Exponential functions 31
decay curve over the first 6 seconds. From the
graph, find (a) the voltage after 3.4 s, and (b) the
time when the voltage is 150 V.
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 = 250 e
−t
3
250.0 179.1 128.4 91.97
t
4
5
6
e
−t
3
0.2636
0.1889
0.1353
v = 250 e
−t
3
65.90
47.22
33.83
The natural decay curve of v = 250 e
−t
3 is shown in
Fig. 4.4.
250
200
Voltage v (volts)
Time t(seconds)
150
100
80
50
0
1 1.5 2
3 3.4 4
5
6
y 5 250e
t
3
2
Figure 4.4
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 exercise
Exercise 16 Further problems on
exponential graphs
1. Plot a graph of y = 3 e 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.
[3.95, 2.05]
2. Plot a graph of y =
1
2 e −1.5x over a range
x =−1.5 to x = 1.5 and hence determine the
value of y when x =−0.8 and the value of x
when y = 3.5.
[1.65, −1.30]
3. In a chemical reaction the amount of starting
material C cm 3 left after t minutes is given by
C = 40 e −0.006t . Plot a graph of C against t and
determine (a) the concentration C after 1 hour,
and (b) the time taken for the concentration to
decrease by half.
[(a) 28 cm 3 (b) 116 min]
4. The rate at which a body cools is given by
θ = 250 e
−0.05t 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. Hence determine (a) the temperature
after 25 minutes, and (b) the time when the
temperature is 195
◦ C.
[(a) 70 ◦ C (b) 5 min]
4.4 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
decay curve over the first 6 seconds. From the
graph, find (a) the voltage after 3.4 s, and (b) the
time when the voltage is 150 V.
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 = 250 e
−t
3
250.0 179.1 128.4 91.97
t
4
5
6
e
−t
3
0.2636
0.1889
0.1353
v = 250 e
−t
3
65.90
47.22
33.83
The natural decay curve of v = 250 e
−t
3 is shown in
Fig. 4.4.
250
200
Voltage v (volts)
Time t(seconds)
150
100
80
50
0
1 1.5 2
3 3.4 4
5
6
y 5 250e
t
3
2
Figure 4.4
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 exercise
Exercise 16 Further problems on
exponential graphs
1. Plot a graph of y = 3 e 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.
[3.95, 2.05]
2. Plot a graph of y =
1
2 e −1.5x over a range
x =−1.5 to x = 1.5 and hence determine the
value of y when x =−0.8 and the value of x
when y = 3.5.
[1.65, −1.30]
3. In a chemical reaction the amount of starting
material C cm 3 left after t minutes is given by
C = 40 e −0.006t . Plot a graph of C against t and
determine (a) the concentration C after 1 hour,
and (b) the time taken for the concentration to
decrease by half.
[(a) 28 cm 3 (b) 116 min]
4. The rate at which a body cools is given by
θ = 250 e
−0.05t 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. Hence determine (a) the temperature
after 25 minutes, and (b) the time when the
temperature is 195
◦ C.
[(a) 70 ◦ C (b) 5 min]
4.4 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