Problems on Definition of a Set:

Which of the following collections are set?

• The collection of easy topics of mathematics

The term used ‘easy topic’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of even natural numbers

It is a collection of well-defined objects. It is a set.

• The collection of rich people in the world

The term used ‘rich people’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of all books in Asiatic Library

It is a collection of well-defined objects. It is a set.

• The collection of clever students of your class

The term used ‘clever students in your class’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of numbers divisible by 3

It is a collection of well-defined objects. It is a set.

• a, e, i, o, u

It is a collection of well-defined objects. It is a set.

• The collection of English alphabet

It is a collection of well-defined objects. It is a set.

• The collection of happy persons in your country

The term used ‘happy person in your country’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of diagrams in your science book

It is a collection of well-defined objects. It is a set.

• 1/2, 1/3, 1/4, 1/5, 1/6

It is a collection of well-defined objects. It is a set.

• 2, 4, 8, 16, 32

It is a collection of well-defined objects. It is a set.

• The collection of the best TV channels

The term used ‘best TV channels’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of colours of rainbow

It is a collection of well-defined objects. It is a set.

• The collection of tasty foods

The term used ‘tasty foods’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of odd positive integers

It is a collection of well-defined objects. It is a set.

• The collection of poor people in Africa

The term used ‘poor people’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• The collection of whole numbers less than 10

It is a collection of well-defined objects. It is a set.

• Collections of vowels of English alphabet

It is a collection of well-defined objects. It is a set.

• A Collection of numbers multiple of 7

It is a collection of well-defined objects. It is a set.

• A collection of beautiful girls in your area

The term used ‘beautiful girls’ is a relative term and choice may vary from person to person. Hence it is not a collection of well-defined objects. It is not a set.

• A collection of integers

It is a collection of well-defined objects. It is a set.

Problems on Writing Set in Roster Form:

Write the following sets in Roster Form.

• A = {x| x ∈ I, -3 ≤ x ≤ 3}

Set in roster form is A = {-3, -2, -1, 0, 1, 2, 3}

• B = {x| x ∈ I, x ∉ W}

Set in roster form is B = {….., -3, -2, -1}

• C = {x| x ∈ I, x ∉ N}

Set in roster form is C = {….., -3, -2, -1, 0}

• D = {x| x ∈ I, x² < 10}

Set in roster form is D = {-3, -2, -1, 0, 1, 2, 3}

• E = {x| x ∈ I, -3 < x ≤ 3}

Set in roster form is A = {-2, -1, 0, 1, 2, 3}

• F = {x| x = (n/(n²-1)), n ∈ N, 2 < x ≤ 4}

n can take values 3 and 4

When n = 3, x = (n/(n²-1)) = (3/(3²-1)) = 3/8

When n = 4, x = (n/(n²-1)) = (4/(4²-1)) = 4/15

Set in roster form is F = {3/8, 4/15}

• G = {x| x is odd prime number less than 10}

Set in roster form is G = {2, 3, 5, 7}

Note: 1 is not a prime number

• H = {x| x is even multiple of 5 less than 60}

Set in roster form is H = {10, 20, 30, 40, 50}

• J = {x| x ∈ W, x ∉ N}

Set in roster form is J = {0}

• K = {x| x is types of triangle based on lengths of sides of triangle}

Set in roster form is K = {Equilateral triangle, Isosceles triangle, Scalene triangle}

• L = {x| x is types of triangle based on measures of angles of triangle}

Set in roster form is L = {Acute angled triangle, Right-angled triangle, Obtuse angled triangle}

• M = {x| x is a vowel of English Alphabet}

Set in roster form is M = {a, e, i, o, u}

• P = {x| x isgreatest two digit, three digit, and four digit number}

Set in roster form is P = {99, 999, 9999}

• S = {x| x ∈ N, 3x – 1 < 8}

When x = 1, 3x – 1 = 3(1) – 1 = 2 < 8, 1 ∈ P

When x = 2, 3x – 1 = 3(2) – 1 = 5 < 8, 2 ∈ P

When x = 3, 3x – 1 = 3(3) – 1 = 8 , 3 ∉ P

Set in roster form is S = {1, 2}

• T = {x| x is divisor of 24}

Set in roster form is T = {1, 2, 3, 4, 6, 8, 12, 24}

• V = {x| x is prime divisor of 24}

Set in roster form is V = {2, 3}

• Y = {x| x is a vowel in the word LOGARITHM}

Set in roster form is Y = {O, A, I}

• A = {x| x ∈ I, x² – 9 = 0}

x² – 9 = 0

x² = 9

x = 3 or x = -3

Set in roster form is A = {-3, 3}

• B = {x| x is non negative integer}

Set in roster form is B = {0, 1, 2, 3, ……….}

• C = {x| x is two-digit natural number such that the sum of its digit is 7}

Set in roster form isC = {16, 25, 34, 43, 52, 61, 70}

• D = {x| x is a square of prime number less than 10}

Set in roster form is D = {4, 9, 25, 49}

Problems on Writing Set in Set Builder Form:

Write the following sets in the Set-Builder Form.

• A = {9, 16, 25, 36, …..,81}

Set-builder form is A = {x| x = n², n ∈ N, 3 ≤ n ≤ 9}

• B = {8, – 8}

Set-builder form is B = {x| x ∈ I, x² – 64 = 0}

• C = {15, 24, 33, 42, 51, 60}

Set-builder form is C = {x| x is two digi natural number whose sum of digits is 6}

• D = {-4, 4}

Set-builder form is D = {x| x ∈ I, x² – 16 = 0}

• E = {1, 8, 27, 64, 125}

Set-builder form is E = {x| x = n³, n ∈ N, 1 ≤ n ≤ 5}

• F = {3, 9, 15, 21, 27, …….}

Set-builder form is F = {x| x = 3n, n is odd natural number}

• G = {1/2, 2/5, 3, 10, 4/17, 5/26}

Set-builder form is G = {x| x = (n/(n² + 1)), n ∈ N, 1 ≤ n ≤ 5}

• H = {a, a + 2, a + 4, a + 6, …………..}

Set-builder form is H = {x| x = a + 2n, n ∈ W}

J = {1, 2, 3, 4, 6, 12}

Set-builder form is J = {x| x is divisor of 12}

• K = {x – 3, x – 6, x – 9, x – 12, …………..}

Set-builder form is K = {x| x = x – 3n, n ∈ N}

• L = {2, 5, 8, 11, 14, ……….}

Set-builder form is L = {x| x = 3n – 1, n ∈ N}

• M = {17, 26, 35, 44, 53, 62, 71, 80}

Set-builder form is M = {x| x is two digit natural number whose sum of digits is 8}

• N = {2, 3, 5, 7, 11, 13}

Set-builder form is N = {x| x is natural prime mumber less than 15}

• P = {1, 2, 3, 6, 9, 18}

Set-builder form is P = {x| x is divisor of 18}

• Q = {5, 10, 15, 20, 25}

Set-builder form is Q = {x| x = 5n, n ∈ N, 1 ≤ n ≤ 5}

• S = {31, 33, 35, 37, 39}

Set-builder form is S = {x| x is odd number, x ∈ N, 30 ≤ x ≤ 40}

• T = {-2, -1, 0, 1, 2}

Set-builder form is T = {x| x ∈ I, -2 ≤ x ≤ 2}

• V = {2³, 2⁴, 2⁵, ……..}

Set-builder form is V = {x| x = 2ⁿ, n ∈ N, n ≥ 3}

• Y = {3², 3³, 3⁴, ……..}

Set-builder form is Y = {x| x = 3ⁿ, n ∈ N, n ≥ 2}

The post Roster Form and Set-Builder Form of Sets appeared first on The Fact Factor.

The basic operations on sets are:

• Union of sets
• Intersection of sets
• A complement of a set
• Set difference
• Cartesian product of sets.

Union of Sets:

If A and B are two sets, the union of A and B (written as A ∪ B) the set of all objects which either belong to A or to B or to both of them.

 A ∪ B = {x|x ∈ A or x ∈ B]

Venn Diagram for Union of Sets:

Example:

If A {1, 2, 3, 4, 5, 6} and B  {2, 4, 6, 8, 9}

then A ∪ B  {1, 2, 4, 5, 6, 8, 9}.

Here the elements 2, 4, and 6, which belong to both sets, are written only once. We can show A ∪ B by a Venn Diagram as follows:

Properties of Union of Sets:

Commutative Property of Union of Sets:

The Commutative Property for Union says that the order of the sets in which we do the operation does not change the result. Thus if A and B are two sets, then

A ∪ B = B ∪ A

Example:

Let A = {x : x is a whole number between 4 and 8} and

B = {x : x is an even natural number less than 10}.

A = {5, 6, 7}, B = {2, 4, 6, 8}

A ∪ B = {5, 6, 7} ∪ {2, 4, 6, 8} = {2, 4, 5, 6, 7, 8} ……….. (1)

B ∪ A = {2, 4, 6, 8} ∪ {5, 6, 7} = {2, 4, 5, 6, 7, 8} ……….. (2)

From (1) and (2) we get

A ∪ B = B ∪ A

Associative Property of Union of Sets:

The Associative Property for Union says that how the sets are grouped does not change the result. Thus, if A, B, and C are three sets, then

A ∪ (B ∪ C) = (A ∪ B) ∪ C

Example:

Let A = {x | x is a whole number between 4 and 8} and

B = {x | x is an even natural number less than 10}.

C = {x | x is an odd natural number less than 10}.

A = {5, 6, 7}, B = {2, 4, 6, 8}, C = {1, 3, 5, 7, 9}

A ∪ (B ∪ C) = {5, 6, 7} ∪ [ {2, 4, 6, 8} ∪ {1, 3, 5, 7, 9}]

A ∪ (B ∪ C) = {5, 6, 7} ∪ {1, 2, 3, 4, 5, 6, 7, 8, 9}

A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} ……….. (1)

(A ∪ B) ∪ C = [{5, 6, 7} ∪ {2, 4, 6, 8}] ∪ {1, 3, 5, 7, 9}

(A ∪ B) ∪ C = {2, 4, 5, 6, 7, 8} ∪ {1, 3, 5, 7, 9}

(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9} ……….. (2)

From (1) and (2) we get

A ∪ (B ∪ C) = (A ∪ B) ∪ C

Identity Property for Union:

The Identity Property for Union says that the union of a set and the empty set is the set itself. Thus, if A is a set, then

A ∪ ∅ = ∅ ∪ A = A

Example:

Let A = {3, 7, 11} and

B = {x | x is a natural number less than 0}.

B = {}

Then A ∪ B = {3, 7, 11} ∪ { } = {3, 7, 11} = A

The empty set is the identity element for the union of sets.

Intersection of Sets:

If A and B are two sets, the intersection of A and B (written as A ∩ B) the set of all objects which belong to A and to B.

 A ∩ B = {x|x ∈ A and x ∈ B}

Venn Diagram for Intersection of Sets:

Example:

If A {1, 2, 3, 4, 5, 6} and B  {2, 4, 6, 8, 9}

then A ∩ B  {2, 4, 6}.

We can show A ∩ B by a Venn Diagram as follows:

Properties of Intersection of Sets:

Commutative Property of Intersection of Sets:

The Commutative Property for Intersection says that the order of the sets in which we do the operation does not change the result. Thus if A and B are two sets, then

A ∩ B = B ∩ A

Example:

Let A = {x : x is a whole number between 4 and 8} and

B = {x : x is an even natural number less than 10}.

A = {5, 6, 7}, B = {2, 4, 6, 8}

A ∩ B = {5, 6, 7} ∪ {2, 4, 6, 8} = {6} ……….. (1)

B ∩ A = {2, 4, 6, 8} ∪ {5, 6, 7} = {6} ……….. (2)

From (1) and (2) we get

A ∩ B = B ∩ A

Associative Property of Intersection of Sets:

The Associative Property for Union says that how the sets are grouped does not change the result. Thus, if A, B, and C are three sets, then

A ∩ (B ∩ C) = (A ∩ B) ∩ C

Example:

Let A = {x | x is a whole number between 4 and 8} and

B = {x | x is an even natural number less than 10}.

C = {x | x is 3 multiple natural number less than 10}.

A = {5, 6, 7}, B = {2, 4, 6, 8}, C = {3, 6, 9}

A ∩ (B ∩ C) = {5, 6, 7} ∩ [ {2, 4, 6, 8} ∩ {3, 6, 9}]

A ∩ (B ∩ C) = {5, 6, 7} ∩ {6} = {6} ……….. (1)

A ∩ (B ∩ C) = [{5, 6, 7} ∩ {2, 4, 6, 8}] ∩ {3, 6, 9}

A ∩ (B ∩ C) = {6} ∩ {3, 6, 9} = {6} ……….. (2)

From (1) and (2) we get

A ∩ (B ∩ C) = (A ∩ B) ∩ C

Identity Property for Intersection:

The Identity Property for Intersection says that the intersection of a set and the empty set is an empty set. Thus, if A is a set, then

A ∩ ∅ = ∅ ∩ A = ∅

Example:

Let A = {3, 7, 11} and

B = {x | x is a natural number less than 0}.

B = {}

Then A ∩ B = {3, 7, 11} ∩ { } = { }

Distributive Properties:

Distributive Property of Union over Intersection:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Example:

Let A = {a, n, t}, B = {t, a, p}, and C = {s, a, p}

A ∪ (B ∩ C) = {a, n, t} ∪ [{t, a, p} ∩ {s, a, p}]

A ∪ (B ∩ C) = {a, n, t} ∪ {a, p} = {p, a, n, t} ……….. (1)

(A ∪ B) ∩ (A ∪ C) = [{a, n, t} ∪ {t, a, p}] ∩ [{a, n, t} ∪ {s, a, p}]

(A ∪ B) ∩ (A ∪ C) = {a, n, t, p} ∩ {a, n, t, p, s}

(A ∪ B) ∩ (A ∪ C) = {p, a, n, t} ……….. (2)

From (1) and (2)

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Distributive Property of Intersection over Union:

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Example:

Let A = {a, n, t}, B = {t, a, p}, and C = {s, a, p}

A ∩ (B ∪ C) = {a, n, t} ∩ [{t, a, p} ∪ {s, a, p}]

A ∩ (B ∪ C) = {a, n, t} ∩ {t, a, p, s} = {a, t} ……….. (1)

(A ∩ B) ∪ (A ∩ C) = [{a, n, t} ∩ {t, a, p}] ∪ [{a, n, t} ∩ {s, a, p}]

(A ∩ B) ∪ (A ∩ C) = {a, t} ∪ {a}

(A ∩ B) ∪ (A ∩ C) = {a, t} ……….. (2)

From (1) and (2)

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Complement of a Set:

Concept of Universal Set:

Whenever a number of sets A, B, C …. are under consideration, they are being thought of as subsets of some set X. Such a set X is called universal set.

For example, if A  {1, 2, 3, 4, 5}, B  {1, 5, 9, 12, 17} and C  {1, 2, 3, ….100} are sets under consideration, we can think of A, B, C as being subsets of the universal set N of natural numbers.

Complement of a Set:

Suppose A is a set which is the subset of a universal set X or U. Then the set of all the elements of X which do not belong to A is called the complement of A with respect to X and is denoted as A’ or A^c or X – A.

In the set builder notation: A’ = {x| x ∉ X and x ∈A}, where A is a subset of the universal set X.

For example, if N is the universal set and

A = {2x| x  ∈ N}i.e. A = {2, 4, 6, 8 …},

Then A’ ={1, 3, 5, 7, 9 ….} or A’ = {(2x-1)|x ∈ N}

A Venn diagram showing A’ is given in the figure. Here X is the universal set and the shaded portion represents A’.

In a Venn diagram showing complements, it is customary (though not compulsory) to draw the universal set X using a rectangle or square and the sets A, B, C, …. using circles or triangles.

Properties of Complement of a Set:

Complement law:

The union of a set A and its complement A’ gives the universal set U of which, A and A’ are a subset. i.e. A ∪ A’ = U

Example:

If A = {1 , 2 , 3 } and X = {1 , 2 , 3 , 4 , 5, 6 }, then A’ = {4 , 5, 6}.

Now, A ∪ A’ = { 1 , 2 , 3 , 4 , 5} = U

Disjoint Law:

A set and its complement, are disjoint sets or the intersection of a set A and its complement A’ gives the empty set ∅. i.e. A ∩ A’ = ∅

Example:

If A = {1 , 2 , 3 } and X = {1 , 2 , 3 , 4 , 5, 6 }, then A’ = {4 , 5, 6}.

Now, A ∩ A’ = { } = ∅

Law of Double Complementation: 

If we take the complement of the complemented set A’ then, we get the set A itself. i.e. (A’)’ = A

Example:

If A = {1 , 2 , 3 } and X = {1 , 2 , 3 , 4 , 5, 6 }, then A’ = {4 , 5, 6}

Now, (A’)’ = {1 , 2 , 3 } = A

Complement of Empty Set:

The complement of an empty set gives us a universal set.

If X = {1 , 2 , 3 , 4 , 5, 6 }, then ∅’ = {1 , 2 , 3 , 4 , 5, 6 } = X

Complement of Universal Set:

The complement of the universal set gives us the empty set.

Example:

If X = {1 , 2 , 3 , 4 , 5, 6 }, then X’ = { } = ∅

Demorgan’s law (I):

(A ∪ B)’ = A’ ∩ B’ (De Morgan’s law)

Example:

If A = {1 , 2 , 3, 7 }, B = {2, 4, 6} and X = {1 , 2 , 3 , 4 , 5, 6, 7 },

then A’ = {4 , 5, 6} and B’ = {1, 3, 5, 7}

A ∪ B = {1 , 2 , 3, 7 } ∪ {2, 4, 6} = {1, 2, 3, 4, 6, 7}

(A ∪ B)’ = {5} …………. (1)

A’ ∩ B’ = {4 , 5, 6} ∩ {1, 3, 5, 7} = {5} …………….. (2)

From (1) and (2)

(A ∪ B)’ = A’ ∩ B’

Demorgan’s law (II):

(A ∩ B)’ = A’ ∪ B’ (De Morgan’s law)

Example:

If A = {1 , 2 , 3, 7 }, B = {2, 4, 6} and X = {1 , 2 , 3 , 4 , 5, 6, 7 },

then A’ = {4 , 5, 6} and B’ = {1, 3, 5, 7}

A ∩ B = {1 , 2 , 3, 7 } ∩ {2, 4, 6} = {2}

(A ∩ B)’ = {1, 3, 4, 5, 6, 7} …………. (1)

A’ ∪ B’ = {4 , 5, 6} ∪ {1, 3, 5, 7} = {1, 3, 4, 5, 6, 7} …………….. (2)

From (1) and (2)

(A ∩ B)’ = A’ ∪ B’

Summary of Results:

• A ∪ A’ = A’ ∪ A = ∪ (Complement law)
• (A ∩ A’) = ϕ (Complement law)
• (A’)’ = A (Law of complementation)
• (ϕ’ = ∪ (Complement of empty set)
• ∪’ = ϕ (Complement of universal set)
• (A ∪ B) = A’ ∩ B’ (De Morgan’s law)
• (A ∩ B)’ = A’ ∪ B’ (De Morgan’s law)

Difference of Sets:

If A and B are two sets, then the difference of two sets A and B is equal to the set which consists of elements present in A but not in B. It is represented by A – B.

The difference of sets (A – B) is represented by the Venn diagram as follows:

Example:

If A = {1, 2, 3, 4, 5, 6, 7} and B = {5, 6, 7} are two sets.

Then, the difference of set A and set B is given by;

A – B = {1, 2, 3, 4}

Properties of Difference of Sets:

The difference of set A and set B is equal to the intersection of set A with the complement of set B. i.e. A − B = A ∩ B^’

Example:

If A = {1,2, 5,6,7}, B = {5, 6, 7} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8}

Then, the difference of set A and set B is given by;

A – B = {1,2} …………….. (1)

A ∩ B^’ = {1,2, 5,6,7} ∩ {1, 2, 3, 4, 8} = {1, 2} ……….. (2)

From (1) and (2)

A − B = A ∩ B^’

Cartesian Product:

Ordered Pair:

In Coordinate Geometry, when we say that a point has a coordinates (4, 5), it means that its x coordinate is 4 and y coordinate is 5.The pair of numbers 4 and 5 is written in a definite order in a bracket. Such a representation is an ordered pair. The ordered pair (5, 4) is different from the ordered pair (4, 5) as the order of the two elements is changed.

Hence in general, an ordered pair (x, y) represents two objects x and y which must be taken in that order.

Two ordered pairs (x, y) and (a, b) are equal if and only if x = a and y = b.

Cartesian Product of Sets:

The Cartesian product of two non-empty sets A and B is denoted by A×B and defined as the “collection of all the ordered pairs (a, b) such that a ∈ A and b ∈ B. a is called the first element and b is called the second element of the ordered pair (a, b).

A×B = {(a, b) : a ∈ A, b ∈ B}

Example 1:

If A = {a, b, c} and B = {1, 2}, then

A × B = {a, b, c} × {1, 2}

A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}.

We note that since A has 3 elements (which is written as n(A) = 3) and B has two elements (which is written as n(B) =2). A x B has 3 x 2 = 6 elements, or n(A x B) = 6.

Example 2:

If A = {a, b, c}, then

A × A = {a, b, c} × {a, b, c}

 A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.

Example 3:

R set of Real numbers. Hence R x R represents all possible ordered pairs of real numbers which are the coordinates of all points on the usual geometric plane. Hence R x R, (which is also written as R²) represents the usual coordinate plane.

The post Operations on Sets appeared first on The Fact Factor.

• N = Set of natural numbers = {1, 2, 3, 4, …},
• I = Set of integers = {…., -3, -2, -2, 0, 1, 2, 3, ……}
• Q = Set of rational numbers and
• R = Set of real numbers.

Elements of a Set:

If an object ‘a’ belongs to set A, this is written as ‘a ∈ A’ and read as ‘a belongs to set A’. The symbol ‘∈’  is a Greek letter ‘epsilon’ and is used to denote “belongs to”. Obviously, ∉ will mean “does not belong to”.

Example 1:

A  = {1, 2, 3, 4, 5}

Then ‘3 belongs to set A’ can be written as 3 ∈ A similarly ‘8 does not belong to set A’ is written as 8 ∉ A.

Example 2:

B   = Set of all the days in a week = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Then ‘Friday belongs to set B’ can be written as Friday ∈ A. Similarly, ‘January does not belong to B’ can be written as January ∉ B.

Number of Elements in a Set:

If A is a set then n(A) denotes the total number of elements in it provided it is finite.

Order or Cardinality of Set:

The order of a set defines the number of elements a set is having. It describes the size of a set. The order of set is also known as the cardinality of the set. The size of set whether it is is a finite set or an infinite set said to be set of finite order or infinite order, respectively.

B   = Set of all the days in a week = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

This set has 7 elements, hence the order of set B or cardinality of set B is 7.  It has finite order.

N = Set of natural numbers = {1, 2, 3, 4, ……..}

In this set there are infinite number of elements, hence the set has infinite order.

Representation of Sets:

Sets can be represented in two ways:

1. Roster Method or Tabular form
2. Set Builder Notation or Rule Method

Roster Method or Tabular Form:

In roster form, all the elements of the set are listed, separated by commas and enclosed between curly braces { }. Here each element of the set is listed.

Consider set A = {1, 2, 3, 4, 5}.

In this case, we are given a list of objects that belong to A. An object which is not included in this list will not be in A. Elements are separated by commas. Such a method of specifying a set is called the Roster method or Tabular method.

N = {1, 2, 3, 4, 5, ….} gives the set of natural numbers. Since it is impossible to list all the elements (natural numbers), the list is not completely given but the dots indicate that similar numbers (natural numbers) are included in this set.

“Set Builder Notation” or “Rule Method”:

The set A = {1, 2, 3, 4, 5) can also be written as

A = The set of natural numbers from 1 to 5 including 1 and 5.

Hence we could have stated that : “x is a member of set A if x is a natural number and 1 less than equal to x  less than equal to 5”. This long sentence can be written, in short as

A = {x| x ∈ N, 1 ≤ x ≤ 5}

which is read as: A is the set of objects x such that x Î N and 1 ≤ x  ≤ 5.

In general, if a set X contains objects having a property P in common, we write X in set builder notation as :

X = {x| x has property P}.

The set builder notation is very important as, in writing down many sets, where the roster method cannot be used.

For example, E =  {x| x ∈  Q, 1 ≤ x ≤ 5} cannot be written by roster method but this can be clearly written and understood in the set-builder notation.

Let us write down some sets in the set builder notation and convert them into the roster method.

A = {x| x ∈ I, -3 ≤ x ≤ 3} means A = {-3, -2, -1, 0, 1, 2, 3}.

B = {y| y ∈ N, 2 < x ≤ 7} means B = {3, 4, 5, 6, 7}.

C = {z| z is a vowel in English alphabet} means C = {a, e, i, o, u}

Types of Sets:

Singleton Set:

A set containing only one element is a singleton set,

Examples:

• the set {a} is a singleton set.
• The set A = set of all integers which are neither positive nor negative.” is a singleton set, as A = {0}. n(A) = 1, thus the order of the set is 1.

Null Set or Empty Set:

A set containing no element is called a null set [or an empty set]. It is usually denoted as f or {  }.

Examples:

• the set of all real numbers whose square is negative is a null set,
• the set {x | x Î N, 3 < x < 4} is a null set,

Note: The set {0} is not a null set as it contains one element, namely zero.

Finite Set:

A set which consists of a definite number of elements is called a finite set.

Example:,A = {1,2,3,4,5,6,7,8,9,10}

In this set there are 10 elements which can be counted. Thus n(A) = 10. The order of set A is 10. It is a finite set.

Infinite Set:

A set which is not finite is called an infinite set.

Example: N = set of all the natural numbers.

N = {1,2,3,4,5,6,7,8,9……}

In this set number of elements cannot be counted. Set N is of an infinite order. Hence set N is infinite set.

Equivalent Sets:

If the number of elements is the same for two different sets, then they are called equivalent sets. It is represented as:  n(A) = n(B)

Where, A and B are two different sets with the same number of elements.

Example: If A = {1, 2, 3, 4} and B = {Red, Blue, Green, Black}

In set A, there are four elements i.e. n(A) = 4 and in set B also there are four elements i.e. n(B) = 4. Thus n(A) = n(B). Therefore, set A and set B are equivalent.

Note: The elements of the two sets may be the same or different, but the number of elements in the set is equal i.e. their order should be the same.

Equal Sets: 

The two sets A and B are said to be equal if they have exactly the same elements, the order of elements do not matter.

Example: A = {1, 2, 3, 4} and B = {4, 3, 2, 1}

Mathematically, A = B

If A = {x| x is a letter in the word LET} = {L, E, T}, and

B = [{x| x is a letter in the word TELE} = {T, E, L},

Mathematically, A = B

A = {x| x is a letter in the word WOLF} = {W, O, L, F}

B = {x| x is a letter in the word FOLLOW} = {F, O, L, W}

Mathematically, A = B

In general sets, A and B are equal if and only if A ⊆ B and B ⊆ A

Disjoint Sets: 

The two sets A and B are said to be disjoint if the set does not contain any common element.

Example: Set A = {1, 2, 3, 4} and set B = {5, 6, 7, 8} are disjoint sets, because there is no common element between them.

Subset:

Consider the sets A = {1, 2, 3}, B = {1, 2, 3, 4, 5}.

We note that every element of set A is an element of set B. This can be expressed ordinarily as: ‘A is a part of B.’ We express this in set theory as ‘A  is a subject of B’ and write it as A ⊆ B.

We can define A ⊆ B as follows: We can say that A ⊆ B, if and only if every element of set A is an element of set B’. This in notation can be written as: A ⊆B if and only if, for every x ∈ A, it is true that x ∈ B.

Example:

If A = {x| x is a letter in the word LET} = {L, E, T} and B = [{x| x is a letter in the word LETTER} = {L, E, T, R},

A ⊆ B

Note: If A = {1, 2, 3} then {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3} are subsets of A.

Proper Subset:

If A and B are two sets for which A ⊆ B is true but B ⊆ A is not true we say that A is a proper subset of B and write it as A ⊂ B. Since B ⊆ A is not true, B must have at least one element not belonging to A.

Hence we can give the definition: We say that A ⊂ B if and only if

1. every element of A is an element of B, and
2. B has at least one element in it which does not belong to A.

Note: In usual notation N ⊂ W ⊂ I ⊂ Q ⊂ R

Superset:

If set A is a subset of set B and all the elements of set B are the elements of set A, then A is a superset of set B. It is denoted by A ⊃ B.

Example: If Set A = {1, 2, 3, 4} is a subset of B = {1, 2, 3, 4}. Then A is superset of B.

Universal Set:

A set which contains all the sets relevant to a certain condition is called the universal set. It is the set of all possible values. 

Example: If A = {1, 2, 3} and B = {2, 3, 4, 5}, then universal set here will be:

U = {1, 2, 3, 4, 5}

Power Set:

In set theory, the power set of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set. It is denoted by P(A). 

If A = {1, 2, 3} then {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3} are subsets of A.

Then power set of A = P(A) = {{}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

The number of elements in the power set of A = 2n, where n is the order of set A.

Venn Diagram:

A set can be represented by a closed figure like a circle, a triangle, etc. Such a representation is called the Venn diagram. The points inside the figure represent the members of the set.

Summary of Terminology and Notations of Sets:

The post Introduction to Concept of Sets appeared first on The Fact Factor.

Expand (a + b)²

Solution:

Expanding Binomially

(a + b)² = 2C₀a² + 2C₁a^2-1b¹ + 2C₂b²

(a + b)² = (1)a² + (2)a¹b¹ + (1)b²

(a + b)² = a² + 2ab+ b²

Note: (a – b)² = a² – 2ab+ b²

Example 02:

Expand (a + b)³

Solution:

Expanding Binomially

(a + b)³ = 3C₀a³ + 3C₁a^3-1b¹ + 3C₂a^3-2b² + 3C₃b³

(a + b)³ = (1)a³ + (3)a²b¹ + (3)a¹b² + (1)b³

(a + b)³ = a³ + 3a²b + 3ab² + b³

Note: (a – b)³ = a³ – 3a²b + 3ab² – b³

Example 03:

Expand (a + b)⁴

Solution:

Expanding Binomially

(a + b)⁴ = 4C₀a⁴ + 4C₁a^4-1b¹ + 4C₂a^4-2b² + 4C₃a^4-3b³ + 4C₄b⁴

(a + b)⁴ = (1)a⁴ + (4)a³b¹ + (6)a²b² + (4)a¹b³ + (1)b⁴

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Note: (a – b)⁴ = a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴

Example 04:

Expand (a + b)⁵

Solution:

Expanding Binomially

(a + b)⁵ = 5C₀a⁵ + 5C₁a^5-1b¹ + 5C₂a^5-2b² + 5C₃a^5-3b³ + 5C₄a^5-4b⁴ + 5C₅b⁵

(a + b)⁵ = (1)a⁵ + (5)a⁴b¹ + (10)a³b² + (10)a²b³ + (5)a¹b⁴ + (1)b⁵

(a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Note: (a – b)⁵ = a⁵ – 5a⁴b + 10a³b² – 10a²b³ + 5ab⁴ – b⁵

Example 05:

Expand (a + b)⁶

Solution:

Expanding Binomially

(a + b)⁶ = 6C₀a⁶ + 6C₁a^6-1b¹ + 6C₂a^6-2b² + 6C₃a^6-3b³ + 6C₄a^6-4b⁴ + 6C₅a^6-5b⁵ + 6C₆b⁶

(a + b)⁶ = (1)a⁶ + (6)a^6-1b¹ + (15)a^6-2b² + (20)a^6-3b³ + (15)a^6-4b⁴ + (6)a^6-5b⁵ + (1)b⁶

(a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶

Note: (a – b)⁶ = a⁶ – 6a⁵b + 15a⁴b² – 20a³b³ + 15a²b⁴ – 6ab⁵ + b⁶

Example 06:

Expand (2x + 3y)⁴

Solution:

Expanding Binomially

(2x + 3y)⁴ = 4C₀(2x)⁴ + 4C₁(2x)^4-1(3y)¹ + 4C₂(2x)^4-2(3y)² + 4C₃(2x)^4-3(3y)³ + 4C₄(3y)⁴

(2x + 3y)⁴ = (1) (2x)⁴ + (4) (2x)³(3y)¹ + (6) (2x)²(3y)² + (4) (2x)¹(3y)³ + (1) (3y)⁴

(2x + 3y)⁴ = 16x⁴ + (4) (8x³) (3y) + (6) (4x²) (9y²) + (4) (2x)(27y³) + (1) (81y⁴)

(2x + 3y)⁴ = 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

(2x – 3y)⁴ = 16x⁴ – 96x³y + 216x²y² – 216xy³ + 81y⁴

Example 07:

Expand (3x – 2y)⁴

Solution:

Expanding Binomially

(3x – 2y)⁴ = 4C₀(3x)⁴ – 4C₁(3x)^4-1(2y)¹ + 4C₂(3x)^4-2(2y)² – 4C₃(3x)^4-3(2y)³ + 4C₄(2y)⁴

(3x – 2y)⁴ = (1) (3x)⁴ – (4) (3x)³(2y)¹ + (6) (3x)²(2y)² – (4) (3x)¹(2y)³ + (1) (2y)⁴

(3x – 2y)⁴ = (1) (81x⁴) – (4) (27x³) (2y) + (6) (9x²) (4y²) – (4) (3x)(8y³) + (1) (16y⁴)

(3x – 2y)⁴ = 81x⁴ – 216x³y + 216x²y² – 96xy³ + 16y⁴

(3x + 2y)⁴ = 81x⁴ + 216x³y + 216x²y² + 96xy³ + 16y⁴

Example 08:

Expand (x² – 2y)⁵

Solution:

Expanding Binomially

(x² – 2y)⁵= 5C₀(x²)⁵ – 5C₁(x²)^5-1(2y)¹ + 5C₂(x²)^5-2(2y)² – 5C₃(x²)^5-3(2y)³ + 5C₄(x²)^5-4(2y)⁴ – 5C₅(2y)⁵

(x² – 2y)⁵= (1) (x¹⁰) – (5) (x⁸)(2y) + (10) (x⁶)(4y²) – (10) (x⁴)(8y³) + (5) (x²) (16y⁴) – (1) (32y⁵)

(x² – 2y)⁵= x¹⁰ – 10x⁸y + 40x⁶y² – 80x⁴y³ + 80x²y⁴ – 32y⁵

Example 09:

Expand (2x² + 3)⁴

Solution:

Expanding Binomially

(2x² + 3)⁴ = 4C₀(2x²)⁴ + 4C₁(2x²)^4-1(3)¹ + 4C₂(2x²)^4-2(3)² + 4C₃(2x²)^4-3(3)³ + 4C₄(3)⁴

(2x² + 3)⁴ = (1)(2x²)⁴ + (4)(2x²)³(3)¹ + (6)(2x²)²(3)² + (4)(2x²)¹(3)³ + (1)(3)⁴

(2x² + 3)⁴ = (16x⁸) + (4)(8x⁶)(3) + (6)(4x⁴) (9) + (4)(2x²) (27) + 81

(2x² + 3)⁴ = 16x⁸ + 96x⁶ + 216x⁴ + 216x² + 81

Example 10:

Expand (2x + y)⁴

Solution:

Expanding Binomially

(2x + y)⁴ = 4C₀(2x)⁴ + 4C₁(2x)^4-1(y)¹ + 4C₂(2x)^4-2(y)² + 4C₃(2x)^4-3(y)³ + 4C₄(y)⁴

(2x + y)⁴ = (1) (2x)⁴ + (4) (2x)³(y)¹ + (6) (2x)²(y)² + (4) (2x)¹(y)³ + (1) (y)⁴

(2x + y)⁴ = 16x⁴ + (4) (8x³) (y) + (6) (4x²) (y²) + (4) (2x) (y³) + (1) (y⁴)

(2x + y)⁴ = 16x⁴ + 32x³y + 24x²y² + 8xy³ + y⁴

Example 11:

Expand (x – 2y)⁴

Solution:

Expanding Binomially

(x – 2y)⁴ = 4C₀(x)⁴ – 4C₁(x)^4-1(2y)¹ + 4C₂(x)^4-2(2y)² – 4C₃(x)^4-3(2y)³ + 4C₄(2y)⁴

(x – 2y)⁴ = (1) (x⁴) – (4) (x³) (2y) + (6) (x²) (4y²) – (4) (x)(8y³) + (1) (16y⁴)

(x – 2y)⁴ = x⁴ – 8x³y + 24x²y² – 32xy³ + 16y⁴

Example 12:

Expand (x + 1)⁶

Solution:

Expanding Binomially

(x + 1)⁶  = 6C₀(x)⁶ + 6C₁(x)^6-1(1)¹ + 6C₂(x)^6-2(1)² + 6C₃(x)^6-3(1)³ + 6C₄(x)^6-4(1)⁴ + 6C₅(x)^6-5(1)⁵ + 6C₆(1)⁶

(x + 1)⁶ = (1)(x)⁶ + (6)(x)⁵(1)¹ + (15)(x)⁴(1)² + (20)(x)³(1)³ + (15)(x)²(1)⁴ + 6(x)¹(1)⁵ + (1)(1)⁶

(x + 1)⁶ = x⁶ + 6x⁵+ 15x⁴+ 20x³ + 15x²+ 6x + 1

Example 12:

Expand (x – 1)⁶

Solution:

Expanding Binomially

(x – 1)⁶  = 6C₀(x)⁶ – 6C₁(x)^6-1(1)¹ + 6C₂(x)^6-2(1)² – 6C₃(x)^6-3(1)³ + 6C₄(x)^6-4(1)⁴ – 6C₅(x)^6-5(1)⁵ + 6C₆(1)⁶

(x – 1)⁶ = (1)(x)⁶ – (6)(x)⁵(1)¹ + (15)(x)⁴(1)² – (20)(x)³(1)³ + (15)(x)²(1)⁴ – 6(x)¹(1)⁵ + (1)(1)⁶

(x – 1)⁶ = x⁶ – 6x⁵+ 15x⁴– 20x³ + 15x² – 6x + 1

Example 12:

Expand (3x² + 2y)⁵

Solution:

Expanding Binomially

(3x² + 2y)⁵ = = 5C₀(3x²)⁵ + 5C₁(3x²)^5-1(2y)¹ + 5C₂(3x²)^5-2(2y)² + 5C₃(3x²)^5-3(2y)³ + 5C₄(3x²)^5-4(2y)⁴ + 5C₅(2y)⁵

(3x² + 2y)⁵ = (1) (3x²)⁵ + (5) (3x²)⁴(2y)¹ + (10) (3x²)³(2y)² + (10) (3x²)²(2y)³ + (5) (3x²)¹(2y)⁴ + (1) (2y)⁵

(3x² + 2y)⁵ = (1) (243x¹⁰) + (5) (81x⁸) (2y) + (10) (27x⁶) (4y²) + (10) (9x⁴)(8y³) + (5) (3x²) (16y⁴) + (1) (32y⁵)

(3x² + 2y)⁵ = 243x¹⁰ + 810x⁸y + 1080x⁶y² + 720x⁴y³ + 240x²y⁴ + 32y⁵

Example 13:

Expand (x + 1/x)⁶

Solution:

Expanding Binomially

 (x + 1/x)⁶= 6C₀(x)⁶ + 6C₁(x)^6-1(1/x)¹ + 6C₂(x)^6-2(1/x)² + 6C₃(x)^6-3(1/x)³ + 6C₄(x)^6-4(1/x)⁴ + 6C₅(x)^6-5(1/x)⁵ + 6C₆(1/x)⁶

(x + 1)⁶ = (1)(x)⁶ + (6)(x)⁵(1/x)¹ + (15)(x)⁴(1/x)² + (20)(x)³(1/x)³ + (15)(x)²(1/x)⁴ + 6(x)¹(1/x)⁵ + (1)( 1/x)⁶

(x + 1)⁶ = x⁶ + 6x⁵(1/x) + 15x⁴(1/x²) + 20x³(1/x³) + 15x²(1/x⁴) + 6x(1/x⁵) +  1/x⁶

(x + 1)⁶ = x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ +  1/x⁶

Example 14:

Expand (2x – 1/x)⁵

Solution:

Expanding Binomially

(2x – 1/x)⁵ = 5C₀(2x)⁵ – 5C₁(2x)^5-1(1/x)¹ + 5C₂(2x)^5-2(1/x)² – 5C₃(2x)^5-3(1/x)³ + 5C₄(2x)^5-4(1/x)⁴ – 5C₅(1/x)⁵

(2x – 1/x)⁵ = (1)(2x)⁵ – (5)(2x)⁴(1/x)¹ + (10)(2x)³(1/x)² – (10)(2x)²(1/x)³ + (5)(2x)¹(1/x)⁴ – (1)(1/x)⁵

(2x – 1/x)⁵ = 32x⁵ – (5)(16x⁴) (1/x) + (10)(8x³) (1/x²) – (10)(4x²) (1/x³) + (5)(2x) (1/x⁴) – (1)(1/x⁵)

(2x – 1/x)⁵ = 32x⁵ – 80x³ + 80x – 40/x + 10/x³ – 1/x⁵

Example 15:

Expand (x² + 1/x²)⁵

Solution:

Expanding Binomially

(x² + 1/x²)⁵ = 5C₀(x²)⁵ – 5C₁(x²)^5-1(1/x²)¹ + 5C₂(x²)^5-2(1/x²)² – 5C₃(x²)^5-3(1/x²)³ + 5C₄(x²)^5-4(1/x²)⁴ – 5C₅(1/x²)⁵

(x² + 1/x²)⁵ = (1)( x²)⁵ – (5)( x²)⁴(1/x²)¹ + (10)( x²)³(1/x²)² – (10)( x²)²(1/x²)³ + (5)( x²)¹(1/x²)⁴ – (1)( 1/x²)⁵

(x² + 1/x²)⁵ = x¹⁰ – (5)( x⁸)(1/x²) + (10)( x⁶)(1/x⁴) – (10)( x⁴)(1/x⁶) + (5)( x²) (1/x⁸) – 1/x¹⁰

(x² + 1/x²)⁵ = x¹⁰ – 5 x⁶ + 10x² – 10/x² + 5/x⁶ – 1/x¹⁰

Example 6:

Expand (2x/3 – 3/2x)⁴

Solution:

Expanding Binomially

The post Binomial Theorem appeared first on The Fact Factor.

In the last article, we have studied to solve problems on the use of laws of logarithms to prove given logarithmic expression. In this article, we shall study to solve more problems on these laws to prove given relation using given logarithmic expression.

Example 01:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 02:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 03:

Solution:

Given

Dividing both sides by xy

Proved as required.

Example 04:

Solution:

Given

Proved as required.

Example 05:

Solution:

Given

Proved as required.

Example 06:

Solution:

Given

Proved as required.

Example 07:

Solution:

Given

Proved as required.

Example 08:

Given 2log₂(x + y) = 3 + log₂x + log ₂y, Show that x² + y² = 6xy

Solution:

Given 2log₂(x + y) = 3 + log₂x + log ₂y

log₂(x + y)² = 3 log₂2 + log₂x + log ₂y

log₂(x + y)² =  log₂2³ + log₂x + log ₂y

log₂(x + y)² =  log₂8 + log₂x + log ₂y

log₂(x + y)² =  log₂8xy

 (x + y)² =  8xy

x² + 2xy + y² = 8xy

x² + y² = 6xy

Proved as required.

Example 09:

If a² + b² = 3ab, show that

Solution:

a² + b² = 3ab

Adding 2ab on both sides

a² + 2ab + b² = 3ab + 2ab

(a + b)² = 5ab

Proved as required.

Example 10:

If a² -12ab+ 4b² = 0, show that

Solution:

a² -12ab+ 4b² = 0

a² + 4b² = 12ab

Adding 4ab on both sides

a² + 4ab + 4b² = 12ab + 4ab

(a + 2b)² = 16ab

Proved as required

Example 11:

If a² + b² = 7ab, show that

Solution:

a² + b² = 7ab

Adding 2ab on both sides

a² + 2ab + b² = 7ab + 2ab

(a + b)² = 9ab

Proved as required.

Example 12:

If b² = ac, prove that log a + log c = 2 log b

Solution:

b² = ac

Taking log on both sides

Log b² = log (ac)

2 log b = log a +log c

Log a + log c = 2 log b

Proved as required.

In the next article, we shall study to solve more problems on these laws to prove given logarithmic relation.

Click Here for More Subtopics in Logarithms

Click Here for More Topics in Mathematics

The post Use of the Laws of Logarithms: Set – III appeared first on The Fact Factor.

In this article, we shall study the laws of logarithm and their proofs. Laws of logarithm are very important in mathematics and every student should have confidence in using them.

Laws of Product:

log_a(mn) = log_am + log_an, where a, m, n are positive real numbers with a ≠ 1

Proof:

Let log_am = x and log_an = y   ……. (1)

By the definition of logarithm, we have

a^x = m and a^y = n

Now mn = a^x . a^y

mn = a^{x + y}

By definition of logarithm

log_a (mn) = x + y

Substituting values of x and y from relation (1) we get

log_a (mn) = log_am + log_an (Proved)

Corollary:

log_a(mnp) = log_am + log_an + log_ap

Proof:

L.H.S. =  log_a (mnp)

∴ L.H.S. = log_a (mn) + log_a (p)  (Law of product)

∴ L.H.S. = log_a m + log_a n + log_a p  (Law of product)

log_a(mnp) = log_am + log_an + log_ap (Proved)

Law of Quotient:

log_a(m/n) = log_am – log_an, where a, m, n are positive real numbers with a ≠ 1, n ≠ 0

Proof:

Let log_am = x and log_an = y   ……. (1)

By the definition of logarithm, we have

a^x = m and a^y = n

Now m/n = a^x / a^y

m/n = a^{x – y}

By definition of logarithm

log_a (m/n) = x – y

Substituting values of x and y from relation (1) we get

log_a (m/n) = log_am – log_an (Proved)

Corollary:

log_a(mn/pq) = log_am + log_an – log_ap – log_aq

Proof:

L.H.S. = log_a(mn/pq)

∴ L.H.S. = log_a(mn) – log_a(pq)   (Law of quotient)

∴ L.H.S. = (log_am + log_an) – (log_ap + log_aq)  (Law of product)

∴ L.H.S. = log_am + log_an – log_ap – log_aq

log_a(mn/pq) = log_am + log_an – log_ap – log_aq (Proved)

Law of Exponent:

log_amⁿ = n. log_am where a, m, n are positive real numbers with a ≠ 1

Proof:

Let log_am = x     ……. (1)

By the definition of logarithm, we have

a^x = m

mⁿ = (a^x)ⁿ = a^nx

By definition of logarithm

log_amⁿ  = n x

Substituting value of x from relation (1) we get

log_amⁿ  = n log_am  (Proved)

Corollary:

log_a(x^py^q/z^rw^s) = (p log_ax + q log_ay) – (r log_az + s log_aw)

Proof:

L.H.S. = log_a(x^py^q/z^rw^s)

∴ L.H.S. = log_a(x^py^q) – log_a(z^rw^s)  (Law of quotient)

∴ L.H.S. = (log_ax^p + log_ay^q) – (log_az^r + log_aw^s)  (Law of product)

∴ L.H.S. = (p log_ax + q log_ay) – (r log_az + s log_aw)  (Law of exponent)

∴ L.H.S. = p log_ax + q log_ay – r log_az – s log_aw 

log_a(x^py^q/z^rw^s) = (p log_ax + q log_ay) – (r log_az + s log_aw) (Proved)

Logarithm of 1 to Any Base

Log_a1 = 0 i.e. Logarithm to any base is always zero.

Proof:

Let log_a1 = x ………. (1)

By definition of logarithm

a^x = 1

a^x = a⁰

x = 0

Substituting this value in equation (1) we get

log_a1 = 0 (Proved)

Logarithm of Any Number to the Same Base:

log_aa = 1 i.e. Logarithm of any number to the same base is 1

Proof:

Let log_aa = x ………. (1)

By definition of logarithm

a^x = a

a^x = a¹

x = 1

Substituting this value in equation (1) we get

log_aa = 1 (Proved)

Important Property:

where a and m are positive real numbers with a ≠ 1

Proof:

Let log _am = x

By definition of logarithm

a^x = m

Substituting in equation (1)

(Proved)

Change of Base Law:

where a, b, and m are positive real numbers with a ≠ 1 and b ≠ 1

Proof:

Let log_am = x, log_bm = y and log_ba = z    ……. (1)

By the definition of logarithm, we have

a^x = m  ……….. (2)

b^y = m  ……….. (3)

b^z = a  ……….. (4)

From equations (2) and (3)

a^x = b^y    ……….. (5)

Substituting value of a from equation (4) in (5) we get

(b^z)^x = b^y

b^zx = b^y

zx = y

x = y/z

Substituting the values of x, y, and z from equation (1) we get

(Proved)

Corollary:

Proof:

We have

Substitute m = b

(Proved)

In the next few articles, we shall study how to use the laws of logarithm to evaluate or simplify given logarithmic expression.

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The post Laws of Logarithms appeared first on The Fact Factor.

In this article, we shall study the concept of logarithms and interconversion between exponential form and logarithmic form.

Laws of Indices:

• a^m × aⁿ = a^{m + n}
• a^m ÷ aⁿ = a^{m – n}           (a ≠ 0)
• (a^m)ⁿ = a^mn
• (ab)^m = a^m × b^m
• (a ÷ b)^m = a^m ÷ b^m    (b ≠ 0)
• a^-m  = 1/a^m                   (a ≠ 0)
• a⁰ = 1

Definition of Logarithm:

If m = a^x, where a > 0 and m > 0 then x is called the logarithm of m to the base a and is written as x = log_am and read as a log of m to the base a where m, a, x are real numbers.

Notes:

• The logarithm of a negative number and zero are not defined.
• Logarithm to the base 10 are called common or Briggsian logarithms.
• Logarithms to the base e, where e is an irrational number whose value is e = 2.7182… are called natural or Naperian logarithms.

Conversion from the Exponential Form into Logarithmic Form:

Conversion from the Logarithmic Form into the Exponential Form:

In the next article, we shall study to solve the problems based on the definition of a logarithm.

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The post Concept of Logarithm appeared first on The Fact Factor.

In the last article, we have studied the laws of logarithms and their proofs. In this article, we shall study to solve problems on laws of logarithms (law of product, Law of quotient, Law of exponent, etc.) to evaluate or simplify given logarithmic expression.

Laws of Logarithms

• Log a + Log b = Log (ab) (Law of Product)
• Log a – log b = log (a/b)    b ≠ 0 (law of Quotient)
• Log a^m = m Log a (Law of exponent)
• Log (1) = 0
• Log_aa = 1
• 
• 
• 

Example 01:

• Simplify log₁₀5 + 2 log₁₀4

Solution:

log₁₀5 + 2 log₁₀4

= log₁₀5 + log₁₀4² (Law of exponent)

= log₁₀5 + log₁₀16

= log₁₀(5 x 16) (Law of product)

= log₁₀80

Ans: log₁₀5 + 2 log₁₀4 = log₁₀80

Example 02:

• Simplify 2 log 7 – log 14

Solution:

2 log 7 – log 14

= log 7² – log 14 (Law of exponent)

= log 49 – log 14

= log (49/14) (Law of quotient)

= log (7/2)

Ans: 2 log 7 – log 14 = log (7/2)

Example 03:

• Simplify log₁₀3 + log ₁₀2 – 2log₁₀5

Solution:

log₁₀3 + log ₁₀2 – 2log₁₀5

= log₁₀3 + log ₁₀2 –log₁₀5² (Law of exponent)

= log₁₀3 + log ₁₀2 –log₁₀25

= log₁₀((3 x 2)/25) (Law of product and quotient)

= log₁₀(6/25)

Ans: log₁₀3 + log ₁₀2 – 2log₁₀5 = log₁₀(6/25)

Example 04:

• Simplify log 5 + log 3 – Log2

Solution:

log 5 + Log 3 – Log2

= log ((5 x 3)/2) (Law of product and quotient)

= log (15/2)

Ans: log 5 + Log 3 – Log2 = log (15/2)

Example 05:

• Simplify ½ log 9 + 1/3 log 27

Solution:

½ log 9 + 1/3 log 27

=  log 9^½ + log 27^1/3(Law of exponent)

=  log 3 + log 3

= 2 log 3

= log 3² (Law of exponent)

= log 9

Ans: ½ log 9 + 1/3 log 27 = log 9

Example 06:

• Simplify 2 log₁₀4 – ½ log₁₀16 + 1

Solution:

2 log₁₀4 – ½ log₁₀16 + 1

= log₁₀4² – log₁₀16^½ + log₁₀10 (Law of exponent)

= log₁₀16 – log₁₀4 + log₁₀10

= log₁₀((16 x 10)/4) (Law of product and quotient)

= log₁₀ (40)

Ans: 2 log₁₀4 – ½ log₁₀16 + 1 = log₁₀ (40)

Example 07:

Simplify 2log3 – ½ log 16 + log 12

Solution:

2log3 –  ½ log 16 + log 12

= log3² – log 16^½ + log 12 (Law of exponent)

= log9 – log 4 + log 12

= log ((9 x 12)/4) (Law of product and quotient)

= log (9 x 3)

= log 27

Ans: 2log3 –  ½ log 16 + log 12 = log 27

Example 08:

Simplify ½ log₅36 + 2log₅7 – ½ log ₅12

Solution:

½ log₅36 + 2log₅7 – ½ log ₅12

= log₅36^½ + log₅7² –log ₅12^½ (Law of exponent)

= log₅6 + log₅49 –log ₅2√3

= log₅((6 x 49)/ 2√3) (Law of product and quotient)

= log₅((3 x 49)/ √3)

= log₅((√3 x √3 x 49)/ √3)

= log₅(49√3)

Ans: ½ log₅36 + 2log₅7 – ½ log ₅12 = log₅(49√3)

Example 09:

• Simplify 2 log 3 + 3 log 2

Solution:

2 log 3 + 3 log 2

= log 3² + log 2³ (Law of exponent)

= log 9 + log 8

= log (9 x 8) (Law of product)

= log 72

Ans: 2 log 3 + 3 log 2 = log 72

Example 10:

• Simplify 2 log 5 + 3 log 4 – 4 log 2

Solution:

2 log 5 + 3 log 4 – 4 log 2

= log 5² + log 4³ – log 2⁴ (Law of exponent)

= log 25 + log 64 – log 16

= log ((25 x 64)/16) (Law of product and quotient)

= log (25 x 4)

= log 100

Ans: 2 log 5 + 3 log 4 – 4 log 2 = log 100

Example 11:

• Simplify log₁₀2 + 3

Solution:

log₁₀2 + 3

= log₁₀2 + 3 x 1

= log₁₀2 + 3 x log₁₀10 (Number and base same rule)

= log₁₀2 + log₁₀10³ (Law of exponent)

= log₁₀2 + log₁₀1000

= log₁₀(2 x 1000) (Law of product)

= log₁₀(2000)

Ans: log₁₀2 + 3 = log₁₀(2000)

Example 12:

• Simplify log (x² – 3x + 2) – log (x – 1) + log (x – 2)

Solution:

log (x² – 3x + 2) – log (x – 1) + log (x – 2)

= log (x – 2)(x – 1) – log (x – 1) + log (x – 2)

= log ((x – 2)(x – 1)(x – 2))/(x – 1)) (Law of product and quotient)

= log (x – 2)(x – 2)

= log (x – 2)²

= 2 log (x – 2) (Law of exponent)

Ans: log (x² – 3x + 2) – log (x – 1) + log (x – 2) = log (x – 2)²

Example 13:

Simplify

Solution:

Example 14:

Simplify

Solution:

Example 15:

Simplify

Solution:

Example 16:

Simplify

Solution:

Example 17:

Simplify

Solution:

Example 18:

Simplify

Solution:

Example 19:

Simplify

Solution:

Example 20:

Simplify

Solution:

Example 21:

Simplify

Solution:

Example 22:

Simplify

Solution:

Example 23:

Simplify

Solution:

In the next article, we shall study to solve more problems on the laws of logarithms to prove given logarithmic expression.

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The post Use of the Laws of Logarithms: Set – I appeared first on The Fact Factor.

In the last article, we have studied the concept of logarithm and interconversion between exponential form and logarithmic form. In this article, we shall study to solve problems based on the definition of logarithm.

Definition of Logarithm:

If m = a^x, where a > 0 and m > 0 then x is called the logarithm of m to the base a and is written as x = log_am and read as a log of m to the base a where m, a, x are real numbers.

Note the change in position of the terms during conversion from one form to another.

Example 01:

• Find the value of log₂32

Solution:

Let log₂32 = x

By definition of logarithm

∴ 2^x = 32

∴ 2^x = 2⁵

∴ x = 5

Ans: log₂32 = 5

Example 02:

• Find the value of log₁₀100000

Solution:

Let log₁₀100000 = x

By definition of logarithm

∴ 10^x = 100000

∴ 10^x = 10⁵

∴ x = 5

Ans: log₁₀100000 = 5

Example 03:

• Find the value of log₅3125

Solution:

Let log₅3125 = x

By definition of logarithm

∴ 5^x= 3125

∴ 5^x = 5⁵

∴ x = 5

Ans: log₅3125 = 5

Example 04:

• Find the value of log₅125

Solution:

Let log₅125 = x

By definition of logarithm

∴ 5^x = 125

∴ 5^x = 5³

∴ x = 3

Ans: log₅125 = 3

Example 05:

Find the value of log _1/2  8

Solution:

Let  log _1/2  8  = x

By definition of logarithm

∴ (1/2)^x = 8

∴ (2^-1)^x = 2³

∴ 2^-x = 2³

∴ -x = 3

∴ x = -3

Ans: log _1/2  8 = -3

Example 06:

• Find the value of log₈16

Solution:

Let log₈16 = x

By definition of logarithm

∴ (8)^x = 16

∴(2³)^x = 2⁴

∴ 2^3x = 2⁴

∴ 3x = 4

∴ x = 4/3

Ans: log₈16 = 4/3

Example 07:

• Find the value of log₆₄4

Solution:

Let log₆₄4 = x

By definition of logarithm

∴ 64^x = 4

∴ (2⁶)^x = 2²

∴ 2^6x = 2²

∴ 6x = 2

∴ x = 1/3

Ans: log₆₄4 = 1/3

Example 08:

• Find the value of log₁₆32

Solution:

Let log₁₆32 = x

By definition of logarithm

∴ 16^x = 32

∴ (2⁴)^x = 2⁵

∴ 2^4x = 2⁵

∴ 4x = 5

∴ x = 5/4

Ans: log₁₆32 = 5/4

Example 09:

• Find the value of log₉(1/81)

Solution:

Let Log₉(1/81) = x

By definition of logarithm

∴ 9^x = 1/81

∴ (3²)^x = 1/3⁴

∴ 3^2x = 3^-4

∴ 2x = -4

 ∴ x = -2

Ans: log₉(1/81) = -2

Example 10:

• Find the value of log₂(1/256)

Solution:

Let Log₂(1/256) = x

By definition of logarithm

∴ 1/256 = 2^x

∴ 1/2⁸ = 2^x

∴ 2^-8 = 2^x

∴ x = -8

Ans: log₂(1/256) = -8

Example 11:

• Find the value of log₈₁ 27

Solution:

Let log₈₁ 27 = x

By definition of logarithm

∴ 81^x = 27

∴ (3⁴)^x = 3³

∴ 3^4x = 3³

∴ 4x = 3

∴ x = ¾

Ans: log₈₁ 27 = 3/4

Example 12:

• Find the value of log₁₀0.001

Solution:

Let log₁₀0.001 = x

By definition of logarithm

∴ 10^x = 0.001

∴ 10 ^x = 1/1000

∴ 10 ^x = 1/10³

∴ 10 ^x = 10^-3

∴ x = -3

Ans: log₁₀0.001 = – 3

Example 13:

• Find the value of log₅0.008

Solution:

Let log₅0.008 = x

By definition of logarithm

∴ 5^x = 0.008

∴ 5^x = 8/1000

∴ 5^x = 1/125

∴ 5^x = 1/5³

∴ 5^x = 5^-3

∴ x = -3

Ans: log₅0.008 = -3

Example 14:

• Find the value of

Solution:

Example 15:

• Find the value of

Solution:

Example 16:

• Find the value of

Solution:

Example 17:

• Find the value of

Solution:

Example 18:

• Find the value of

Solution:

Example 19:

• Find the value of

Solution:

Example 20:

• Find the value of log₂(log_xx²)

Solution:

Log₂(log_xx²) = Log₂(2log_xx) = Log₂(2(1))

= Log₂2 = 1

Ans: Log₂(log_xx²)  = 1

Example 21:

• Find the value of log₅(log_xx²)

Solution:

Log₂(log₃3) = Log₅(1) = 0

Ans: Log₅(log_xx²) = 0

Example 22:

Find the value of

Solution:

Example 23:

Find the value of

Solution:

In the next article, we shall see laws of the logarithm, and problems based on them.

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The post Problems Based on Definition of Logarithm appeared first on The Fact Factor.

In the last article, we shall study problems based on the change of base rule. In this article, we shall study the use of definition and the laws of logarithms to find the value of ‘x’ (solve for x).

Example 01:

Solve for x, if log₅x = 2

Solution:

Given: log₅x = 2

∴ x = 5² = 25

∴ x = 25

Example 02:

Solve for x, if log₂(1/2) = x

Solution:

Given: log₂(1/2) = x

∴ log₂2^-1 = x

∴ x = -1 Log22

∴ x = – 1(1)

∴ x = -1

Example 03:

Find x, if log₄(x – 2) = 2

Solution:

Given: log₄(x – 2) = 2

∴ x – 2 = 4²

∴ x – 2 = 16

∴ x = 18

Example 04:

Find x, if log₅(x² – 5x + 11) = 1

Solution:

Given: log₅(x² – 5x + 11) = 1

∴ x² – 5x + 11 = 5¹

∴ x² – 5x + 6 = 0

∴ (x – 3)(x – 2) = 0

∴ x – 3 = 0 or x – 2 = 0

∴ x = 3 or x = 2

Both x = 3 and x = 2 satisfy given relation

x = 3 or x = 2

 Example 05:

Find x, if log₂x – log₂(x – 1) = 5

Solution:

Given: Log₂x – log₂(x – 1) = 5

∴ x = 32x – 32

∴31x = 32

∴ x = 32/31

Example 06:

Find x, if log x + log(x + 1) = log 6

Solution:

Given: log x + log(x + 1) = log 6

∴ log x(x + 1) = log 6

∴x(x + 1) = 6

∴ x² + x – 6 = 0

∴ (x + 3)(x – 2) = 0

∴ x = -3 or x = 2

x = -3 is not possible because log of negative number not defined.

∴ x = 2

Example 07:

Find x, if log (1 + x) – log (1 – x) = 1

Solution:

Given: log (1 + x) – log (1 – x) = 1

Example 08:

Find x, if log₃(x + 6) = 2

Solution:

Given: log₃(x + 6) = 2

∴ x + 6 = 3² = 9

∴ x = 3

Example 09:

Find x, if log (x + 3) + log (x – 3) = log 16, find x

Solution:

Given: log (x + 3) + log (x – 3) = log 16

∴ log [(x + 3)(x – 3)] = log 16

∴ log (x² – 3²) = log 16

∴ x² – 9 = 16

∴ x²  = 25

∴ x = ± 5

∴ x = -5 is not possible because the log of negative numbers not defined.

∴ x = 5

Example 10:

Find x, if log (3x + 2) + log (3x – 2) = log 5

Solution:

Given: log (3x + 2) + log (3x – 2) = log 5

∴ log [(3x + 2)(3x – 2)] = log 5

∴ log ((3x)² – 2²) = log 16

∴ 9x² – 4 = 5

∴ 9x²  = 9

∴ x²  = 1

∴ x = ± 1

x = -1 is not possible because log of negative number not defined.

∴  x = 1

Example 11:

Find x, if log (3x + 2) – log (3x – 2) = log 5

Solution:

Given: log (3x + 2) – log (3x – 2) = log 5

∴ 3x + 2 = 15x – 10

∴ 12 = 12x

∴x = 12/12 = 1

Example 12:

Find x, if

Solution:

Given:

∴ log(x + 6) = 2 log x

∴ log(x + 6) = log x²

∴ x + 6 = x²

∴ x² – x – 6 = 0

∴ (x – 3)(x + 2) = 0

∴ x – 3 = 0 and x + 2 = 0

∴ x = 3 or x = -2

x = -2 is not possible because log of negative number not defined.

∴ x = 3

Example 13:

Find x, if log₃x + log₃4 = 2

Solution:

Given: log₃x + log₃4 = 2

∴ log₃4x = 2

∴ 4x = 3² = 9

∴ x = 9/4

Example 14:

Find x, if log₂(x² + 7) = 4

Solution:

Given: log₂(x² + 7) = 4

∴ x² + 7 = 2⁴

∴ x² + 7 = 16

∴ x² = 9

∴ x = ±3

both x = 3 and x = -3 satisfy the given relation

∴ x = 3 and x = -3

Example 15:

Find x, if log₂x + log₂(x + 2) = 3

Solution:

Given: log₂x + log₂(x + 2) = 3

∴ log₂x(x + 2) = 3

∴ x(x + 2) = 2³

∴ x² + 2x = 8

∴ x² + 2x – 8 = 0

∴ (x + 4)(x – 2) = 0

∴  x + 4 = 0 or x – 2 = 0

∴ x = -4 and x = 2

x = -4 is not possible because the log of negative numbers not defined.

∴ x = 2

In the next article, we shall study to solve more problems on the laws of the logarithm to find the value of x.

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The post Use of the Laws of Logarithms: Set – VI appeared first on The Fact Factor.