116 Basic Engineering Mathematics
Problem 22. Solve the equation 2 x = 5, correct
to 4 significant figures
Taking logarithms to base 10 of both sides of 2
x
= 5
gives
log 10 2
x
= log 10 5
i.e.
x log 10 2 = log 10 5
by the third law of logarithms
Rearranging gives x =
log 10 5
log 10 2
=
0.6989700 ...
0.3010299 ...
= 2.322, 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 2.7 = 34.68,
correct to 4 significant figures
Taking logarithms to base 10 of both sides gives
log 10 x
2.7
= log 10 34.68
2.7 log 10 x = log 10 34.68
Hence,
log 10 x =
log 10 34.68
2.7
= 0.57040
Thus,
x = antilog 0.57040 = 10
0.57040
= 3.719,
correct to 4 significant figures.
Now try the following Practice Exercise
Practice Exercise 61 Indicial equations
(answers on page 346)
In problems 1 to 8, solve the indicial equations for
x, each correct to 4 significant figures.
1. 3 x = 6.4
2 . 2 x = 9
3. 2 x−1 = 3 2x−1
4. x 1.5 = 14.91
5. 25.28 = 4.2 x
6. 4 2x−1 = 5 x+2
7. x −0.25 = 0.792
8. 0.027 x = 3.26
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.
15.4 Graphs of logarithmic functions
A graph of y = log 10 x is shown in Figure 15.1 and a
graph of y = log e x is shown in Figure 15.2. Both can
be 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
(a) 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 logarithms.
Hence, log a 1 = 0. In the above graphs it is seen
that log 10 1 = 0 and log e 1 = 0.
(b) log a a = 1
Let log a a = x then a x = a from the definition of
a logarithm.
If a x = a then x = 1.
Problem 22. Solve the equation 2 x = 5, correct
to 4 significant figures
Taking logarithms to base 10 of both sides of 2
x
= 5
gives
log 10 2
x
= log 10 5
i.e.
x log 10 2 = log 10 5
by the third law of logarithms
Rearranging gives x =
log 10 5
log 10 2
=
0.6989700 ...
0.3010299 ...
= 2.322, 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 2.7 = 34.68,
correct to 4 significant figures
Taking logarithms to base 10 of both sides gives
log 10 x
2.7
= log 10 34.68
2.7 log 10 x = log 10 34.68
Hence,
log 10 x =
log 10 34.68
2.7
= 0.57040
Thus,
x = antilog 0.57040 = 10
0.57040
= 3.719,
correct to 4 significant figures.
Now try the following Practice Exercise
Practice Exercise 61 Indicial equations
(answers on page 346)
In problems 1 to 8, solve the indicial equations for
x, each correct to 4 significant figures.
1. 3 x = 6.4
2 . 2 x = 9
3. 2 x−1 = 3 2x−1
4. x 1.5 = 14.91
5. 25.28 = 4.2 x
6. 4 2x−1 = 5 x+2
7. x −0.25 = 0.792
8. 0.027 x = 3.26
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.
15.4 Graphs of logarithmic functions
A graph of y = log 10 x is shown in Figure 15.1 and a
graph of y = log e x is shown in Figure 15.2. Both can
be 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
(a) 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 logarithms.
Hence, log a 1 = 0. In the above graphs it is seen
that log 10 1 = 0 and log e 1 = 0.
(b) log a a = 1
Let log a a = x then a x = a from the definition of
a logarithm.
If a x = a then x = 1.