Basic Engineering Mathematics

114 Basic Engineering Mathematics

= log 4 × 3 by the first and second 1
25 log25 − log125 + log625
laws of logarithms
2
= log 12 = log 0.48 3 log5
25 log52 − log 53 + 1 log 54
=2
Problem 16. Write (a) log 30 (b) log 450 in terms
of log 2, log 3 and log 5 to any base 3 log5
2 log5 − 3 log5 + 4 log5
(a) log30 = log(2 × 15) = log(2 × 3 × 5) =2

= log2 + log 3 + log5 3 log5
by the first law
of logarithms = 1 log5 = 1
3 log5 3

Problem 19. Solve the equation
log(x − 1) + log(x + 8) = 2 log(x + 2)

(b) log450 = log(2 × 225) = log(2 × 3 × 75) LHS = log(x − 1) + log(x + 8) = log(x − 1)(x + 8)
from the first
= log(2 × 3 × 3 × 25)
= log(2 × 32 × 52) law of logarithms
= log2 + log32 + log52
= log(x2 + 7x − 8)
by the first law
of logarithms RHS = 2 log(x + 2) = log(x + 2)2
from the first
i.e. log450 = log 2 + 2 log 3 + 2 log 5
by the third law of logarithms law of logarithms

= log(x2 + 4x + 4)

8 × √4 5 Hence, log(x2 + 7x − 8) = log(x2 + 4x + 4)
Problem 17. Write log
in terms of from which, x2 + 7x − 8 = x2 + 4x + 4
81
log 2, log 3 and log 5 to any base i.e. 7x − 8 = 4x + 4

log 8 × √4 5 = log 8 + log √4 5 − log 81 i.e. 3x = 12
81
and x = 4
by the first and second laws
of logarithms Problem 20. Solve the equation 1 log 4 = log x
2

= log 23 1 − log 34 11
log 4 = log 4 2 from the third law of
+ log 5 4
2
by the laws of indices logarithms

8 × √4 5 1 √
i.e. log = 3 log2 + log5 − 4 log3 = log 4 from the laws of indices
81 4
1 log4 = log x
by the third law of logarithms Hence, 2√
becomes log 4 = log x

Problem 18. Evaluate i.e. log 2 = log x
log25 − log125 + 1 log625
2 from which, 2=x
3 log5
i.e. the solution of the equation is x = 2.

Logarithms 115

Problem 21. Solve the equation Simplify the expressions given in problems
log x2 − 3 − log x = log2 12 to 14.
12. log27 − log9 + log81
log x2 − 3 − log x = log x2 − 3 from the second
x 13. log64 + log32 − log128
law of logarithms
14. log8 − log4 + log32

Hence, log x2 − 3 = log2 Evaluate the expressions given in problems 15
x and 16.

from which, x2 −3 1 log16 − 1 log8
Rearranging gives =2 15. 2 3

x log 4
x2 − 3 = 2x
log9 − log3 + 1 log81
and x2 − 2x − 3 = 0 16. 2

2 log3

Factorizing gives (x − 3)(x + 1) = 0

from which, x = 3 or x = −1 Solve the equations given in problems 17 to 22.
17. log x4 − log x3 = log5x − log2x
x = −1 is not a valid solution since the logarithm of a 18. log2t 3 − logt = log16 + logt
negative number has no real root. 19. 2 logb2 − 3 logb = log8b − log4b

Hence, the solution of the equation is x = 3. 20. log(x + 1) + log(x − 1) = log3
1
Now try the following Practice Exercise
21. log 27 = log(0.5a)
Practice Exercise 60 Laws of logarithms 3
(answers on page 346)
22. log(x2 − 5) − log x = log 4
In problems 1 to 11, write as the logarithm of a
single number. 15.3 Indicial equations

1. log2 + log3 2. log3 + log5 The laws of logarithms may be used to solve
certain equations involving powers, called indicial
3. log3 + log4 − log6 equations.

4. log7 + log21 − log49 For example, to solve, say, 3x = 27, logarithms to a base
of 10 are taken of both sides,
5. 2 log2 + log3 6. 2 log2 + 3 log5

7. 2 log5 − 1 log81 + log36 i.e. log10 3x = log10 27
2
and x log10 3 = log10 27
8. 1 log8 − 1 log81 + log27 by the third law of logarithms
32
Rearranging gives x = log10 27 = 1.43136 . . .
9. 1 log4 − 2 log3 + log45 log10 3 0.47712 . . .
2
= 3 which may be readily
10. 1 log16 + 2 log3 − log18
4 checked.

Note, log 27 is not equal to log 27

11. 2 log2 + log5 − log10 log3 3

116 Basic Engineering Mathematics

Problem 22. Solve the equation 2x = 5, correct Thus, x = antilog 0.57040 = 100.57040
to 4 significant figures
= 3.719,
Taking logarithms to base 10 of both sides of 2x = 5 correct to 4 significant figures.
gives
Now try the following Practice Exercise
log10 2x = log10 5
i.e. x log10 2 = log10 5 Practice Exercise 61 Indicial equations
(answers on page 346)
by the third law of logarithms

Rearranging gives x = log10 5 = 0.6989700 ... In problems 1 to 8, solve the indicial equations for
log10 2 0.3010299 ... x, each correct to 4 significant figures.

= 2.322, correct to 4

significant figures. 1. 3x = 6.4 2. 2x = 9

Problem 23. Solve the equation 2x+1 = 32x−5 3. 2x−1 = 32x−1 4. x1.5 = 14.91
correct to 2 decimal places
5. 25.28 = 4.2x 6. 42x−1 = 5x+2

Taking logarithms to base 10 of both sides gives 7. x−0.25 = 0.792 8. 0.027x = 3.26

log10 2x+1 = log10 32x−5 9. The decibel gain n of an amplifier is given
i.e. (x + 1) log10 2 = (2x − 5) log10 3
by n = 10 log10 P2 , where P1 is the power
x log10 2 + log10 2 = 2x log10 3 − 5 log10 3 P1
x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771) input and P2 is the power output. Find the
power gain P2 when n = 25 decibels.
P1

i.e. 0.3010x + 0.3010 = 0.9542x − 2.3855

Hence, 2.3855 + 0.3010 = 0.9542x − 0.3010x 15.4 Graphs of logarithmic functions

2.6865 = 0.6532x A graph of y = log10 x is shown in Figure 15.1 and a
graph of y = loge x is shown in Figure 15.2. Both can
from which x = 2.6865 = 4.11, be seen to be of similar shape; in fact, the same general
0.6532 shape occurs for a logarithm to any base.
In general, with a logarithm to any base, a, it is noted
correct to 2 decimal places. that

Problem 24. Solve the equation x2.7 = 34.68, (a) loga 1 = 0
correct to 4 significant figures Let loga = x then ax = 1 from the definition of the
logarithm.
Taking logarithms to base 10 of both sides gives If ax = 1 then x = 0 from the laws of logarithms.

log10 x2.7 = log10 34.68 Hence, loga 1 = 0. In the above graphs it is seen
that log101 = 0 and loge 1 = 0.
2.7 log10 x = log10 34.68
(b) loga a = 1
Hence, log10 x = log10 34.68 = 0.57040 Let loga a = x then ax = a from the definition of
2.7 a logarithm.
If ax = a then x = 1.

Logarithms 117

y y
0.5 2

0 1 2 3x 1
20.5
x 3 2 1 0.5 0.2 0.1 0 1 2 3 4 5 6x
y 5 log10x 0.48 0.30 0 2 0.30 2 0.70 2 1.0 21

21.0 x 6 5 4 3 2 1 0.5 0.2 0.1
Figure 15.1 y 5 logex 1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30

22

Figure 15.2

Hence, loga a = 1. (Check with a calculator that If ax = 0, and a is a positive real number, then
log10 10 = 1 and loge e = 1.)
x must approach minus infinity. (For example,
(c) loga 0 → −∞ check with a calculator, 2−2 = 0.25, 2−20 =
Let loga 0 = x then ax = 0 from the definition of a 9.54 × 10−7, 2−200 = 6.22 × 10−61, and so on.)
logarithm. Hence, loga 0 → −∞

Chapter 16

Exponential functions

16.1 Introduction to exponential e0.25 − e−0.25
functions 5 e0.25 + e−0.25

An exponential function is one which contains ex , e =5 1.28402541 . . . − 0.77880078 . . .
being a constant called the exponent and having an 1.28402541 . . . + 0.77880078 . . .
approximate value of 2.7183. The exponent arises from
the natural laws of growth and decay and is used as a =5 0.5052246 . . .
base for natural or Napierian logarithms. 2.0628262 . . .
The most common method of evaluating an exponential
function is by using a scientific notation calculator. Use = 1.2246, correct to 4 decimal places.
your calculator to check the following values.
Problem 3. The instantaneous voltage v in a
e1 = 2.7182818, correct to 8 significant figures,
e−1.618 = 0.1982949, correct to 7 significant figures, capacitive circuit is related to time t by the equation
v = V e−t/CR where V , C and R are constants.
e0.12 = 1.1275, correct to 5 significant figures, Determine v, correct to 4 significant figures,
e−1.47 = 0.22993, correct to 5 decimal places, when t = 50 ms, C = 10 μF, R = 47 k and
e−0.431 = 0.6499, correct to 4 decimal places, V = 300 volts

e9.32 = 11159, correct to 5 significant figures, v = V e−t/CR = 300e(−50×10−3)/(10×10−6×47×103)
e−2.785 = 0.0617291, correct to 7 decimal places. Using a calculator, v = 300e−0.1063829...

Problem 1. Evaluate the following correct to 4 = 300(0.89908025 . . .)
decimal places, using a calculator: = 269.7 volts.

0.0256(e5.21 − e2.49) Now try the following Practice Exercise

0.0256(e5.21 − e2.49) Practice Exercise 62 Evaluating
= 0.0256(183.094058 . . . − 12.0612761 . . .) exponential functions (answers on page 347)
= 4.3784, correct to 4 decimal places.
1. Evaluate the following, correct to 4 significant
Problem 2. Evaluate the following correct to 4 figures.
decimal places, using a calculator: (a) e−1.8 (b) e−0.78 (c) e10

e0.25 − e−0.25 2. Evaluate the following, correct to 5 significant
5 e0.25 + e−0.25 figures.
(a) e1.629 (b) e−2.7483 (c) 0.62e4.178

DOI: 10.1016/B978-1-85617-697-2.00016-8

Exponential functions 119

In problems 3 and 4, evaluate correct to 5 decimal = 2.71828
places.
i.e. e = 2.7183, correct to 4 decimal places.
3. (a) 1 e3.4629 (b) 8.52e−1.2651
7 The value of e0.05, correct to say 8 significant figures, is
5e2.6921 found by substituting x = 0.05 in the power series for
ex . Thus,
(c) 3e1.1171
e0.05 = 1 + 0.05 + (0.05)2 + (0.05)3 + (0.05)4
4. (a) 5.6823 e2.1127 − e−2.1127 2! 3! 4!
e−2.1347 (b) (0.05)5
+ +···
2 5!

4 e−1.7295 − 1 = 1 + 0.05 + 0.00125 + 0.000020833
(c) e3.6817
+ 0.000000260 + 0.0000000026
5. The length of a bar, l, at a temperature, θ,
is given by l = l0eαθ , where l0 and α are i.e. e 0.05 = 1.0512711, correct to 8 significant figures.

constants. Evaluate l, correct to 4 significant In this example, successive terms in the series grow
smaller very rapidly and it is relatively easy to deter-
figures, where l0 = 2.587, θ = 321.7 and mine the value of e0.05 to a high degree of accuracy.
α = 1.771 × 10−4. However, when x is nearer to unity or larger than unity,
a very large number of terms are required for an accurate
6. When a chain of length 2L is suspended from result.
two points, 2D metres apart on√the same hor- If, in the series of equation (1), x is replaced by −x,
izontal level, D = k ln L + L2 + k2 . then
k
Evaluate D when k = 75 m and L = 180 m. e−x = 1 + (−x) + (−x)2 + (−x )3 + · · ·
2! 3!
16.2 The power series for ex
i.e. e−x = 1 − x + x 2 − x 3 + · · ·
The value of ex can be calculated to any required degree 2! 3!
of accuracy since it is defined in terms of the following
power series: In a similar manner the power series for ex may
be used to evaluate any exponential function of the form
ex = 1 + x+ x 2 + x 3 + x 4 + ··· (1) aekx , where a and k are constants.
2! 3! 4! In the series of equation (1), let x be replaced by kx.
Then
(where 3! = 3 × 2 × 1 and is called ‘factorial 3’).
The series is valid for all values of x. aekx = a 1 + (kx ) + (kx )2 + (kx )3 + · · ·
2! 3!
The series is said to converge; i.e., if all the terms are
added, an actual value for ex (where x is a real number) Thus, 5e2x = 5 1 + (2x ) + (2x)2 + (2x)3 + · · ·
2! 3!
is obtained. The more terms that are taken, the closer
will be the value of ex to its actual value. The value of 4x2 8x3
= 5 1 + 2x + + + · · ·
the exponent e, correct to say 4 decimal places, may be
determined by substituting x = 1 in the power series of 26
equation (1). Thus, i.e. 5e2x = 5 1 + 2x + 2x2 + 4 x3 + · · ·

e1 = 1 + 1 + (1)2 + (1)3 + (1)4 + (1)5 + (1)6 3
2! 3! 4! 5! 6!
+ (1)7 + (1)8 + · · · Problem 4. Determine the value of 5e0.5, correct
7! 8! to 5 significant figures, by using the power series
for ex
= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833
+ 0.00139 + 0.00020 + 0.00002 + · · ·

120 Basic Engineering Mathematics

From equation (1), Hence,
ex = 1 + x + x2 + x3 + x4 + ···
2! 3! 4! ex (x2 − 1)

(0.5)2 (0.5)3 = x2 x3 x4 x5 (x2 − 1)
(2)(1) (3)(2)(1) 1+x + + + + +···
2! 3! 4! 5!
Hence, e0.5 = 1 + 0.5 + +

i.e. (0.5)4 (0.5)5 = x2 + x3 + x4 + x5 +···
Hence, + (4)(3)(2)(1) + (5)(4)(3)(2)(1) 2! 3!

+ (0.5)6 − 1+ x + x2 + x3 + x4 + x5 +···
(6)(5)(4)(3)(2)(1) 2! 3! 4! 5!

= 1 + 0.5 + 0.125 + 0.020833

+ 0.0026042 + 0.0002604 Grouping like terms gives

+ 0.0000217 ex (x2 − 1)

e0.5 = 1.64872, correct to 6 significant = −1 − x + x2 − x2 + x3 − x3
figures 2! 3!

5e0.5 = 5(1.64872) = 8.2436, correct to 5 + x4 − x4 + x5 − x5 +···
significant figures.

2! 4! 3! 5!

Problem 5. Determine the value of 3e−1, correct = −1 − x + 1 x2+ 5 x3+ 11 x4+ 19 x5
to 4 decimal places, using the power series for ex 2 6 24 120

Substituting x = −1 in the power series when expanded as far as the term in x5.

gives ex = 1 + x + x2 + x3 + x4 + ··· Now try the following Practice Exercise
2! 3! 4!
Practice Exercise 63 Power series for ex
e−1 = 1 + (−1) + (−1)2 + (−1)3 (answers on page 347)
2! 3!
(−1)4 1. Evaluate 5.6e−1, correct to 4 decimal places,
+ +··· using the power series for ex .
4!
2. Use the power series for ex to determine, cor-
= 1 − 1 + 0.5 − 0.166667 + 0.041667 rect to 4 significant figures, (a) e2 (b) e−0.3
and check your results using a calculator.
− 0.008333 + 0.001389
3. Expand (1 − 2x)e2x as far as the term in x4.
− 0.000198 + · · · 4. Expand (2ex2 )(x1/2) to six terms.

= 0.367858 correct to 6 decimal places

Hence, 3e−1 = (3)(0.367858) = 1.1036, correct to 4
decimal places.

Problem 6. Expand ex (x2 − 1) as far as the term 16.3 Graphs of exponential functions
in x5
Values of ex and e−x obtained from a calculator, correct
The power series for ex is to 2 decimal places, over a range x = −3 to x = 3, are
shown in Table 16.1.
ex = 1 + x + x2 + x3 + x4 + x5 + ···
2! 3! 4! 5!

Exponential functions 121

Table 16.1

x −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 0.5 1.0 1.5 2.0 2.5 3.0

ex 0.05 0.08 0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48 7.39 12.18 20.09

e−x 20.09 12.18 7.39 4.48 2.72 1.65 1.00 0.61 0.37 0.22 0.14 0.08 0.05

Figure 16.1 shows graphs of y = ex and y = e−x . y y ϭ 2e 0.3x
5
y 3.87 4
20 3

y ϭ e Ϫx y ϭ ex
16

12 2
1.6

1

8

Ϫ3 Ϫ2 Ϫ1 0 1 2 3x
Ϫ0.74 2.2

4

Ϫ3 Ϫ2 Ϫ1 0 1 2 3 x Figure 16.2

Figure 16.1 A graph of 1 e−2x is shown in Figure 16.3.
3
Problem 7. Plot a graph of y = 2e0.3x over a
range of x = −2 to x = 3. Then determine the value From the graph, when x = −1.2, y = 3.67 and when
of y when x = 2.2 and the value of x when y = 1.6 y = 1.4, x = −0.72

A table of values is drawn up as shown below. y ϭ 1 e Ϫ2x y
3 7
x −3 −2 −1 0 1 2 3
2e0.3 x 0.81 1.10 1.48 2.00 2.70 3.64 4.92 6

A graph of y = 2e0.3x is shown plotted in Figure 16.2. 5
From the graph, when x = 2.2, y = 3.87 and when
y = 1.6, x = −0.74 4
3.67
Problem 8. Plot a graph of y = 1 e−2x over the
3 3

range x = −1.5 to x = 1.5. Determine from the 2
graph the value of y when x = −1.2 and the value 1.4
of x when y = 1.4
1
A table of values is drawn up as shown below.
Ϫ1.5 Ϫ1.0 Ϫ0.5 0.5 1.0 1.5 x
x −1.5 −1.0 −0.5 0 0.5 1.0 1.5 Ϫ1.2 Ϫ0.72
1 e−2x 6.70 2.46 0.91 0.33 0.12 0.05 0.02
3 Figure 16.3

Problem 9. The decay of voltage, v volts, across

a capacitor at time t seconds is given by
v = 250e−t/3. Draw a graph showing the natural

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

122 Basic Engineering Mathematics

A table of values is drawn up as shown below.

t 01 2 3 2. Plot a graph of y = 1 e−1.5x over a range
2
e−t /3 1.00 0.7165 0.5134 0.3679
x = −1.5 to x = 1.5 and then determine the
v = 250e−t/3 250.0 179.1 128.4 91.97 value of y when x = −0.8 and the value of x
when y = 3.5
t 4 5 6
e−t /3 0.2636 0.1889 0.1353 3. In a chemical reaction the amount of starting
v = 250e−t/3 65.90 47.22 33.83 material C cm3 left after t minutes is given
by C = 40e−0.006 t. Plot a graph of C against
The natural decay curve of v = 250e−t/3 is shown in t and determine
Figure 16.4.
(a) the concentration C after 1 hour.
Voltage v (volts) 250 v 5 250e 2t /3
(b) the time taken for the concentration to
200 decrease by half.

150 4. The rate at which a body cools is given by
θ = 250e−0.05 t where the excess of tempera-
100 ture of a body above its surroundings at time t
80 minutes is θ ◦C. Plot a graph showing the nat-
50 ural 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

0 1 1.5 2 3 3.4 4 5 6 Logarithms having a base of e are called hyperbolic,
Figure 16.4 Time t (seconds) Napierian or natural logarithms and the Napierian
logarithm of x is written as loge 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:

From the graph, ln 4.328 = 1.46510554 . . . = 1.4651, correct to 4
(a) when time t = 3.4 s, voltage v = 80 V, and
(b) when voltage v = 150 V, time t = 1.5 s. 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

Now try the following Practice Exercise ln 0.00042 = −7.77526, correct to 6 significant

Practice Exercise 64 Exponential graphs figures
(answers on page 347)
ln e3 = 3
1. Plot a graph of y = 3e0.2x over the range ln e1 = 1
x = −3 to x = 3. Hence determine the value
of y when x = 1.4 and the value of x when From the last two examples we can conclude that
y = 4.5
loge ex = x

This is useful when solving equations involving expo-
nential functions. For example, to solve e3x = 7, take

Napierian logarithms of both sides, which gives

Exponential functions 123

ln e3x = ln 7 Since loge eα = α, then ln 4 = 3x
9
i.e. 3x = ln 7
from which x = 1 ln 7 = 0.6486, Hence, x = 1 ln 4 = 1 (−0.81093) = −0.2703,
39 3
3
correct to 4 decimal places. correct to 4 significant figures.

Problem 13. Given 32 = 70 1 − e− t ,
2

Problem 10. Evaluate the following, each correct determine the value of t , correct to 3 significant
1
figures
to 5 significant figures: (a) ln 4.7291
2 Rearranging 32 = 70 1 − e− t gives
2
ln 7.8693 3.17 ln 24.07
(b) 7.8693 (c) e−0.1762

(a) 1 ln 4.7291 = 1 (1.5537349 . . .) = 0.77687, 32 = 1 − e− t
22 2
correct to 5 significant figures. 70

and

(b) ln 7.8693 = 2.06296911 . . . = 0.26215, correct e− t =1− 32 = 38
7.8693 7.8693 2
70 70
to 5 significant figures.

3.17 ln24.07 3.17(3.18096625 . . .) Taking the reciprocal of both sides gives
e−0.1762 0.83845027 . . .
(c) = = 12.027, t 70

correct to 5 significant figures. e2 = 38

ln e2.5 Taking Napierian logarithms of both sides gives
Problem 11. Evaluate the following: (a) lg 100.5
t 70
5e2.23 lg 2.23
(b) ln 2.23 (correct to 3 decimal places) ln e 2 = ln 38

ln e2.5 2.5 i.e. t = ln 70
(a) lg 100.5 = 0.5 = 5 2 38

5e2.23 lg 2.23 from which, t = 2 ln 70 = 1.22, correct to 3 signifi-
(b) ln 2.23 38

= 5(9.29986607 . . .)(0.34830486 . . .) cant figures.
(0.80200158 . . .)
Problem 14. Solve the equation
= 20.194, correct to 3 decimal places. 2.68 = ln 4.87 to find x
x

Problem 12. Solve the equation 9 = 4e−3x to find From the definition of a logarithm, since
x, correct to 4 significant figures 2.68 = ln 4.87 then e2.68 = 4.87

Rearranging 9 = 4e−3x gives 9 = e−3x xx
4
Rearranging gives x = 4.87 = 4.87e−2.68
Taking the reciprocal of both sides gives i.e. e2.68

x = 0.3339,

4 = 1 = e3x correct to 4 significant figures.
9 e−3x

Taking Napierian logarithms of both sides gives Problem 15. Solve 7 = e3x correct to 4
4
ln 4 = ln(e3x )
9 significant figures