122 Basic Engineering Mathematics
A table of values is drawn up as shown below.
t
0
1
2
3
e
−t /3
1.00
0.7165 0.5134 0.3679
v = 250e −t /3 250.0 179.1
128.4
91.97
t
4
5
6
e −t /3
0.2636
0.1889
0.1353
v = 250e −t /3
65.90
47.22
33.83
The natural decay curve of v = 250e −t /3 is shown in
Figure 16.4.
250
200
150
Voltage
v (volts)
100
80
50
Time t (seconds)
v 5 250e 2t /3
0
1 1.5 2
3 3.4 4
5
6
Figure 16.4
From the graph,
(a) when time t = 3.4 s, voltage v = 80 V, and
(b) when voltage v = 150 V, time t = 1.5 s.
Now try the following Practice Exercise
Practice Exercise 64 Exponential graphs
(answers on page 347)
1. Plot a graph of y = 3e 0.2x over the range
x = −3 to x = 3. Hence determine the value
of y when x = 1.4 and the value of x when
y = 4.5
2. Plot a graph of y =
1
2
e −1.5x over a range
x = −1.5 to x = 1.5 and then determine the
value of y when x = −0.8 and the value of x
when y = 3.5
3. In a chemical reaction the amount of starting
material C cm 3 left after t minutes is given
by C = 40e −0.006 t . Plot a graph of C against
t and determine
(a) the concentration C after 1 hour.
(b) the time taken for the concentration to
decrease by half.
4. The rate at which a body cools is given by
θ = 250e −0.05 t where the excess of temperature of a body above its surroundings at time t
minutes is θ
◦ C. Plot a graph showing the natural decay curve for the first hour of cooling.
Then determine
(a) the temperature after 25 minutes.
(b) the time when the temperature is 195 ◦ C.
16.4 Napierian logarithms
Logarithms having a base of e are called hyperbolic,
Napierian or natural logarithms and the Napierian
logarithm of x is written as log e x or, more commonly,
as ln x. Logarithms were invented by John Napier, a
Scotsman (1550–1617).
The most common method of evaluating a Napierian
logarithm is by a scientific notation calculator. Use your
calculator to check the following values:
ln 4.328 = 1.46510554 ... = 1.4651, correct to 4
decimal places
ln 1.812 = 0.59443, correct to 5 significant figures
ln 1 = 0
ln 527 = 6.2672, correct to 5 significant figures
ln 0.17 = −1.772, correct to 4 significant figures
ln 0.00042 = −7.77526, correct to 6 significant
figures
ln e
3
= 3
ln e 1 = 1
From the last two examples we can conclude that
log e e
x
= x
This is useful when solving equations involving exponential functions. For example, to solve e 3x = 7, take
Napierian logarithms of both sides, which gives
A table of values is drawn up as shown below.
t
0
1
2
3
e
−t /3
1.00
0.7165 0.5134 0.3679
v = 250e −t /3 250.0 179.1
128.4
91.97
t
4
5
6
e −t /3
0.2636
0.1889
0.1353
v = 250e −t /3
65.90
47.22
33.83
The natural decay curve of v = 250e −t /3 is shown in
Figure 16.4.
250
200
150
Voltage
v (volts)
100
80
50
Time t (seconds)
v 5 250e 2t /3
0
1 1.5 2
3 3.4 4
5
6
Figure 16.4
From the graph,
(a) when time t = 3.4 s, voltage v = 80 V, and
(b) when voltage v = 150 V, time t = 1.5 s.
Now try the following Practice Exercise
Practice Exercise 64 Exponential graphs
(answers on page 347)
1. Plot a graph of y = 3e 0.2x over the range
x = −3 to x = 3. Hence determine the value
of y when x = 1.4 and the value of x when
y = 4.5
2. Plot a graph of y =
1
2
e −1.5x over a range
x = −1.5 to x = 1.5 and then determine the
value of y when x = −0.8 and the value of x
when y = 3.5
3. In a chemical reaction the amount of starting
material C cm 3 left after t minutes is given
by C = 40e −0.006 t . Plot a graph of C against
t and determine
(a) the concentration C after 1 hour.
(b) the time taken for the concentration to
decrease by half.
4. The rate at which a body cools is given by
θ = 250e −0.05 t where the excess of temperature of a body above its surroundings at time t
minutes is θ
◦ C. Plot a graph showing the natural decay curve for the first hour of cooling.
Then determine
(a) the temperature after 25 minutes.
(b) the time when the temperature is 195 ◦ C.
16.4 Napierian logarithms
Logarithms having a base of e are called hyperbolic,
Napierian or natural logarithms and the Napierian
logarithm of x is written as log e x or, more commonly,
as ln x. Logarithms were invented by John Napier, a
Scotsman (1550–1617).
The most common method of evaluating a Napierian
logarithm is by a scientific notation calculator. Use your
calculator to check the following values:
ln 4.328 = 1.46510554 ... = 1.4651, correct to 4
decimal places
ln 1.812 = 0.59443, correct to 5 significant figures
ln 1 = 0
ln 527 = 6.2672, correct to 5 significant figures
ln 0.17 = −1.772, correct to 4 significant figures
ln 0.00042 = −7.77526, correct to 6 significant
figures
ln e
3
= 3
ln e 1 = 1
From the last two examples we can conclude that
log e e
x
= x
This is useful when solving equations involving exponential functions. For example, to solve e 3x = 7, take
Napierian logarithms of both sides, which gives