.058]
5. 25.28 =4.2
x
[2.251]
6. 4 2x−1 = 5 x+2
[3.959]
7. x −0.25 = 0.792
[2.542]
8. 0.027 x = 3.26
[−0.3272]
9. The decibel gain n of an amplifier is given by:
n = 10 log 10
P 2
P 1
where P 1 is the power input and P 2 is the
power output. Find the power gain
P 2
P 1
when
n =25 decibels.
[316.2]
3.4 Graphs of logarithmic functions
A graph of y = log 10 x is shown in Fig. 3.1 and a graph
of y = log e x is shown in Fig. 3.2. Both are seen to be
of similar shape; in fact, the same general shape occurs
for a logarithm to any base.
In general, with a logarithm to any base a, it is noted
that:
(i) log a 1 = 0
Let log a = x, then a x = 1 from the definition of
the logarithm.
If a x = 1 then x = 0 from the laws of indices.
Hence log a 1 =0. In the above graphs it is seen
that log 10 1 = 0 and log e 1 = 0
Rearranging gives
x =
log 10 27
log 10 3
=
1.43136 ...
0.4771 ...
= 3
which may be readily checked
Note,
log8
log2
is not equal to lg
8
2
Problem 22. Solve the equation 2
x
= 3, correct to
4 significant figures.
Taking logarithms to base 10 of both sides of 2 x = 3
gives:
log 10 2 x = log 10 3
i.e.
x log 10 2 = log 10 3
Rearranging gives:
x =
log 10 3
log 10 2
=
0.47712125 ...
0.30102999 ...
= 1.585, correct to 4 significant figures
Problem 23. Solve the equation 2 x+1 = 3 2x−5
correct to 2 decimal places.
Taking logarithms to base 10 of both sides gives:
log 10 2 x+1 = log 10 3 2x−5
i.e.
(x + 1) log 10 2 = (2x − 5) log 10 3
x log 10 2 + log 10 2 = 2x log 10 3 − 5 log 10 3
x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771)
i.e. 0.3010x + 0.3010 = 0.9542x − 2.3855
Hence
2.3855 + 0.3010 = 0.9542x − 0.3010x
2.6865 = 0.6532x
from which x =
2.6865
0.6532
= 4.11, correct to
2 decimal places
Problem 24. Solve the equation x 3.2 = 41.15,
correct to 4 significant figures.
Taking logarithms to base 10 of both sides gives:
log 10 x 3.2 = log 10 41.15
3.2 log 10 x = log 10 41.15
Hence log 10 x =
log 10 41.15
3.2
= 0.50449
Thus x = antilog 0.50449 =10 0.50449 = 3.195 correct to
4 significant figures.
Now try the following exercise
Exercise 13 Indicial equations
Solve the following indicial equations for x, each
correct to 4 significant figures:
1. 3 x = 6.4
[1.690]
2. 2
x
= 9
[3.170]
3. 2 x−1 = 3 2x−1
[0.2696]
4. x 1.5 = 14.91
[6.058]
5. 25.28 =4.2
x
[2.251]
6. 4 2x−1 = 5 x+2
[3.959]
7. x −0.25 = 0.792
[2.542]
8. 0.027 x = 3.26
[−0.3272]
9. The decibel gain n of an amplifier is given by:
n = 10 log 10
P 2
P 1
where P 1 is the power input and P 2 is the
power output. Find the power gain
P 2
P 1
when
n =25 decibels.
[316.2]
3.4 Graphs of logarithmic functions
A graph of y = log 10 x is shown in Fig. 3.1 and a graph
of y = log e x is shown in Fig. 3.2. Both are seen to be
of similar shape; in fact, the same general shape occurs
for a logarithm to any base.
In general, with a logarithm to any base a, it is noted
that:
(i) log a 1 = 0
Let log a = x, then a x = 1 from the definition of
the logarithm.
If a x = 1 then x = 0 from the laws of indices.
Hence log a 1 =0. In the above graphs it is seen
that log 10 1 = 0 and log e 1 = 0