Exponential functions 31
decay curve over the first 6seconds. From the
graph, find (a) the voltage after 3.4s, and (b) the
time when the voltage is 150V.
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 ofv=250e
−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
(^23)
Figure 4.4
From the graph:
(a) when timet=3.4s, voltagev=80Vand
(b) when voltagev=150V, timet=1.5s.
Now try the following exercise
Exercise 16 Further problemson
exponential graphs
- Plot a graph of y=3e^0.^2 x over the range
x=−3tox=3. Hence determine the value
ofywhenx= 1 .4 and the value ofxwhen
y= 4 .5. [3.95, 2.05]
- Plot a graph of y=^12 e−^1.^5 x over a range
x=− 1 .5tox= 1 .5 and hence determine the
value ofywhenx=− 0 .8 and the value ofx
wheny= 3 .5. [1.65,−1.30] - In a chemical reaction the amount of starting
materialCcm^3 left aftertminutes is given by
C=40e−^0.^006 t.PlotagraphofCagainsttand
determine (a) the concentrationCafter 1 hour,
and (b) the time taken for the concentration to
decrease by half.
[(a) 28cm^3 (b) 116min] - The rate at which a body cools is given by
θ=250e−^0.^05 twhere the excess of tempera-
ture of a body above its surroundings at
timetminutes isθ◦C. Plot a graph showing
the natural decay curve for the firsthour of
cooling. Hence determine (a) the temperature
after 25 minutes, and (b) the time when the
temperature is 195◦C.
[(a) 70◦C(b)5min]
4.4 Napierian logarithms
Logarithms having a base of ‘e’ are calledhyperbolic,
Napierianornatural logarithmsand the Napierian
logarithm ofxis written as logex, or more commonly
as lnx. Logarithms were invented by John Napier, a
Scotsman (1550–1617).
The most common method of evaluating a Napierian
logarithmisbyascientificnotationcalculator.Useyour
calculator to check the following values:
ln4. 328 = 1. 46510554 ...
= 1. 4651 ,correct to 4 decimal places
ln1. 812 = 0. 59443 ,correct to 5 significant figures
ln1= 0
ln527= 6. 2672 ,correct to 5 significant figures
ln0. 17 =− 1. 772 ,correct to 4 significant figures
ln0. 00042 =− 7. 77526 ,correct to 6 significant
figures
lne^3 = 3
lne^1 = 1