INDICES AND LOGARITHM

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FUNDAMENTALS OF INDICES AND LOGARITHMS

1.Introduction

Indices (sing. index) together with logarithms1are central to many scientific and
engineering processes. Most of the equations we encounter, either in simplified or
complex forms, are based on them. In fact they are part of our daily activities, e.g.
finance, economics and natural phenomenon, although we use them unknowingly. So
here we will take a look at how to simplify exponential and logarithmic expressions and
how to solve their equations using certain rules that will be discussed shortly. However,
graphs of their functions will not be covered here as these will be appropriately dealt
with in another book. Readers are reminded that this part of the book focuses on the
background regarding the topic and therefore few examples will be provided. For
comprehensive and varied examples, do refer to the Worked Examples part (pp. 19 -71).

2.What is Indices?

Well, indices is when a number is expressed in the form where is called the base
and the index. The index, i.e. , could also be referred to as a power or exponent; they
all essentially mean the same. Nonetheless you probably would have found out that one
of the terms (index, exponent or power) is used more than the others. In general, power
is the most frequently used as such, is read as ‘ raised to the power of ’, ‘ raised
to the power ’ or simply ‘ to the power . What does this imply? It means that has
been multiplied by itself times. For example, if and then one can write
read as ‘five to the power three’ or ‘five cubed’. This implies , which equals
.

So what do you think about ? It is in value but it is written in
index form as . Now let us work in the reverse order and find out what is. This
can be written as a multiple of tens, i.e.
and it is the same as (a billion). From this example, it is obvious that we
need indices to express numbers such as in the example not only because it saves space
and time but more importantly it is sometimes easy to remember and work with. So
is a better way of writing a cumbersome 1 000 000 000. Calculation involving indices
can be carried out on a scientific calculator, commonly via a button marked or
something similar.

1 Precisely, we mean indicial and logarithmic functions.

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The next question that might come to mind is whether any number can be written in
index form. The answer is yes. Any rational number can be expressed in this form just

like any integer can be written in the form

where . For instance, can be

expressed in both index and fractional forms as and

respectively, but this is

generally omitted. This is the same for any other number.

3.Laws of Indices

Undoubtedly one would need to carry out calculations in index form because it is a
useful and compact way to express numbers. There are certain laws2 that govern these
operations and they will be discussed in this section.

3.1.Fundamental laws

There are three fundamental laws

Law (1)

Multiplication Law

This law can be written as

For example, . This can alternatively be expressed as

( )( )

The following should be noted about this rule:

i)The terms must have the same base otherwise the law cannot be used. For
example because 5 and 6 are not like base.

ii)The sign between the terms must be a multiplication and not an addition. In
other words, but . This can easily be proven
because and .

iii)The power can either be same or different.

iv)The terms can be two or more.

Law (2)

Quotient (or Division) Law

This law can be written as

2 They are also referred to as theorems, laws and properties. I will use them interchangeably in this work.

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or

For example, .Again, this can alternatively be expressed as

The following should be noted about this rule:

i)The terms must have the same base otherwise the law cannot be used. For
example .

ii)The sign between the terms must be a division and not a subtraction. In other
words, but . Again as previously performed
and and .

iii)The power can either be the same or different.

iv)The base is any real number excluding zero. i.e. { }.

v)The terms can be two or more.

Law (3)

Power Law

This law states that

( )

For example, ( ).Alternatively, this can be shown using Law 1 as

( ) ()

[] []

3.2.Special or derived laws

Other laws of indices include

Law (4)

Zero Power Law

This law can be written as

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It is simply put as ‘anything’3 to the power of zero is 1. Although this rule might look
odd, it is logical and well established. This can be derived from Law 2. Given that

If m=n, then

But

Law (5)

Negative Power Law

This law can be written as

Provided that, it implies that a number to the power of a negative index is the
reciprocal of that number to the same but with a positive index. Sometimes one might
need to carry out some operations that require a change from a positive index to a
negative index. Let us account for this and re-state the rule as ‘whenever the reciprocal
of a number with an index is taken, the sign of the index changes’. For instance,

( )

or

( )

or

( )

3 This excludes zero.

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In each of the above three cases, we have taken the reciprocal of the left-hand side to
obtain the corresponding right-hand side. So in (a) and (c) the sign of the index changed
from negative to positive whilst in (b) it changed from positive three (+3) to negative
three (-3).

This negative index rule can be derived from Law 2 (division rule) and Law 4 (power of
zero law)4 as follows:

Using Laws 4,

Using Laws 1,

( )

Using Law 2, the right-hand side of equation (i) can be written as

this implies that

Using Laws 4,

( )

Equating equations (i) and (ii), therefore

Law (6)

Fractional Power (or Root) Law

This law can be written as

√

This simply means that ‘anything’ raised to the power of is equal to the
root of ‘the same thing’5.If it is not clear, this will shortly be put this into context. But

4 Or simply from the division rule since Law 4 is also a special case of Law .

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remember that a power is the inverse of a root just like a whole number is the opposite

of a fraction. For example,

√

and

√

. We know that the answers are and

respectively. One can however obtain the same answers using Laws 2 and 3. How?
Here we go:

( )

using Law 1

( )

( )

using Law 3

(

)

Similarly,

( )

using Law 1

( )

( )

using Law 3

(

)

What about a situation when the numerator of the fractional index is not 1? Yes, there is
a rule for this case which states that

(√

)

or

5 Note that √ √

but 2 is generally omitted. It is read as ‘square root of x’. Similarly √

is the ‘cube root of x’.

While √

and √

the ‘fourth root of x’ and ‘fifth root of x’ respectively. So what are √

and √

?

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√

This is a combination of Laws 3 and 4 because

(

)

(

)

(√

)

For example,

(√

)

. You can double-check this with a calculator or

move to the Worked Examplessection for much more.

Law (7)

Same Power Law

So far we have introduced rules for dealing with terms that share a common base. We
will now look at rules to be used when the bases of the terms are not the same. In
general, the operations are carried out as you would normally do with one exception.
This exclusion is that when the index of the terms is the same then the following rules
should be applied:

( )

( )

or

( ) i.e.

[

]

Also,

( )

and (

)

The rules above are very simple and can easily be verified. Let us illustrate this using
these two examples:

( )

( )

and

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( )

( )

Conversely, it can be said that if there are terms in a bracket which is raised to the
power of , then the power applies to each term in the bracket provided the operation
between them is either multiplication or division. Consequently,

( )

and

( )

Finally, have you wondered if there is a rule relating to powers of one (1), i.e. ? I
guess not. But whatever the case, there is no special rule for numbers or expressions
raised to power of 1 because this is the default state of all numbers. To save time and
space, it is an unspoken rule as previously mentioned. Remember that , means
and respectively but 1 is always left unwritten.

Before proceeding to logarithms, you may wish to go through the Worked Exampleson
indices first. These are contained in sections 1- 4 (pp. 19 – 38).

4.What is Logarithm?

Logarithm is a derived term from two Greek words, namely: logos(expression) and
arithmos(number) (Singh, 2011). Thus, logarithm is a technique of expressing numbers.
In fact, it is a system of evaluating multiplication, division, powers and roots by
appropriately converting them to addition and subtraction. This concept is primarily
attributed to a Scotsman from Edinburgh, John Napier (1550 – 1617) (Bird, 2010) who
developed it.

Technically, the logarithm6 of a number to a given base is the value of the power to
which the base must be raised in order to produce the number. Let (for ) and
be the number7 and its associated base respectively and equals to ‘the logarithm to
base of ’ or ‘the logarithm of to base’, then we can write this as

6 This is commonly shortened to or written as log and sometimes as lg.
7 must be a positive real number excluding , i.e. { }. This is because raised to the power of
anything is and so will only be valid for since .

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For example, the logarithm of 100 to base 10 is 2 because, if the base 10 is raised to the
power of 2 we will get the number (100). In other words, . It is evident that
indices and logarithm are related and one can change from one form to the other, i.e.
one is the inverse of the other. This relationship can be written as

Logarithms are classified according to the value of their base. Essentially, there are two
(or three) types.

a)Common Logarithm: This is a logarithm to the base of 10, i.e. . In general,

when the base is 10, it is usually omitted. In other words, is simply written as
. Common logarithm is also called Briggsian named after Henry Briggs.

b)Natural Logarithm: This is logarithm to the base of an irrational number denoted as

, where or to 4 decimal places. Why ‘natural’? It is
possible that this logarithm is classified as natural due to the behaviour of certain
natural phenomena (e.g. radioactive decay, charging-discharging a capacitor,
frequency response, biological functions, interest rate, etc.) being dependent on the
functions of . Natural logarithm is also called hyperbolic or Napierian logarithm,
named after its inventor John Napier. Furthermore, the logarithmic ratio unit, neper
(abbreviated as nep), is also used in his honour8. Usually, natural logarithm is
written as (read as ‘’) instead of as one might have expected.

c) The third category (if we may say) is any other logarithm with a base other than 10

and .

5.Laws of Logarithm

Like indices, there are certain laws governing the operation of logarithms and these will
be discussed under the following headings.

5.1.Fundamental laws

Essentially, there are three main laws of logarithms.

Law (1)

Addition-Product Law

This rule can be written as

( )

8 This is when the base is , often with current ratio. However, when the base is , usually for power ratio
(attenuation or amplification), the unit of logarithm ratio is decibels ().

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In other words, the sum of logs of numbers to the same base is equal to the log of their
products and vice versa. The following should be noted about the rule:

i)The logs must have the same base otherwise the law cannot be used. For example
.

ii)The sign between the terms must be addition and not multiplication. In other
words, but . This can
simply be proven as with the indices.

Law (2)

Subtraction-Quotient Law

This rule can be written as

(

)

The following should be noted about the rule:

i)The logs must have the same base otherwise the law cannot be used. For example
.

ii)The coefficient of the log must be 1.

iii)The sign between the terms must be subtraction and not division. In other words,

(

) but . Again,

this can be proven.

Law (3)

Power Law

This rule can be written as

There are three special formulae or properties resulting from the above power law,
namely:

, ( )

and ( )

5.2.Special or derived laws

Other laws of logarithms include

Law (4)

Unity Law (or Log of Unity Law)

This rule can be written as

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This rule states that the logarithm of unity (1) to any base is zero. Yes, it is true because
the power that any number can be raised to produce one is actually a zero. This is in
accordance with the power of zero law in indices. From this it can be said that

{

One may be prompted to ask about the logarithm of zero. Well, it has been said that
logarithms are only defined (or valid) for numbers greater than zero. This is because
given that

a)When , we have

There is no value for to satisfy the above condition. Let us put this in practice by
taking and ask ourselves if there is a power which can be raised to produce
zero? Could this be 1? No, because . Did you say power zero? No, we just talked
about a rule of power zero, so . You may start wondering whether a lower power
would work. What is less than a zero? The answer is a negative number. In this case, let
us take and evaluate,

√

At least by now we know the answer is not zero. Actually the answer is . What
else do you have in mind, try it and let us know if you have obtained a zero.

This said, you may want to try negative infinity, i.e. . So let us work this out.

The above is simply an approximation! Therefore, it can be said that the limiting value
of logarithm of zero to any base is negative infinity. That is

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b)When , i.e. is a negative, we have

( )

Remember is a positive number; now what is the positive number9 that when it is
raised to a certain power (positive, negative, whole number or fraction) the answer will
be a negative number. Again, there is none.

Law (5)

(Logarithm to the) Same Base Law

This rule can be written as

The logarithm of any number to the same base is 1. This is because .

Law (6)

Change of Base Law

This law can be written as

The logarithm of a number to a certain base is the same as the logarithm of the number

divided by the logarithm of the base such that both are given a new but same base. For

example,

You can check this out with a calculator; they will all be equal.

There is a special application of this rule when one need to multiply two or more logs
together. This states that

9 Here we mean only real numbers because a complex number such as when squared gives a negative number.
In other words ( ). For further details on complex numbers, refer to ‘Complex Numbers Explained with
Worked Examples’ by the same author, which is available online.

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( )( ) ( )( )

This is because

( )( )

( ) ( )

For instance

Can one imagine evaluating the above logs separately before multiplying? Without a
calculating aid, this will of course prove difficult if not impossible. Using a calculator,
the following was obtained.

Using this rule, we can also show that

6.Indicial-Logarithmic Equation

Indicial equations are equations involving powers, where either the base or the
exponent is the unknown variable to be found. Indicial equations can generally be
solved using any known method of solving polynomials – linear or otherwise – in
conjunction with the laws of indices and logarithms discussed above.

The method of solving an indicial equation will be determined by the nature of the
equation, which can be broadly grouped into two.

a)When the unknown is the base: this is often solved by simplifying the expressions,

and a suitable method of solving polynomials is subsequently applied. Kindly refer
to section 10 (pp 47 – 51) for examples.

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b)When the unknown is the index: in this case, if the base of the two sides of the

equation can be made the same then equate their index otherwise the log of both
sides must be taken (sections 11 and 12, pp 51 – 58). In the latter case, one need to
apply the one-to-one property of logarithm which states that if

In both aforementioned instances, there must be only a single term on each side of the
equation before the index or base can be equated. Refer to the Worked Examplespart of
this book for further clarification.

7.Scientific Notation

Scientific notation (also known as ‘standard form’) is a way of representing a number
such that

8.Logarithm-Antilogarithm Table
8.1.Mathematical tables

Mathematical tables10 are used to find the square (or square root), cube (or cube root),
power (or root) and logarithm (or antilogarithm11) of numbers. Sine, cosine and
tangent of angles are some other functions that can be obtained using a mathematical
table.

Since we are dealing with logarithm here, we will show how a log (logarithm) table can
be used to perform calculations involving multiplications, divisions, powers and roots.
It should be mentioned that a log table is used in conjunction with another table, called
antilogarithm table. Antilog table, as it is commonly called, is the opposite of log table.
The tables are meant for the common log, i.e. to base 10, but the procedure and
principle is applicable to any base.

How is it possible to represent logs of numbers in a single table? Yes, it is doable correct
to four significant figures. One of the techniques used to achieve this is explained
below. Supposing we know that

10 With the proliferation of calculating devices, it is not very common nowadays but it still remains a useful
invention. Addition and subtraction of numbers is possible but could one imagine multiplication and division
without a calculating device such as a calculator?
11 Anti-logarithm is the opposite of logarithm to a certain base. For instance, if is then the antilog of to
base is . In other words, the antilog of is . Therefore, the antilog (2) and antilog (3) to base 10 are
and respectively.

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we can therefore find the value of for example. This is accomplished using
relevant laws of logarithms and the scientific notation. Hence, the above log can be
resolved as

( )

What about ? Well, the same principle is applicable.

( )

It can therefore be said that have the same log value except that
they differ in their leading integer or digit. This also applies to any number. The above
illustration shows that a cell in the table can be used to represent a group of numbers.

8.2.Components of a number

In general, given that can be expressed in scientific form as and that
, then ( ) . In this case, the fractional part or is
called the mantissa and the integer part is referred to as the characteristicof .
Whilst mantissa is always positive, the characteristic can either be positive or negative.
When the characteristic is negative it is written as ̅(read as ‘bar n’). For example, if a
number has a mantissa and characteristic of and respectively, it will be
written as ̅ and read as ‘bar three point three, zero, one, zero’. You however
need to remember that ̅is equivalent to . It should be
further noted that the values in the log tables are for mantissa and the characteristics are
simply multipliers and are obtained from the number itself. This is easy when the latter
is expressed in scientific form.

8.3.When to use the log-antilog table

The log tables are typically used for multiplication and divisions rather than for
addition and subtraction of numbers. With multiplication, the logs of the multiplier and

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multiplicand are added and the antilog of the result is taken to obtain the answer. On
the other hand, when dividing numbers, the log of the divisor is subtracted from the log
of the dividend and followed by taking the antilog of the resulting value.

We can also evaluate powers and roots of numbers using log-antilog tables. Remember
that power is the number of times a number is multiplied so in this case, we multiple
the log of that number by the index. Similarly, a root which is the inverse of power is
computed by division. In each case one need to take the antilog of the result to obtain
the solution.

8.4.How to use the log-antilog table

Looking at log-antilog tables, there are three distinct parts:

(a)The first column: This is numbered from 10 to 99 and 0.00 to 0.99 for log-table

and antilog-table respectively. They are used for the first two digits of the
number whose log or antilog12 is to be found.

(b)A set of 10-column: These ten columns are numbered from 0 to 9 and are used

for the third digit. For example, the log of 234 is placed on the row numbered ‘23’
but under the fifth column, i.e. the column numbered ‘4’.

(c)Mean difference: This is a set of 9-column, known as the ‘mean difference’,

which can be found written above the columns. They are numbered from 1 to 9
and used for the fourth number. However, the number found here is added to
that obtained in (b) above. What about a fifth number? Since the table is a ‘four-
figure’ table, it implies that one can only evaluate up to four digits. We therefore
need to round off numbers to four significant figures before finding their logs in
the log tables. Also, if the number whose log is to be determined is a two-digit or
three-digit number, assume there is / are one and two zeros respectively to the
right of the number. 2-digit or 3-digit numbers have no value in the mean
difference section.

In summary, to use the log table do the following:

1)Determine the characteristic of the number as described above.
2)Locate the row corresponding to the first two digits in the log table and move

across the table until you get to the column number corresponding to the third
digit of the number whose log you are looking for. Note down the value in this
cell of the table.

12 In the case of antilog table, the first two digits are those immediate to the left of the decimal point. The integer
to the right of the decimal is a multiplier.

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3)Stay on the same row and move until you get to the section under the mean

difference. Now read the value in the cell of the column number (in the mean
difference section) corresponding to the fourth digit of the number whose log is
to be found.

4)Add the numbers obtained in 2 and 3 together. This gives the mantissa of the log

of the number you are looking for.

5)The log of the number is now the characteristic obtained in 1 dot(.) the mantissa

obtained in 4. For instance, given a number whose characteristic and mantissa
are 3 and 4567 respectively, then the log of the number is ‘3.4567’.

6)Repeat 1 to 5 for each number.
7)If two numbers are multiplied, add the logs but subtract the logs if there is a

division between them. For root and power, divide and multiply the log
respectively by .

8)The resulting value from 7 above should contain two parts: the integer and

decimal part.

9)Repeat steps 2 to 5 on the decimal part (mantissa) obtained in 7. You should

however use antilog-table in this case.

10)Know that the integer part from 7 is simply a multiplier. Thus, provided that the

number obtained from the antilog table in 9 is written with a dot after the first
digit, we can therefore write the final answer as , where and are the
answer from the antilog table and the integer from 7 respectively.

8.5.Illustration of the use of log-antilog table

To further clarify this, the process of using log-antilog tables will be illustrated by
multiplying by .

Number Log

What to do

Step 1.

Remember that can be written in standard
form as so its characteristic is . Write
this down and insert a dot after it.

In the log table, we now need to locate 40 in the first
column. Since the third digit is 1 (one), move across
this row until you reach the column labeled ‘1’ and
read off the number in the cell formed by the
intersection of ‘row 40’ and ‘column 1’. This is 6031.

The next step is to stay on the same row and move

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until you get to the column labeled ‘5’ in the ‘mean
difference’ column section. This is because the
fourth digit in the number whose log we are looking
for is 5. Coincidentally, the number in the mean
difference section is 5, though this is not always the
case.

Finally, add the two results as .
This implies that the mantissa is 0.6036. Therefore,
the log of 401.5 is 2.6036, i.e. characteristicdot (.)
characteristic.

Step 2.

Do the same as described above for 2.67. However,
since there is no fourth number here, the value in the
‘mean difference’ section of the table is taken to be
zero. Thus, the log of 2.67 is 0.4265.

Remember also that and that is
why its characteristic is zero.

Step 3.

Since the two numbers have been multiplied, one
should add their logs.

Step 4.

Antilog

We want the actual result and not log so the antilog
of the number in step 3 should be obtained. This
number contains two parts: the integer 3 and the
decimal .0301.

Find the antilog of .0301 in the antilog table the same
way we found the logs of the two numbers. This
should give you 1072.

Step 5.

Answer

The final answer is .

Note that although a calculator would give a more
accurate answer as 1072.005, which is 1072 correct to
4 significant figures.

END OF FUNDAMENTALS OF INDICES AND LOGARITHMS

AND

BEGINNING OF WORKED EXAMPLES

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WORKED EXAMPLES

Section 1: Basic Applications of Laws of

Indices (Numbers Only)

INTRODUCTION
In this section of the Worked Examples, questions
are based on the laws of indices involving
numbers and are structured such that one, two or
more laws can be used at a time. The law(s) to be
used is stated in the Hint box. Sometimes
different combination of the laws can be
employed to solve a single question but only one
approach is shown. Nonetheless, the author
provides hints and/or a prevailing alternative
method to a question. No calculators will be
required in this and many of the subsequent
sections. However, one can be used to confirm
answers.

1)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 1

of indices exclusively.

(a)

Solution

(b)

Solution

( )

(c)

⁄

⁄

Solution

⁄

⁄(⁄ ⁄ )

(d)

Solution

(

)

2)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying Law 2

of indices exclusively.

(a)

Solution

(b)

Solution

(

)

(c)

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Solution

( )

(d)

⁄⁄

Solution

⁄⁄

(

)

3)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 3

of indices exclusively.

( )

(a)( )

Solution

( )

(b)( )

Solution

( )

(c)( )

Solution

( )

(d)

Solution

( )

NOTE

Initially it does not look like Law 3 can be applied

as presented in the Hint box but we have

managed to change 216 so that the question looks

like the format presented. Good. This technique

will be utilized in most cases.

4)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying Law 4

of indices exclusively.

(a)

Solution

(b)

Solution

(c)

Solution

(d)( )

Solution

( )

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(e)(

)

Solution

(
)

(f)(

)

Solution

(
)

(g)(

)

Solution

(

)

5)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 5

of indices exclusively.

(a)

Solution

(b)

Solution

(c)( )

Solution

( )

( )

(d)

Solution

(e)

Solution

NOTE

In general,

so

as before.

(f)(

)

Solution

(
)

(g)(

)

Solution

(
)

(h)

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Solution

(i)(

)

Solution

(
)

6)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying Law

6(a) of indices exclusively.

√

(a)

Solution

⁄ √

√

NOTE

In general,

√

Hence, for ease of evaluation one can express the

radicand (the number under the root symbol) in

index form when determining roots of numbers.

(b)

Solution

⁄ √

√

(c)

Solution

√

√

(d)

Solution

⁄ √

√

(e)(

)

Solution

(
)

⁄

√

√(

)

(f)(

)

Solution

(
)

⁄

√

√

√(

)

(g)( )

Solution

( )

√( )

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23

√( )

(h)

⁄

Solution

⁄ (

)

⁄

√

√

√

√

√

NOTE

In general,

√

√
√

and

√ √

√

7)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law

6(b) of indices exclusively.

(√

)

(a)

Solution

⁄ (√

)

(√ )

(b)

Solution

(√

)

(√ )

(c)

Solution

(√ )

(√ )

(i)(

)

Solution

(
)

⁄

(√

)

(√

)

(

)

(d)(

)

Solution

(

)

(

)

(√

)

(√

)

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24

(e)

Solution

(√

)

(√ )

(f)

⁄

Solution

⁄ (

)

⁄

(√

)

(√

)

8)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying Laws

3 and 5 of indices exclusively.

( ) and

(a)( )

Solution

( )

(b)( )

Solution

( )

(c)()

Solution

( ) ( )

(d)[(

)

]

Solution

[(

)

]

(

)

( )

[( )]

( )

9)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Laws

5 and 6 of indices exclusively.

and

(√

)

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25

(a)(

)

⁄

Solution

(
)

⁄

( )

⁄

(√

)

(√ )

(b)(

)

Solution

(

)

(

)

(

)

√

√

(c)

Solution

(

)

(

)

√

√

NOTE

For this type of questions, one does not simplify the

fraction, i.e.

unless it is known that the simplified

form can be expressed in the power that would

eliminate the root.

(d)(

)

Solution

(
)

(

)

(

)

(√

)

(√

)

NOTE

Unlike with the previous question, the fraction here

needs to be simplified first before proceeding with the

root operation.

10)Without using a calculator, evaluate the

following.

Hint

Since laws 1- 6 have been tried independently (for

numbers) and with commonly occurring

combinations, I hope we have developed necessary

confidence. In the following questions, we will be

applying the relevant rule(s) of indices including Law

7.

(a)

Solution

( )

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NOTE

Here, we have expressed the base in an index form and

applied Law 3. This is an alternative way to using the

fractional power law and it is a very useful way of

dealing with situations where roots might prove

difficult to determine. We will employ this

extensively in subsequent questions.

(b)(

)

Solution

(

)

(

)

[

]

[

]

(

)

(c)(

)

Solution

(

)

(

)

(

)

(d)( )

Solution

( )

( )

(e)√

Solution

√

(

)

(

)

( )

NOTE

This last question can obviously be approached in

various ways using different law combinations.

(f)√

Solution

√

( )

⁄

(

)

( )

(g)

Solution

(h) ( )

Solution

( )

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(i)( )

Solution

( )

[( )]

(j)()

Solution

( )

[( )]

( )

( )

NOTE

The key rule to know here is that

( ) {

(k)(

)

(

)

(

)

Solution

(
)

(

)

(

)

NOTE

Remember that one cannot apply Law 1 when there is

addition (+), so the above question should be simplified

numerically.

Section 2: Basic Applications of Laws of

Indices (Numbers and Letters)

INTRODUCTION
This section is similar in structure and approach
to section 1 except that numbers and letters will
be used here. Again no calculators are required.

11)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 1

of indices exclusively.

(a)

Solution

(b)

Solution

( )

(c)

Solution

(d)

Solution

( )

( )

( )

(e)

Solution

( )

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NOTE

As shown above, it is evident that the index can be

exclusively numbers, wholly letters or a combination

of both and the laws will still be valid.

12)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying Law 2

of indices exclusively.

(a)

Solution

NOTE

Remember that,

(b)

Solution

( )

(c)

Solution

(d)

Solution

( ) ( )

( )( )

( )( )

NOTE

Remember that,

( )

(e)

()

Solution

( )

13)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 3

of indices exclusively.

( )

(a)()

Solution

( )

(b)()

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Solution

( )

NOTE

Remember that the square of a negative number or

expression gives a positive number or expression.

Alternatively, we can write

( ) ( )

( ) ( )

(c)()

Solution

( ) ( )

( )

(d)()

Solution

( ) ( )

( )

(e)()

Solution

( ) ( )

(f)()

Solution

( )

( )

(

) (

) (

)

14)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Law 5

of indices exclusively.

(a)

Solution

(b)

Solution

(c)

Solution

(d)(

)

Solution

(
)

(e)

Solution

(f)()

Solution

( ) (

)

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15)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Laws

1 and 3 of indices exclusively.

and ( )

(a) ( )

Solution

( )

(b)()()

Solution

( ) ( ) ( ) ( )

(c)()()

Solution

( ) ( ) ( ) ( )

(d)()()

Solution

( ) ( )

( ) ( )

(e)()

( )

Solution

( )

( )

(

) (

)

16)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Laws

1 and 4 of indices exclusively.

and

(a)

Solution

(b)

Solution

(c)

Solution

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31

(d)

Solution

(e)

Solution

17)Without using a calculator, evaluate the

following.

Hint

In the following questions, we will be applying Laws

3 and 5 of indices exclusively.

( ) and

(a)()

Solution

( )

(b)()

Solution

( )

( )

(c)( )

Solution

( )

( )

(d)(

)

Solution

(
)

( )

(e)(

)

Solution

(
)

(

)

18)Without using a calculator, evaluate the

following.

Hint

Since the laws (1- 6) have been used independently

(for numbers and letters) and with common

combinations, I hope we have developed necessary

confidence. In the following questions, we will be

applying the relevant rule(s) of indices including Law

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7.

(a) ( )

Solution

( )

(b)()()

Solution

( ) ( )

(c)()()

Solution

( ) ( )

( )

(d)()()

Solution

( ) ( )

( ) ( )

( )

NOTE

We can consider the expression in the brackets as

a unit and simply evaluate it as:

( ) ( ) ( )

as before.

(e)

Solution

{

} {

}

{

} { ( )}

(f)

()

()

Solution

( )

( ) ( )

(g)

()

Solution

( )

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(h)√

Solution

√

( )

⁄

( )

(

)(

)

(i)()

()

Solution

( )

( )

[( )( )]

[( )( )( )]

[( )( )( )]

[( )]

(j)()

()

Solution

( )

( )

( )

(

)

( )

( )

NOTE

Until now we have only indirectly applied Law 7

(Same Power Law). The last two questions clearly

underscore the relevance of this rule (though one can

still do without it in these questions).

Section 3: Advanced Applications of

Laws of Indices

INTRODUCTION
This section provides advanced questions on
indices. As a result, readers should be familiar
with laws of indices, and should in addition be
able to handle complex algebraic expressions. If
you have not done so, sections 1 and 2 are good
starting points for beginners.

19)Simplify the following.

(a)()

()

()

Solution

( )

( )

( )

[( )( )]

( )

( )

We have employed difference of two squares to

obtain the expression above. Now we can carry on

with the application of laws of indices.

( )

( )

( )

( )

( )

( )

( ) ( )

(b)(

)

(

)

()

Solution

(

)

(

)

()

(

) (

) (

)

(

) (

)

(

) (

)

(

) (

)

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(c)√

(

)

Solution

√

(

)

(

)

(

)

( )(

)(

)

(d)√

√

Solution

√

√

(

)

(

)

(

)(

)( )

(e)( ) √

( )

Solution

( ) √

( )

[( )( )] ( )

( )

( ) ( ) ( )

( )

( )

( )

( )

( )

[( )( )]

( )

√( )

(f)()()

()

Solution

()()

()

( ) {( ) ( )}

( ) {

( )}

( )

( )

( )

( )

Section 4: Applications of Laws of

Indices (Simplification)

INTRODUCTION
In this section, as the title suggests, laws of indices
will be used to re-write an expression in its
simplified form, either prescribed in the question
or otherwise. Sometimes, the process will involve
substituting the value of a variable.

20)Simplify the following, giving each answer in

the form .

Hint

Questions 20 and 21 are similar except while

multiplication and division are used between the

terms in the former, the latter has addition and

subtraction between its terms. In both cases the

final expression must be expressed in power of

3.

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35

(a)

Solution

( ) ( )

(b)(

) ( )

Solution

(
)

( ) ( ) ( )

(c)( ) ( )

Solution

( ) ( )

( )

NOTE

Alternatively, one can re-arrange the expression

in the brackets such that

( ) ( ) ( ) ( )

( )

( )

as before.

(d)

Solution

( )

(e)

( )

Solution

( )

(f)√

√

Solution

√

√

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

21)Express each of the following in the form .

(a)

Solution

()

( )

(b)

Solution

()

( )

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(c)

Solution

( ) ( )

( )

( ) ( )

(d)

Solution

()

( )

( )( )

( )( )

22)Without using a calculator, evaluate the

following, leaving the answers in the form

√

, where , and are real numbers. State

the value of , and .

Hint

Note that the first two questions can also be solved

using surd but I guess it will be easier to use indices.

Law 7 should be applied here.

(a)√

√

Solution

√ √

( )

( )

( )

( )

√

√

(b)√

√

Solution

√ √

( )

( )

( )

( )

√

√

NOTE

This can also be dealt with using rules or

properties of surds. See ‘Surds Explained with

Worked Examples’ by Shefiu S. Zakariyah.

(c)

Solution

( )

(

)

(

)

(

)

√

(d)

Solution

( )

( )

( )

√

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23)If , without using a calculator find the

value of each of the following.

Hint

You may first simplify the expression and then

substitute for or vice versa.

(a)()

Solution

( ) ( )

(b)

Solution

(c)

()

Solution

( ) ( )

( ) ( )

(d)√

()

Solution

√

()

√ ( )

√

√()

24)If without using a calculator find the

value of each of the following.

Hint

This is similar to the previous question except that

the substituting value is negative and this would

need to be diligently dealt with. Refer to the rule on

page 27.

(a)

Solution

( )

[( )]

( )

(b)

Solution

( )

( )

[( )]

[( )]

[( )]

[ (

)]

(c)

Solution

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( )

[( )]

( )

√

(d)(

)

Solution

(
)

(

)

( )

(√

)

√

NOTE

Although it is said that there are no solutions to

questions (c) and (d), what is meant is that the answers

are not real values. In reality, the two questions have

answers, which are complex numbers. For example, the

solution to (c) are ( ) and ( ) , i.e.

. For further details on complex numbers, kindly

refer to ‘Complex Numbers Explained with Worked

Examples’ by Shefiu S. Zakariyah. This and others are

freely available online.

25)Rewrite the following expressions using index

notation without fractions.

(a)√

Solution

√

()

( )

( )

( )

(b)(√)

Solution

(√)

(

)

(c)

√

Solution

√

( )

(d)

√

( √)

Solution

√

(√

)

(

)

26)Without using a calculator simplify (√)

(√)

(√)

(√)

(√)

, giving the

answer in surd form.

Solution

(√)

(√)

(√)

(√)

(√)

√ {(√ )

(√ )

(√ )

(√ )

(√ )}

√ {

(√ )

(√ )

√

√ }

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√ {

√

√

√ }

√ { √ √

√

}

√ { √

√

} √

( √ )

Section 5: Conversions between Indices

and Logarithms (Basic)

INTRODUCTION
In this section, we want to practise the conversion
from an index notation to an equivalent logarithm
form and vice versa. This is exclusively to show
conversion between indices and logarithm and no
final answer is therefore required.

27)Change each of the following index forms into

their equivalent logarithmic forms.

(a)

Solution

(b)

Solution

(c)

Solution

(d)( )

Solution

( )

( )

(e)

Solution

(

)

(f)√

Solution

√

(g)

Solution

(h)

Solution

(i)

Solution

(j)

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Solution

( ) ( )

(k)

Solution

Divide both sides by 2.5 and then convert as

(l)

Solution

Re-arrange the above equation and then convert

as

()

28)Express the following logarithmic forms in

their corresponding index notation.

(a)

Solution

(b)

Solution

(c)

Solution

(d)(

)

Solution

(

)

(e)

(

)

Solution

(
)

(
)

(f)

Solution

(g)

Solution

(h)( )

Solution

( )

Section 6: Applications of Laws of

Logarithms

INTRODUCTION
In this section of the Worked Examples, questions
are based on the laws of logarithms and are
structured such that one out of the three main

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laws or their combinations can be used to solve a
question as stated in the Hint box. The derived
laws will be employed in subsequent sections.

29)Without using a calculator, simplify the

following leaving the final answer in log

form.

Hint

In the following questions, we will be applying Law 1

of logarithms exclusively.

( )

(a)

Solution

( )

(b)√ √

Solution

√ √ (√ √ )

(√ )

(√ )

(c)

Solution

( )

30)Without using a calculator, simplify the

following leaving the final answer in log

form.

Hint

In the following questions, we will be applying Law 2

of logarithms exclusively.

(

)

(a)

Solution

(

)

(b)√

√

Solution

√

√

(√

√
)

(√

)

(√

)

(√ )

(c)

Solution

(

)

(

)

(

)

(√ )

31)Without using a calculator, simplify the

following leaving the final answer in log

form.

Hint

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In the following questions, we will be applying Law 3

of logarithms exclusively.

(a)

Solution

(b)√

Solution

√

( )

(c)√

Solution

√

( )

32)Without using a calculator, simplify the

following.

Hint

In the following questions, we will be applying the

relevant rule(s) of logarithms.

(a)

Solution

( )

( )

( )

( )

(b)

Solution

( )

( )

(

)

[ (

)]

(

)

(c)√ √

√

Solution

√ √

√

( )

( )

( )

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NOTE

Although we have only used numbers to show the three

fundamental laws of logarithm, the same applies when

variables (or letters) are used as would be shown later.

Section 7: Logarithms (simplification)

INTRODUCTION
The first part of the current section requires that
we express a given logarithmic expression in
terms of a log whose value is given. The second
part is to prove the equality of a given problem.

33)Given that , ,

and , find the

value of the following correct to 4 significant

figures.

(a)

Solution

( )

(b)

Solution

( )

( )

( )

Alternatively,

( )

( )

( )

as before.

(c)(

)

Solution

(

)

( )

( )

( )

( )

(

)

(d)

Solution

( )

(e)

Solution

( )

( )

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(f)

Solution

( )

(g)

Solution

(

)

34)Show that

(a)

Solution

LHS

( )

NOTE

Alternatively (though longer),

Let

( )

and

( )

Substitute for in the left-hand side of the

original equation

( ) ( )

( ) (

)

( )

(b) ( )

Solution

LHS

{

( )
( )}

[ ( ) ( )]

( )

( )

( )

Section 8: Finding Anti-logarithms of

Numbers

INTRODUCTION
Typically, logarithm of a number is required.
However, the inverse might be needed
sometimes. This is called anti-logarithm and it is
the focus of this section. This requires a
calculating device. Good, isn’t it?

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45

35)Determine the antilog to the given base of the

following, giving the answer correct to 4

significant figures.

(a) ( )

Solution

Let equals, thus

(b) ( )

Solution

Let equals , thus

(c) ( )

Solution

Let equals , thus

(d)̅ ( )

Solution

Let equals ̅, thus

̅

̅

(e) ( )

Solution

Let equals , thus

(f)̅ ( )

Solution

Let equals ̅, thus

̅

̅

NOTE

As would be shown later, the antilog of numbers

to base 10 can be found from anti-log tables.

Section 9: Conversions between Indices

and Logarithms (Advanced)

INTRODUCTION
This section provides advanced questions on
logarithms and as a result, readers need to be
familiar with laws of logarithms. If you have not
done so, previous sections (5 - 8) are
recommended starting points for beginners.

36)Express in terms of if

(a)( )

Solution

( )

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46

(b)

( )

Solution

( )

( )

37)Rewrite the following in their simplest log

form.

(a)

Solution

[

]

[ ]

[ ]

(b)

( )( )

Solution

( )( )

[

( )( )]

( ) ( )

( )

( )

( )

( )

(c)

( )(√ )

Solution

( )(√ )

[

( )(√ )]

(

) ( ) √

( )

( )

( )

NOTE

Here we have taken the natural log of both sides

of the equation rather than the common log.

Apart from satisfying the instruction in the

question, it is correct to take log to any base when

dealing with this type of problem. We will

employ this when solving equations later in this

book.

(d)

√

Solution

√

[

√

]

√

[ √ √ ]

[

]

[

]

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[

]

38)Express the following without logarithms.

(a)

Solution

( ) ( )

( ) ( )

[

]

Therefore

[

]

(b)

Solution

Therefore

(c)

( )

Solution

( )

( )

[

( )

]

Therefore

[

( )

]

( )

(d) ( ) (

)

Solution

( )

(

)

( ) ( )

{

}

Therefore

{

}

Section 10: Indicial Equations (I)

(Unknown as the Base)

INTRODUCTION
The next three sections deal with equations
involving indices. In this current section, the base
is the variable to solve for. Notes will be added on
the key steps as deemed necessary particularly for
leading questions.

39)Solve the following indicial equations given

that .

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48

(a)

Solution

Raise both sides to the power of 4

(

)

(b)

Solution

Raise both sides to the power of

(

)

( )

(c)

Solution

Divide both sides by 6

Express as a power of so the powers on both

the LHS and RHS cancel each other

(d)

Solution

( )

( )

( )

[( )]

(e)

√

Solution

√

Square both sides

(

)

( √ )

( )

Therefore, either

or

NOTE

Note that the approach shown below will give

only one value .

√

(

)

Divide both sides by

Now square both sides

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49

(

)

(

)

(f)

√

Solution

√

Multiply both sides by

(one could also divide

both sides by

)

Raise both sides to the power of

(

)

( )

( )

(g)

√

Solution

√

(

)

(√

)

( )

Therefore, either

or

40)Solve the following equations.

Hint

Unlike most of the questions we have encountered

so far, the indicial questions here will result in

quadratic equations. So there will always be two

values to the unknown variable. For clarity, I will

provide relevant notes on the steps taken in solving

each problem.

(a) √

Solution

√

Square both sides

( ) (

)

Open the bracket

Collect the like terms

Factorise the above equation

( )( )

This implies that

( )

(b)

Solution

Multiply through by

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(

)

This equals

Now square both sides

( ) (

)

Open the bracket

Collect the like terms

Factorise the above equation

( )( )

Therefore, either

or

(c)

Solution

Re-arrange this to have

(

)

(

) (

)

Therefore, either

Raise both sides to power 5

( )

or

Raise both sides to power 5

( )

NOTE

In this question we could have simplified the

equation as we did in the preceding ones until we

have a quadratic equation of the form

and then solve for the unknown variable ( in this

case). However, this is not always easy and/or

necessary. Once the power of one term is twice

the other then it is a quadratic equation and we

can use any appropriate technique of solving a

quadratic equation. To simplify it, we sometimes

substitute for the terms with fractional power as

will be shown in the next question.

41)By letting

, or otherwise, find the

values of for which

.

Solution

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Now let

and substitute this in the above

equation, which gives

Multiply through by and collect the like terms

( )( )

Therefore, either

Now substitute for

or

Again, substitute for

( )

NOTE

Alternatively,

(

)

(

) (

)

Therefore, either

or

( )

as before.

Section 11: Indicial Equations (II)

(Unknown as the Index)

INTRODUCTION
We are still dealing with indicial equations and in
this current section, the exponent (or index) is the
variable to solve for. Again, notes will accompany
the solution when appropriate.

42)Solve the following indicial equations.

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Hint

In the following questions, if there can be only one

term on each side of the equation such that they are

both in the same base to a certain power then

equate the index - the job is done! It is as easy as

that, so let’s get started.

(a)

Solution

(b)

Solution

( )

Therefore

(c)

Solution

Therefore

(d)

Solution

Therefore

(e)

Solution

( )

( )

Therefore

(f)

Solution

Therefore

(g)

Solution

Therefore, if

then

(h)

Solution

( ) ( )

( )( ) ( )

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Therefore

(i)

Solution

( )

Therefore

(j)

Solution

( )

Therefore

(k)( )

Solution

( )

Therefore

(l)

Solution

( ) ( )

Factorise

( )

( )

Divide both sides by

Therefore

NOTE

Alternatively, we could have solved this problem

by substitution as shown below. This approach

will be used for questions in No43.

Let thus

( ) ( )

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Divide both sides by 6

Substituting back, we have

Therefore

as before.

(m)

Solution

( )

Therefore

(n)

Solution

( )

( )

Therefore

(o)

Solution

( )

( )

Therefore

43)Solve the following equations.

Hint

In the following questions, the simplification will

lead to a quadratic equation and therefore two

solutions to the unknown variable.

(a) ( )

Solution

( )

( ) ( )

Let thus

( )( )

Therefore, either

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or

(b) ( )

Solution

( )

( ) ( )

( ) ( )

Let thus

( )( )

Therefore, either

or

(c) ( )

Solution

( )

( ) ( )

( ) ( )

Let thus

( )( )

Therefore, either

or

(d) ( )

Solution

( )

( )

( )

( ) ( )

Let thus

( )( )

Therefore, either

or

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(e) ( )

Solution

( )

( ) ( )

Let thus

( )( )

Therefore, either

or

(f) ( )

Solution

( )

( ) ( )

Let thus

( )( )

Therefore

( )

Section 12: Indicial Equations (III)

(Involving Logarithms)

INTRODUCTION
In this final category of indicial equations, we will
be dealing with equations where it will not be
practical to express all terms in the same base. The
way out in this case is to employ some rules (or
properties) of logarithms as will be shown shortly.
Whilst the examples use both natural and
common logs, logarithms to any base can be
employed.

44)Solve the following equations, giving the

results to 4 significant figures.

(a)

Solution

We will need to take the logarithm to base 10 of

both sides.

NOTE

Alternatively, we could have solved this problem

by converting the index form to logarithmic form

as

as before.

It also worth noting that since 7 cannot be

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57

expressed in the form of the where is any real

number ( ) the only viable means of

solving the problem is by using logarithmic

relationship.

(b)

Solution

( )

(c)

Solution

( )

( )

( )

( ) ( )

(d)

Solution

This implies

( ) ( )

( )

( ) ( )

( )

(e) ( )

Solution

( )

This implies

( )

( )

( )

Collect the like terms

( )

( ) [ (

)]

( )

(

)

(f)

Solution

( ) ( )

Open the brackets

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Collect the like terms

(

)

Simplify and make the subject

( ) (

)

( )

(

)

NOTE

Alternatively (for ease of evaluation using a

calculator),

(

)

as before.

(g) ( ) ( )

Solution

( ) ( )

This implies

( ) ( )

( ) ( ) ( )

( )

Open the brackets

Collect the like terms

Simplify and make the subject

( )

(

) [ (

)]

(

)

(

)

Section 13: Logarithmic Equations

INTRODUCTION
Another class of equations, known as logarithmic
equations, will be dealt with in this section. In
simple cases, we can convert a logarithmic
equation into an indicial equivalent. In a more
advanced and complex case, the problem is
simplified such that there is only a term on both
sides of the equation each expressed in log form
to the same base. Once this is done, the following
property of logarithm is applied to determine the
variable and the job is done.

45)Solve the following logarithmic equations

leaving the answer correct to 3 significant

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figures.

(a)( )

Solution

( )

Write this in index form

( )

(b) ( )

Solution

( )

Write this in index form

(c)( )

Solution

( )

Write this in index form

( )

√

√

(d) ( )

Solution

( )

Write this in index form

√

(e)

Solution

Write this in index form

(

)

√ √

NOTE

Whilst the values of are and , the

only one that satisfy the original logarithmic

equation is .

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60

However both values satisfy the modified version

of the equation, i.e. .

Alternatively, we can solve the equation as

follows

(

√

)

Therefore,

√

√

46)Solve the following logarithmic equations.

(a)( )

Solution

( )

( )

( )

( )

( )

( )

Thus

NOTE

Alternatively,

( )

change the base of the LHS to 3, we then have

( )

( )

( )

Thus

as before.

(b)( )

Solution

( )

( )

Therefore

( )

( )( )

Either

or

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NOTE

Although the values of are and , the only

one that satisfy the original logarithmic equation

is because log is not valid for negative

numbers. Therefore is not a solution to

this equation.

(c)( ) ( )

Solution

( ) ( )

( )( )

Therefore

( )( )

( )( )

Either

or

NOTE

Again, there is no solution for ( )

( ) as and do not

satisfy the above logarithmic equation.

(d)( ) ( )

Solution

( ) ( )

[

( )

( )
]

Therefore

( )

( )( )

Either

or

(e)( )

Solution

( )

√

√

Either

or

(f)( )

Solution

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( )

( )

Therefore

√

√( ) ( )( )

√

or

(g)( )

Solution

( )

( )

Therefore

( )( )

Either

or

(h)( )

Solution

( )

( )

Therefore

(√ )

( √ )( √ )

Either

√

√

or

√

√

√

NOTE

Alternatively, we can simply evaluate as

√

√

as before.

(i)( )

Solution

( )

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( )

Therefore

( )( )

Either

or

( )

(j)( )

Solution

( )

( )

Therefore

( )( )

Either

or

Section 14: Scientific Notation

INTRODUCTION
We are now through with indices and logarithms
- not absolutely though as we have one last part of
logarithm to deal with. Before then however, we
need to touch on expressing numbers in standard
form and this is the focus of this section.

47)Express the following numbers in standard

form.

(a)

Solution

(b)

Solution

(c)

Solution

(d)

Solution

(e)

Solution

(f)

Solution

(g)

Solution

(h)

Solution

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(i)

Solution

(j)

Solution

48)Evaluate the following, giving the result in

standard form.

(a)

Solution

( )( )

( )( )

( )( )

(b)

Solution

( )( )

( )(

)

( )( )

(c)( )( )

Solution

( )( )

( )( )

( )( )

(d)

( )( )

Solution

( )( )

(

) (

)

(

) (

)

Section 15: Solving Common

Logarithms using Log-Antilog Tables

INTRODUCTION
In this last section of this book, the primary
attention is on the use of log and antilog tables to
solve problems. It should be noted that answers
obtained using these tables are accurate only to
four significant figures; calculators would give
more accurate results. Kindly refer to pages 14 to
18 for a detailed description on their usage as
notes will be kept at minimal.

49)Express the following numbers in standard

form and write down the integer parts of their

logarithms.

Solution

Number

Standard form

Integer part

a)4.7

4.7 x 100

0

b)12.93

1.293 x 10

1

c)795

7.95 x 102

2

d)7 004

7.004 x 1033

e)25 500

2.55 x 1044

f)783 000

7.83 x 105

5

g)2 000 000

2.0 x 106

6

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65

50)Given that , without using

the log table, find the logarithm of the

following numbers.

Hint

Although we will not be using log-antilog tables in

the following questions, it is important to mention

that the tables were constructed based on the

approach used here. This allows several numbers to

be placed in a single cell of the tables.

(a)

Solution

( )

(b)

Solution

( )

(c)

Solution

( )

(d)

Solution

( )

(e)

Solution

( )

̅

NOTE

You may want to refer to section 8.2 on how to convert

a negative number into the characteristic and

mantissa and vice versa.

(f)

Solution

( )

̅

(g)

Solution

( )

̅

(h)

Solution

( )

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66

̅

51)Use a log table to express the following as

powers of 10.

Hint

In the following questions, we need to use our log

table to find the logarithm of the number and

subsequently express it in the power of 10 using the

relationship below.

(a)

Solution

(b)

Solution

(c)

Solution

(d)

Solution

52)Write down the logarithm of the following

numbers.

Hint

The solutions to the questions to follow are

presented in two-column tables. The left column

represents the number while the right column is the

corresponding logarithm. Log tables will be used in

this case.

(a)

Solution

Number Logarithm

80

1.9031

81

1.9085

82

1.9138

83

1.9191

84

1.9243

85

1.9294

(b)

Solution

Number Logarithm

23.0

1.3617

23.1

1.3636

23.2

1.3655

23.3

1.3674

23.4

1.3692

23.5

1.3711

(c)

Solution

Number Logarithm

191.0

2.2810

191.1

2.2812

191.2

2.2814

191.3

2.2817

191.4

2.2819

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67

191.5

2.2821

(d)

Solution

Number Logarithm

7.58

0.8797

75.8

1.8797

758

2.8797

7 580

3.8797

75 800

4.8797

758 000

5.8797

53)Write down the numbers whose logarithms

are given below.

Hint

The solutions to the questions to follow are

presented in two-column tables. The left column

represents the log of a number while the right

column is the number whose log is given in the left

column. In other words, the right column is the

antilog of the corresponding number in the left

column; therefore antilog tables should be used.

(a)

Solution

Logarithm Number

0.8900

7.762

0.8910

7.780

0.8920

7.798

0.8930

7.816

0.8940

7.834

(b)̅̅

Solution

Logarithm

Number

418.3

41.83

4.183

̅ 0.4 183

̅̅̅̅̅̅̅̅̅̅0.041 83

54)Calculate the following. Check your answers

with a calculator wherever possible.

Hint

Now it is time to show how the log-antilog tables

can be used to perform calculations which, without a

calculating device, would have been difficult or

impractical to undertake. Again, we will set this out

in a tabular form but an alternative method will be

shown at the very end. Kindly refer to a detailed

description on how this is carried out on pages 14 -18

particularly the table on pages 17 and18.

(a)

Solution

No

Log

3.46

0.5391

plus

72.41

1.8598

= 2.3989

Antilog = 2505

(b)

Solution

Antilog = 2505

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No

Log

576

2.7604

minus

25.8

1.4116

= 1.3488

Antilog = 2232

(c)

Solution

No

Log

5.503

0.7406

plus

14.64

1.1656

plus

202

2.3054

= 4.2116

Antilog = 1628

(d)

Solution

No

Log

105.7

2.0241

plus

63.44

1.8024

86.23 x 4058

= 3.8265

3.8265

minus

913.6

2.9608

= 0.8657

Antilog = 7340

(e)

Solution

No

Log

3.926

0.5940

3.9262

0.5940 x 2 =1.1880

plus

85.43

1.9317

= 3.1197

Antilog = 1317

(f)(

)

Solution

(
)

No

Log

8.245

0.9162

minus

0.8971

̅

0.9634

=0.9634

(
)

0.9634 x 3

= 2.8902

Antilog = 7766

(

)

(g)√

Solution

√

No

Log

0.3174

̅

Antilog = 7766

Antilog = 2232

Antilog = 1628

Antilog =1317

Antilog = 7340

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69

plus

0.1986

̅

= ̅̅

√

̅

̅

Antilog = 5010

√

(h)√

Solution

√

No

Log

19.2

1.2833

plus

12.3

1.0899

plus

75.4

1.8774

= 4.2506

4.2506

√

4.2506 ÷ 5

= 0.8501

Antilog = 7081

√

(i)√

Solution

√

No

Log

0.2511

̅

0.25115̅

̅̅

√

̅

̅

Antilog = 3161

√

(j)√

Solution

√

No

Log

1989

3.2987

minus

2014

3.3040

̅̅

√

̅

̅

Antilog = 9959

√

(k)

Solution

No

Log

1.840

0.2648

1.8402

0.2648 x 2

= 0.5296

0.5296

1.435

0.1568

plus

1.4353

0.1568 x 3

0.4704

=1.0000

Antilog =1000

(l)√

Solution

Antilog = 5010

Antilog = 9959

Antilog = 3161

Antilog = 10

00

Antilog = 2505

Antilog = 7081

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70

√

No

Log

313

2.4955

plus

99

1.9956

=4.4911

4.4911

minus

95.43

1.9796

=2.5115

√

2.5115÷ 3

=0.8372

Antilog = 6874

√

(m)

Solution

No

Log

41.53

1.6183

plus

70.03

1.8453

41.53 x 70.03

=3.4636

Numerator

3.4636

648.2

2.8117

plus

minus

2.547

0.4060

648.2 x 2.547

=3.2177

Denominator

3.2177

(Numerator) –

=0.2459

(Denominator)

Antilog = 1762

(n)

( )

√

Solution

( )

√

No

Log

57.5

1.7597

57.52

1.7597 x 2 = 3.5194

plus

7.88

0.8965

57.52 x 7.88

= 4.4159

Numerator

4.4159

679.8

2.8324

√

2.8324 ÷ 5

minus

Denominator

= 0.5665

0.5665

( )

√

(Numerator) –

(Denominator)

=3.8494

Antilog = 7069

( )

√

(o)√(

)

Solution

√(

)

No

Log

48.11

1.6822

plus

3.753

0.5743

= 2.2565

Antilog = 6874

Antilog = 7069

Antilog = 1762

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71

Numerator

2.2565

7.003

0.8453

plus

minus

6.099

0.7852

=1.6305

Denominator

1.6305

(Numerator) –

(Denominator)

=0.6260

(
)

0.6260

x 2

=1.2520

√(

)

=.2520

÷ 3

=

0.4173

Antilog = 2614

√(

)

NOTE

There is an alternative way to solve the problems

presented so far and the immediate last question

has been chosen to show this. This could be

applied to all.

√(

)

(

)

Take the log of

(

)

[ ]

Now, look in the log table for the logs of the

numbers above, which gives

[ ]

Simplify this to obtain

[ ]

In other word

(

)

Now look up for 0.4173 in the antilog table. This

gives 2614.

√(

)

55)Using the log and antilog tables, evaluate the

following correct to 3 significant figures. Take

the value of to be .

(a)The area of a rectangle which is 40.5 m long

and 3.4 m wide.

Solution

No Log

40.5 1.6075

plus

3.4 0.5315

= 2.1390

Antilog = 1377

Thus,

( )

(b)The volume of a cube of edge 3.56 mm.

Solution

No

Log

3.56 0.5514

Antilog = 2614

Antilog = 1377

Shefiu S. Zakariyah

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72

3.5630.5514 x 3

= 1.6542

Antilog = 4510

Thus,

( )

(c)The area of a circular disc 11.86 cm in

diameter.

Solution

Where

Therefore,

No

Log

0.4971

5.93

0.7731

5.932

0.7731 x 2

= 1.5462

0.4971+1.5462

=2.0433

Antilog = 1105

Thus,

( )

(d)The volume of a sphere with a radius of 3.5

mm.

Solution

No

Log

4

0.6021

plus

0.4971

3.5

0.5441

3.53

0.5441 x 3 = 1.6323

plus

=2.7315

Numerator

2.7315

3

0.4771

minus

Denominator

0.4771

(Numerator) –

(Denominator)

=2.2544

Antilog = 1797

Thus,

( )

END OF WORKED EXAMPLES

Antilog = 4510

Antilog = 1105

Antilog = 1797

Shefiu S. Zakariyah

[e-mail]

73

Bibliography and Further Reading

1)Bird, J. 2010. Higher Engineering Mathematics. 6th ed. Oxford: Newnes - Elsevier.
2)Croft, A., Davison, R. and Hargreaves, M., 1995. Introduction to Engineering

Mathematics. Harlow: Prentice Hall - Pearson Education.

3)Godman, A. and Talbert, J.F., 1975. Additional Mathematics Pure and Applied in SI

Units. 2nd ed. London: Longman.

4)James, G., Burley, D., Clements, D., Dyke, P., Searl, J. and Wright, J., 2010. Modern

Engineering Mathematics. 4th ed. Harlow: Prentice Hall - Pearson Education.

5)Morris, N.M., 1994. Mastering Mathematics for Electrical and Electronic Engineering.

London: Macmillan Press Ltd.

6)Neil, H. and Quadling, D., 2004. Advanced Mathematics Core 1 & 2. Cambridge:

Cambridge University Press.

7)Singh, K., 2011. Engineering Mathematics through Applications. 2nd ed. Basingstoke:

Palgrave Macmillan.

8)Stroud, K.A. and Booth, D.J., 2013. Engineering Mathematics. 7th ed. London: Palgrave

Macmillan.

9)Taylor, D.C. and Atkinson, I.S., 1981. Essential Mathematics for A Level. Edinburgh:

Thomas Nelson.

10)Tebbutt, P., 1994. Basic Mathematics for Chemists. Chichester: John Wiley & Sons Ltd.
11)Tuttuh Adegun, M.R., Sivasubramaniam, S. and Adegoke, R., 1992. Further

Mathematics Project 1. Ibadan: NPS Educational.

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