Basic Engineering Mathematics, Fifth Edition

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= 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 ofv= 250 e−t/^3 is shown in

Figure 16.4.

250

200

150

Voltage

v (volts)

100

80

50

Time t (seconds)

v 5 250 e^2 t/3

0 1 1.5 2 3 3.4 4 5 6

Figure 16.4

From the graph,

(a) when timet=3.4s,voltagev=80V,and

(b) whenvoltagev=150V,timet=1.5s.

Now try the following Practice Exercise

PracticeExercise 64 Exponential graphs

(answers on page 347)

1. Plot a graph ofy= 3 e^0.^2 x over the range
   x=−3tox=3. Hence determine the value
   ofywhenx= 1 .4 and the value ofxwhen
   y= 4. 5


2. Plot a graph ofy=

1

2

e−^1.^5 x over a range

x=− 1 .5tox= 1 .5 and then determine the

value ofywhenx=− 0 .8 and the value ofx

wheny= 3. 5

3. In a chemical reaction the amount of starting
   materialCcm^3 left aftertminutes is given
   byC= 40 e−^0.^006 t.PlotagraphofCagainst
   tand determine
   (a) the concentrationCafter 1hour.
   (b) the time taken for the concentration to
   decrease by half.


4. The rate at which a body cools is given by
   θ= 250 e−^0.^05 twhere the excess of tempera-
   ture of a body above its surroundings at timet
   minutes isθ◦C. Plota graph showingthe nat-
   ural decay curve for the firsthour 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 ofeare calledhyperbolic,

Napierianornatural logarithmsand the Napierian

logarithm ofxis written as logexor, 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:

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

lne^3 = 3

lne^1 = 1

From the last two examples we can conclude that

logeex=x

This is useful whensolving equations involving expo-

nential functions. For example, to solvee^3 x=7, take

Napierian logarithms of both sides, which gives