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EDEXCEL NATIONAL CERTIFICATE
UNIT 4 MATHEMATICS FOR TECHNICIANS
TUTORIAL 1 - 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
On completion of this unit a learner should: 1 Know how to use algebraic methods
2 Be able to use trigonometric methods and standard formula to determine areas and volumes
3 Be able to use statistical methods to display data
4 Know how to use elementary calculus techniques.
OUTCOME 1- Know how to use algebraic methods
Indices and logarithms: laws of indices (am
x an = a, = am+n, nmn
a , (am
)n = a
laws of logarithms (log A + log B = log AB, log An = n log A,
AlogB logA log )
e.g. common logarithms (base 10), natural logarithms (base e), exponential growth and decay
Linear equations and straight line graphs: linear equations e.g. y = mx + c;
straight line graph (coordinates on a pair of labelled Cartesian axes, positive or negative
gradient, intercept, plot of a straight line);
experimental data e.g. Ohms law, pair of simultaneous linear equations in two unknowns
Factorisation and quadratics: multiply expressions in brackets by a number, symbol or by another expression in a bracket; by
extraction of a common factor
e.g. ax + ay, a(x + 2) + b(x + 2);
by grouping e.g. ax - ay + bx - by,
quadratic expressions e.g. a2 + 2ab + b
roots of an equation e.g. quadratic equations with real roots by factorisation, and by the use of
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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.
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 n
th power of a. There are four laws that help us use this to solve
Law of Multiplication notable results
a x . a
y = a
x + y
Law of Division
Law of Powers
ax . a
y . a
z = a
a . a . a ....n times = an
y = a n then a =
ax . a
x . a
x ....n times = a
x)n = a
Law of Roots
y = a1/n
....n times a1/n
y = (a1/n
WORKED EXAMPLE No.1
Simplify the following
s = a 2 . a
3 . 3 a
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 = x
dF (F = d)
.bbA (A = b
D = a . a 3 (D = a
(S = a2.5
yxS (S = x. y
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, dont worry about where it comes from but you need to know that it possesses special
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
260 (2.415) 70 (1.845) 6 (0.778) 0.5 (-0.301)
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Consider the expression
and try evaluating it for bigger and bigger values of n. For
n = 1, x = 2
n = 10, x = 2.594
n = 100, x =2.705
n = 1000, x = 2.717.
n = , x2.7183 (rounded off)
The value when n = is called the natural number or exponential and denoted with 'e'. For practical
purposes we take e = 2.7183. Further explanation is not given here but this number has very special
properties and a set of logarithms with the base of e was originally devised by Robert Napier and
are used for good reasons. These are called NATURAL LOGARITHMS or sometimes Naperian
Logarithms and they are denoted loge or 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)
Your calculator may allow you to find the logarithms to other bases by programming in the base
number but this wont 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)
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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) = 10
3 = 1000
On a calculator this is usually shown as 10x and is often the second function of the same key as
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.
5. POWER LAWS
If A = (xn) and B = (x
m) then AB = x
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)
Ay then mnlog(B)log(A)
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
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
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)
You know that: n1
n AA hence if log(A)n
1log(y) then AAy n
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WORKED EXAMPLE No.5
Find the fifth root of 600 using logarithms.
594.310 566)antilog(0. y
1 y log 600600y
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.
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 G2
Taking logarithms G(db) = G1(db) + G2(db)
The gains in db are the sum of the individual
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
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
The ratio of the tensions in a pulley belt is given by R = e2.5
Find the value of when R is 5.
5 = e2.5
So take natural logs and ln(5) = 2.5 ln(e) and by definition ln(e) is 1
ln(5) = 2.5 = 1.609 = 1.609/2.5 = 0.644
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WORKED EXAMPLE No. 8
A well known formula used in the analysis of damped vibrations is -1
Where x1 and x2 are the amplitude of two successive vibrations and is the damping ratio.
Calculate when x1= 3 mm and x2 = 0.5 mm respectively. Calculate the amplitude reduction
factor and the damping ratio.
0.2740.075 and 0.07513.298
1 and 1 13.298
12.298-1 so -1
sidesboth square -1
WORKED EXAMPLE No. 9
When a gas is compressed from pressure volume V1 and temperature T1 to a final volume V2
and temperature T2 the relationship is :
where C is a constant. Given 5.2
1 determine C
C85.2 Take logs and log(2.5) = C log(8) C = log(2.5)/log() = 0.398/0.903 = 0.441
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 = e find . (Answer 2.485)
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8. EXPONENTIAL GROWTH AND DECAY
It was shown earlier that the exponential 'e' is the value of
Exponential growths refer to the way things change with time under certain circumstances. Here is
Suppose that you have 100 in a bank account. Every year you add 12% interest to the sum in your
After year 1 you would have 100 + 12% x 100 = 112
Suppose instead you added 6% every 6 months.
After 6 months you would have 100 + 100 x 6% = 106
After a year you would have 106 + 106 x 6% = 112.36 which is more than previously.
Suppose that you added 1% every month. This would take 12 calculations but after a year you
would have 112.68.
If you took it down to 1 second intervals the sum would be 112.75 after 1 year.
If we took smaller and smaller intervals of time we would get no closer to the sum when rounded
off to the nearest penny. This is known as continuous compound interest. It can be shown that the
way to calculate the sum for continuous compound interest is
Sum = (Original sum) x enp
N is the number of years and p is annual rate of interest.
Hence after 1 year compound interest the sum is
100 x e0.12
= 100 x 1.1275 = 112.75
After 40 years it is 100 x e0.12x40
= 12 151
This is called an exponential growth. It starts off growing
slowly but as time goes on it grows quicker and quicker.
If you invested 100 at 12% compound interest it would
grow over the years like this.
In nature and in science many things grow
and decay in this manner. A prime example
is when something fills or empties through a
resistance. For example if you allowed water
to run out of a tank through a hole in the
bottom, the level would fall with time like
this. This is an exponential decay where the
level falls quickly at the start but then falls
slower and slower as time goes on.
An exponential growth follows the mathematical law Tt
Aex and an exponential decay follows the
where A is the starting value, T is a constant and t is time. Further studies will show
that these curves and equations have special properties.
There is one other graph that you should know about. This is when something grows quickly at the
beginning and then slows down exponentially. This is given by the equation
This is illustrated in the following example.
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When an electrical capacitor is charged through a resistance from a battery of voltage Vs, the
voltage 'Vc' across the capacitor grows with time 't' in such a way that
s e-1Vv . T is constant
called the time constant (T = RC) and the larger the value of T the slower the change. The natural
number 'e' is called the exponent in this case hence we have an exponential growth. A plot of the
above equation shows the
If we plotted the voltage 'VR' across the resistor, we would find that since Vs = VR + VC then VR
falls with time so that Tt
and we have an exponential decay.
THE MEANING OF THE TIME CONSTANT
Starting with the equation
If we put t = T we have
SC V 0.633e-1 Ve-1 VV
So the time constant is the time taken for the
voltage to change to 63.3% of the final value.
This is useful when finding T from a graph.
Note in theory that the final voltage is never reached but if we calculate the voltage at t = 4T we
find V = 98.2%VS. We can say effectively that it takes 4T for a capacitor to almost completely
charge or discharge.
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WORKED EXAMPLE No. 10
A capacitor of value 50 F is charged from zero to 100 V through a 5 M resistor. Calculate the
time constant and the time taken for the voltage to rise to 50 V.
T = RC = 50 x 10-6
x 5 x 106 = 250
5.05.01e e-1 0.5 e-1 10050 e-1 VV 250-t
t = 250 x 0.6931 = 173.2 seconds
SELF ASSESSMENT EXERCISE No. 6
1. The graph shows a charging curve for a capacitor and resistance. Work out the time constant T
and determine the capacitance if the resistor value is 6 k.
(Answers 0.12 s and 2 F)
2. A tank of compressed air contains a pressure of 20 bar. When a small leak is made, the pressure
falls with time such that the Tt
and the resulting plot is shown below. Determine the time
constant T in minutes from the graph and calculate how long it takes for the pressure to fall to 10
bar. Confirm this from the graph.
(Answer 2.2 minutes and 1.52 minutes)
3. If 500 is invested at 5% annual compound interest, what is the total sum after 10 years?