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EDEXCEL NATIONAL CERTIFICATE
UNIT 4 – MATHEMATICS FOR TECHNICIANS OUTCOME 1
TUTORIAL 5 - INDICES, LOGARITHMS AND FUNCTION
Determine the fundamental algebraic laws and apply algebraic manipulation techniques to the solution of problems involving algebraic functions, formulae and graphs CONTENTS Algebraic laws and algebraic manipulation techniques Algebraic laws and manipulation: introduction to algebraic expressions and equations, algebraic operations, rules for manipulation and transposition of formulae; manipulation and algebraic solution of algebraic expressions; direct and inverse proportion and constants of proportionality; linear, simultaneous and quadratic equations; graphical solution of simple equations; use of standard formulae to solve surface areas and volumes of regular solids; use of calculator for algebraic problems Indices, logarithms and functions: the index as a power, numerical and literal numbers in index form, laws of indices and logarithms, common and Naperian logarithms, functional notation and manipulation, logarithmic and exponential functions, other simple algebraic functions. Algebra is essential for any student studying Engineering and Technology. Students following the Edexcel National should have a GCSE pass grade in Maths. This probably means that many will not have done any Algebra in their GCSE studies. For this reason the level in this tutorial is set at very basic. It is assumed that the student is familiar with arithmetical operations, the use of brackets and fractions. CONTENTS
• Explain the laws of Indices. • Define a logarithm. • Define an antilogarithm. • Explain base numbers. • Explain basic power laws. • Multiply and divide using logarithms. • Roots using logarithms. • Applications. • Logarithmic graphs.
© D.J.Dunn www.freestudy.co.uk 1
1. INTRODUCTION Before we had electronic calculators, accurate calculations involving multiplication and division were done with the aid of logarithms, pen and paper. To do this, the logarithms of numbers had to be looked up in tables and added or subtracted. This is easier than multiplying and dividing. The greater the accuracy needed, the larger the tables. To do less accurate calculations we used slide rules and these were devices based on logarithms. Although we do not need these today, we do use logarithms widely in mathematics as part of the wider understanding of the relationship between variables so this is an important area work. In order to understand logarithms, it is necessary to understand indices and we should start with this. 2. INDICES In algebra, a way of writing a number or symbol such as 'a' that is multiplied by itself 'n' times is an
n is called the index. For example a3 is the shorthand for a x a x a or a.a.a to avoid use of the multiplication sign. an is called the nth power of a. There are four laws that help us use this to solve problems. Law of Multiplication notable results a x . a y = a x + y
Law of Division
yxy
xa
aa −= →
0xxx
xaa
aa1 === −
xx a
a1 −=
Law of Powers ax . ay . az = a x+y+z
a . a . a ....n times = an → y = a n then a = n√y ax . ax . ax ....n times = anx → (ax)n = anx
Law of Roots y = a1/n + a1/n + a1/n....n times → a1/n = n√y y = (a1/n) n → a1/n = n√a WORKED EXAMPLE No.1
Simplify the following ( )5
332
aaas =
SOLUTION s = a 2 . a 3 . 3 a - 5 s = a 2 + 9 - 5 = a 6
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SELF ASSESSMENT EXERCISE No.1 Simplify the following. C = x 4 . x 3 (C = x7 )
4
5
ddF = (F = d)
6
35
b.bbA = (A = b2 )
D = √a . a 3 (D = a3.5 )
a
aS3
= (S = a2.5 )
2
52
xyyx
=S (S = x. y3)
3. DEFINITION of a LOGARITHM The logarithm of a number is the power to which a base number must be raised in order to produce it. The most commonly used base numbers are 10 and the natural number ‘e’ which has a rounded off value of 2.7183 (also known as Naperian Logarithms). If you have not come across this number yet, don’t worry about where it comes from but you need to know that it possesses special properties. BASE 10 This is usually shown as log on calculators but more correctly it should be written as log10. Since it is the most widely used, it is always assumed that log means with a base of 10. Suppose we want the log of 1000. We should know that 1000 = 103 so the log of 1000 is 3. Similarly : The log of 100 is 2 since 102 is 1000.
The log of 10 is 1 since 101 is 10. The log of 1 is 0 since 100 is 1. The log of 0.1 is -1 since 10-1 is 0.1 … and so on.
This is all well and good if we are finding the log of multiples of 10 but what about more difficult numbers. In general if y = 10n then n is the log of y and without calculators we would have to look them up in tables. You can use your calculator. SELF ASSESSMENT EXERCISE No.2 Use your calculator to find the log of the following numbers. Just enter the number and press the
log button. 260 (2.415) 70 (1.845) 6 (0.778) 0.5 (-0.301)
© D.J.Dunn www.freestudy.co.uk 3
BASE ‘e’ The natural number ‘e’ (2.7183) will not be explained here but it is used for very good reasons. Natural logarithms are written as loge or more likely as ln as it appears on most calculators. If y = en then n is the natural logarithm. You get it from your calculator by simply entering the number and pressing the button marked ln. SELF ASSESSMENT EXERCISE No.3 Use your calculator to find the ln of the following numbers. (answers in red) 260 (5.561) 70 (4.248) 6 (1.792) 0.5 (-0.693) OTHER BASES Your calculator may allow you to find the logarithms to other bases by programming in the base number but this won’t be covered here. Here are some simple examples. WORKED EXAMPLE No.2 Find the log2 of the number 8. Since 23 = 8 then log2(8) = 3. Find log3(81). Since 34 = 81 then log3(81) = 4 SELF ASSESSMENT EXERCISE No.4 Find the following (answers in red) log2(16) (4) log3(27) (3) log5(625) (4) 4. ANTILOGS An Antilog is the number that gives us the logarithm or put another way, the number resulting from raising the base number to the power of the logarithm. For example if the base is 10, Antilog(2) = 102 = 100 Antilog(3) = 103 = 1000 On a calculator this is usually shown as 10x and is often the second function of the same key as log10. If the base is ‘e’ then for example anti-ln(5.561) = e5.561 = 260 and so on. On a calculator this is the button marked ex and is often the second function of the same key as ln.
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5. POWER LAWS If A = (xn) and B = (xm) then AB = xn+m
If x is the base of our logarithms then n = logx(A) and m = log x(B) and log x(AB)= log xA+ log xB This is useful because if we can look up the logs of numbers we can solve multiplication problems by adding the logs. NEGATIVE POWERS (INDICES)
If mnm
n
xxx
BAy −=== then mnlog(B)log(A)
BAloglog(y) −=−==
WORKED EXAMPLE No.3 Solve y = (36.5)(17.72) Taking logs we have log y = log(36.5) + log(17.72) = 1.562 + 1.248 = 2.811 y = antilog(2.811) = 102.811 = 646.78 WORKED EXAMPLE No.4 Solve y = (36.5) ÷ (17.72) Taking logs we have log y = log(36.5) - log(17.72) = 1.562 - 1.248 = 0.314 y = antilog(0.314) = 100.314 = 2.060 Of course we can get the same answers on our calculators without this process but it is very useful to change multiplication into adding and division into subtraction. SIMPLIFYING NUMBERS WITH POWERS (INDEXES) ROOTS
You know that: n1
n AA = hence if log(A)n1log(y) then AAy n
1n ===
WORKED EXAMPLE No.5 Find the fifth root of 600 using logarithms.
( )
594.310 566)antilog(0. y
0.556 2.77851 log(600)
51 y log
600600y
0.556
51
5
===
=⎟⎠⎞
⎜⎝⎛==
==
This can be done directly on a calculator to check the answer but the basic transformation is
very useful in derivations and manipulation of formulae. © D.J.Dunn www.freestudy.co.uk 5
6. DECIBELS The ratio of two numbers can be expressed in decibels. The definitions is G(db) = 10 log(G) where G(db) is the ratio in decibels and R is the actual ratio. This is commonly applied to equipment in which there is a change in POWER such that G = Power Out/Power in and the G is the Gain. The reason for doing this is that if you put two such items in series the overall gain is:
G(over all) = G1 G2Taking logarithms G(db) = G1(db) + G2(db) The gains in db are the sum of the individual gains. 7. PRACTICAL EXAMPLES OF LOGARITHMS WORKED EXAMPLE No.6 An electronic amplifier increases the power of the signal by a factor of 20. What is the gain in
decibels? G(db) = 10 log 20 = 13 db The amplifier is fed into another amplifier with a gain of 5. What is the overall gain in decibels? The gain of the second amplifier is G = 10 log 5 = 7 db The total gain is 13 + 7 = 20 db Check this way:- Overall gain = 20 x 5 = 100 In decibels G = 10 log 100 = 20 db WORKED EXAMPLE No.7
A well known formula used in the analysis of damped vibrations is δ-1
δ 2π xxln
22
1 =⎟⎟⎠
⎞⎜⎜⎝
⎛
Where x1 and x2 are the amplitude of two successive vibrations and δ is the damping ratio. Calculate δ when x1= 3 mm and x2 = 0.5 mm respectively. Calculate the amplitude reduction factor and the damping ratio.
SOLUTION
0.2740.075δ and 0.07513.298
1 δ and 1 13.298δ
12.298δδ-1 so δ-1
39.478δ3.21
sidesboth square δ-1
δ 2π 1.792
1.792 ln60.53ln
xxln
22
222
2
2
2
1
=====
==
=
==⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
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WORKED EXAMPLE No.8 When a gas is compressed from pressure volume V1 and temperature T1 to a final volume V2
and temperature T2 the relationship is :C
2
1
1
2
VV
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛= where C is a constant. Given 5.2
TT
1
2 = and
8VV
2
1 = determine C
SOLUTION Take logs and log(2.5) = C log(8) C = log(2.5)/log() = 0.398/0.903 = 0.441 ( )C85.2 = WORKED EXAMPLE No.9 The ratio of the tensions in a pulley belt is given by R = e2.5µ Find the value of µ when R is 5. SOLUTION 5 = e2.5µ So take natural logs and ln(5) = 2.5µ ln(e) and by definition ln(e) is 1 ln(5) = 2.5µ = 1.609 µ = 1.609/2.5 = 0.644 SELF ASSESSMENT EXERCISE No.5 1. An amplifier has a gain of 32. What is this in decibels? (Answer 15 db) 2. Given y = xn determine the value of n when y = 6 and x = 20 (Answer 0.598) 3. Given 12 = eµ find µ. (Answer 2.485) LOGARITHMIC GRAPHS Logarithms may be used to change to simplify functions by changing them into a straight line graph law. Consider the function y = f(x) = Cxn C is a constant and n a power. Except when n = 1, this a curve when plotted. If we take logarithms we find:- log(y) = φ(x) = log(C) + n log(x) The graph of φ(x) is now a straight line law where log(C) is the intercept and n is the gradient. This is most useful in determining the function from experimental data.
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WORKED EXAMPLE No.10 The graph shows the results of an experiment in which a variables x and y are recorded and
plotted. When log(x) and log(y) are plotted the straight line graph shown is produced. Determine the function f(x).
SOLUTION From the straight line graph we have an intercept of 0.7 and a gradient of (3.7 – 0.7)/1 = 3 φ(x) = 0.7 + 3 log(x) Take antilogs f(x) = 5x3
WORKED EXAMPLE No.11 The graph shows the results of an experiment in which a variables x and y are recorded and
plotted as logs. Determine the function f(x).
SOLUTION From the straight line graph we have an intercept of 0.845. The gradient is (2.345– 0.845)/(-15) = - 0.1 φ(x) = 0.845 – 0.1 log(x) Take antilogs f(x) = 7x-0.1
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SELF ASSESSMENT EXERCISE No.6 Determine the function f(x) for each of the graphs below,
Answers f(x) = 1.5x2 and f(x) = 0.5x-0.5
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