All questions have only one correct answer (answers on page 1158). 1. The relationship between the temperature in
degrees Fahrenheit (F) and the temperature in degrees Celsius (C) is given by: F=9
5C+32 135◦F is equivalent to:
(a) 43◦C (b) 57.2◦C (c) 185.4◦C (d) 184◦C 2. TransposingI=V
R for resistance Rgives:
(a) I - V (b) V
I (c) I
V (d) VI
3. When two resistorsR1andR2are connected in par- allel the formula 1
RT = 1 R1+ 1 R2 is used to deter- mine the total resistance RT. If R1=470and
R2=2.7 k,RT (correct to 3 significant figures) is equal to:
(a) 2.68 (b) 400 (c) 473 (d) 3170 4. Transposing v= fλ to make wavelength λ the
subject gives: (a) v
f (b)v+ f (c) f −v (d) f
v
5. Transposing the formula R=R0(1+αt) for t gives: (a) R−R0 (1+α) (b) R−R0−1 α (c) R−R0 αR0 (d) R R0α
6. The current I in an a.c. circuit is given by:
I=√ V R2+X2
WhenR=4.8, X=10.5 andI=15, the value of voltageV is:
(a) 173.18 (b) 1.30 (c) 0.98 (d) 229.50 7. The heightsof a mass projected vertically upwards
at time t is given by: s=ut−1
2gt
2. When
g=10,t=1.5 ands=3.75, the value ofuis: (a) 10 (b)−5 (c)+5 (d)−10
8. The quantity of heat Q is given by the formula
Q=mc(t2−t1). Whenm=5,t1=20,c=8 and
Q=1200, the value oft2is:
(a) 10 (b) 1.5 (c) 21.5 (d) 50 9. Electrical resistance R=ρ
a ; transposing this
equation forgives: (a)ρa R (b) R aρ (c) a Rρ (d) Ra ρ
10. The solution of the simultaneous equations 3x−2y=13 and 2x+5y= −4 is:
(a) x= −2, y=3 (b) x=1, y= −5 (c) x=3, y= −2 (d) x= −7, y=2 11. A formula for the focal length f of a convex lens
is 1
f =
1
u+
1
v When f =4 andu=6, vis:
(a)−2 (b) 12 (c) 1
12 (d)−
1 2 12. Volume= mass
density. The density (in kg/m 3) when the mass is 2.532 kg and the volume is 162 cm3is: (a) 0.01563 kg/m3 (b) 410.2 kg/m3
(c) 15 630 kg/m3 (d) 64.0 kg/m3
13. P V =m RT is the characteristic gas equation. When P=100×103, V =4.0, R=288 and
T =300, the value ofmis:
(a) 4.630 (b) 313 600 (c) 0.216 (d) 100 592 14. The quadratic equation inxwhose roots are−2 and
+5 is:
(a) x2−3x−10=0 (b) x2+7x+10=0 (c) x2+3x−10=0 (d) x2−7x−10=0 15. The area A of a triangular piece of land of
sides a, b and c may be calculated using
A=√[s(s−a)(s−b)(s−c)] wheres=a+b+c
Section
A
Whena=15 m,b=11 m andc=8 m, the area, correct to the nearest square metre, is:
(a) 1836 m2 (b) 648 m2 (c) 445 m2 (d) 43 m2 16. In a system of pulleys, the effortPrequired to raise
a loadW is given byP=aW+b, whereaandb
are constants. IfW =40 whenP=12 andW =90 whenP=22, the values ofaandbare:
(a) a=5,b=1 4 (b) a=1,b= −28 (c) a=1 3, b= −8 (d) a=1 5, b=4
17. Resistance R ohms varies with temperature t
according to the formula R=R0(1+αt). Given
R=21 , α=0.004 and t=100, R0 has a value of: (a) 21.4 (b) 29.4 (c) 15 (d) 0.067 18. 8x2+13x−6=(x+p)(q x−3). The values ofp andq are: (a) p= −2,q=4 (b) p=3,q=2 (c) p=2, q=8 (d) p=1,q=8
19. The height S metres of a mass thrown verti- cally upwards at time t seconds is given by
S=80t−16t2. To reach a height of 50 metres on the descent will take the mass:
(a) 0.73 s (b) 5.56 s (c) 4.27 s (d) 81.77 s 20. The final lengthl2of a piece of wire heated through
θ◦C is given by the formulal
2=l1(1+αθ). Trans- posing, the coefficient of expansionαis given by: (a)l2 l1− 1 θ (b) l2−l1 l1θ (c)l2−l1−l1θ (d) l1−l2 l1θ
21. The roots of the quadratic equation 8x2+10x−3=0 are: (a)−1 4 and 3 2 (b) 4 and 2 3 (c)−3 2 and 1 4 (d) 2 3 and−4
22. The volumeV2of a material when the temperature is increased is given by V2=V1
1+γ (t2−t1)
. The value oft2whenV2=61.5 cm3,V1=60 cm3,
γ=54×10−6andt1=250 is:
(a) 213 (b) 463 (c) 713 (d) 28 028
23. Current I in an electrical circuit is given by
I= E−e
R+r. Transposing forRgives:
(a) E−e−I r I (b) E−e I+r (c)(E−e)(I+r) (d) E−e I r
24. The roots of the quadratic equation
2x2−5x+1=0, correct to 2 decimal places, are: (a) −0.22 and−2.28 (b) 2.69 and−0.19 (c) 0.19 and−2.69 (d) 2.28 and 0.22 25. Transposingt=2π l g forggives: (a)(t−2π) 2 l (b) 2π t l2 (c) t 2π l (d) 4π2l t2
For a copy of this multiple choice test, go to:
Chapter 15
Logarithms
Why it is important to understand:Logarithms
All types of engineers use natural and common logarithms. Chemical engineers use them to measure radioactive decay and pH solutions, both of which are measured on a logarithmic scale. The Richter scale which measures earthquake intensity is a logarithmic scale. Biomedical engineers use logarithms to mea- sure cell decay and growth, and also to measure light intensity for bone mineral density measurements. In electrical engineering, a dB (decibel) scale is very useful for expressing attenuations in radio propagation and circuit gains, and logarithms are used for implementing arithmetic operations in digital circuits. Logarithms are especially useful when dealing with the graphical analysis of non-linear relationships and logarithmic scales are used to linearise data to make data analysis simpler. Understanding and using logarithms is clearly important in all branches of engineering.
At the end of this chapter, you should be able to:
• define base, power, exponent and index • define a logarithm
• distinguish between common and Napierian (i.e. hyperbolic or natural) logarithms • evaluate logarithms to any base
• state the laws of logarithms • simplify logarithmic expressions • solve equations involving logarithms • solve indicial equations
• sketch graphs of log10xand logex
15.1
Introduction to logarithms
With the use of calculators firmly established, logarith- mic tables are now rarely used for calculation. However, the theory of logarithms is important, for there are sev- eral scientific and engineering laws that involve the rules of logarithms.
From Chapter 7, we know that 16=24
The number 4 is called thepoweror theexponentor theindex. In the expression 24, the number 2 is called thebase.
In another example, we know that 64=82. In this example, 2 is the power, or exponent, or index. The number 8 is the base.
15.1.1 What is a logarithm?
Consider the expression 16=24An alternative, yet equivalent, way of writing this expression is log216=4
This is stated as ‘log to the base 2 of 16 equals 4’ We see that the logarithm is the same as the power or index in the original expression. It is the base in Understanding Engineering Mathematics. 978-0-415-66284-0, © 2014 John Bird. Published by Taylor & Francis. All rights reserved.
Section
A
the original expression that becomes the base of the logarithm.
The two statements 16=24and
log216=4 are equivalent
If we write either of them, we are automatically imply- ing the other.
In general, if a numberycan be written in the formax, then the indexxis called the ‘logarithm ofyto the base ofa’, i.e.
if y=axthenx=logay
In another example, if we write down that 64=82then the equivalent statement using logarithms is log864=2. In another example, if we write down that log327=3 then the equivalent statement using powers is 33=27. So the two sets of statements, one involving powers and one involving logarithms, are equivalent.
15.1.2 Common logarithms
From the above, if we write down that 1000=103, then 3=log101000. This may be checked using the ‘log’ button on your calculator.
Logarithms having a base of 10 are called common logarithmsand log10is usually abbreviated to lg. The following values may be checked using a calculator.
lg 27.5=1.4393... lg 378.1=2.5776... lg 0.0204= −1.6903...
15.1.3 Napierian logarithms
Logarithms having a base ofe(whereeis a mathemat- ical constant approximately equal to 2.7183) are called
hyperbolic, Napierian or natural logarithms, and logeis usually abbreviated to ln. The following values may be checked using a calculator.
ln 3.65=1.2947... ln 417.3=6.0338... ln 0.182= −1.7037...
Napierian logarithms are explained further in Chapter 16.
Here are some worked problems to help understanding of logarithms.
Problem 1. Evaluate log39
Letx=log39 then 3x=9 from the definition of a logarithm,
i.e. 3x=32, from whichx=2
Hence, log39=2
Problem 2. Evaluate log1010
Letx=log1010 then 10x=10 from the definition of a logarithm,
i.e. 10x=101, from whichx=1
Hence, log1010=1(which may be checked using a calculator).
Problem 3. Evaluate log168
Letx=log168 then 16x=8 from the definition of a logarithm, i.e. (24)x=23 i.e. 24x=23from the laws
of indices,
from which, 4x=3 andx=3
4 Hence, log168= 3 4 Problem 4. Evaluate lg 0.001 Letx=lg 0.001=log100.001 then 10x=0.001
i.e. 10x=10−3from which,x= −3 Hence, lg 0.001= −3 (which may be checked
using a calculator)
Problem 5. Evaluate lne
Letx=lne=logee then ex=e
i.e. ex=e1,
from which
x=1
Hence, lne=1(which may be checked
Section
A
Problem 6. Evaluate log3 1 81 Letx=log3 1 81 then 3 x= 1 81 = 1 34=3− 4 from whichx= −4 Hence, log3 1 81= −4
Problem 7. Solve the equation lgx=3 If lgx=3 then log10x=3
and x=103 i.e.x=1000
Problem 8. Solve the equation log2x=5 If log2x=5 then x=25=32
Problem 9. Solve the equation log5x= −2 If log5x= −2 then x=5−2=
1 52=
1 25 Now try the following Practice Exercise
Practice Exercise 61 Laws of logarithms (answers on page 1114)
In Problems 1 to 11, evaluate the given expressions. 1. log1010000 2. log216 3. log5125
4. log2 1
8 5. log82 6. log7343
7. lg 100 8. lg 0.01 9. log48
10. log273 11. ln e2
In Problems 12 to 18, solve the equations.
12. log10x=4 13. lgx=5 14. log3x=2 15. log4x= −2 1 2 16. lgx= −2 17. log8x= −4 3 18. lnx=3
15.2
Laws of logarithms
There are three laws of logarithms, which apply to any base:
(1) To multiply two numbers:
log(A×B)=log A+log B
The following may be checked by using a calculator. lg 10=1
Also, lg 5+lg 2=0.69897...+0.301029...=1 Hence, lg(5×2)=lg 10=lg 5+lg 2
(2) To divide two numbers:
log A B =log A−log B
The following may be checked using a calculator.
ln 5 2 =ln 2.5=0.91629... Also, ln 5−ln 2=1.60943...−0.69314... =0.91629... Hence, ln 5 2 =ln 5−ln 2
(3) To raise a number to a power:
log An=nlog A
The following may be checked using a calculator. lg 52=lg 25=1.39794...
Also, 2 lg 5=2×0.69897...=1.39794... Hence, lg 52=2 lg 5
Here are some worked problems to help understanding of the laws of logarithms.
Problem 10. Write log 4+log 7 as the logarithm of a single number
log 4+log 7=log(7×4) by the first law of logarithms
Section
A
Problem 11. Write log 16−log2 as the logarithm of a single number
log 16−log2=log
16 2
by the second law of logarithms
=log 8
Problem 12. Write 2 log 3 as the logarithm of a single number
2 log 3=log 32 by the third law of logarithms
=log 9
Problem 13. Write1
2log 25 as the logarithm of a single number
1
2log 25=log 25 1
2 by the third law of logarithms
=log√25=log 5
Problem 14. Simplify log 64−log 128 + log 32 64=26,128=27and 32=25
Hence, log 64−log 128+log32
=log 26−log 27+log 25
=6 log 2−7 log2+5 log 2 by the third law of logarithms
=4 log 2
Problem 15. Write1
2log 16+ 1
3log 27−2 log5 as the logarithm of a single number
1 2log 16+ 1 3 log 27−2 log5 =log 16 1 2+log 27 1 3−log 52
by the third law of logarithms
=log√16+log√327−log 25
by the laws of indices = log 4+log 3−log 25
=log
4×3
25
by the first and second laws of logarithms =log 12 25 =log 0.48
Problem 16. Write (a) log30 (b) log450 in terms of log2, log3 and log5 to any base
(a) log 30=log(2×15)=log(2×3×5)
=log 2+log 3+log 5
by the first law of logarithms
(b) log 450=log(2×225)=log(2×3×75)
=log(2×3×3×25)
=log(2×32×52)
=log 2+log 32+log 52 by the first law of logarithms
i.e. log 450=log 2+2 log 3+2 log 5
by the third law of logarithms
Problem 17. Write log
8×√45 81
in terms of log 2, log 3 and log 5 to any base
log
8×√45 81
=log 8+log√45−log 81 by the first and second laws of logarithms
=log 23+log 5 1
4−log 34 by the laws of indices
i.e. log 8×√45 81 =3 log 2+1 4log 5−4 log 3
by the third law of logarithms
Problem 18. Evaluate
log 25−log125+1 2log 625 3 log 5