CLASS 7TH
SUBJECT MATH
EXPONENT.
The larger numbers can be written in shorter form using exponents.Observe the following :
1,00,000= 10 * 10 * 10 * 10 * 10 = 105
The short notation 105 stands for the product 10 * 10 * 10 * 10 * 10. Here ‘10’ is called the base and ‘5’ is the exponent. The number 105 is read as 10 raised to the power of 5 or simply as fifth power of 10. 105 is called the exponential form of 1,00,000.
We can also express 10000 as a power of 10. Since 10000 is 10 multiplied by itself 4 times,
10,000 = 10 * 10 * 10 * 10 = 104
Here, again, 104 is the exponential form of 10000.
Similarly, 10,00,000 = 10 * 10 * 10 * 10 * 10 * 10 = 106
What is the exponential form of 1,00,00,000 ?
In both these examples, the base is 10; in case of 104 the exponent is 4 and in case of 106 the exponent is 6.
We have already used these in writing numbers in an expanded form.
For example : 59432 = 5 * 10000 + 9 * 1000 + 4 * 100 + 3 * 10 + 2
which can be written as 5 * 104 + 9 * 103 + 4 * 102 + 3 * 10 + 2
In all the above given examples, the base is 10. However, it can be any other number also.
For example : 243 = 3 * 3 * 3 * 3 * 3 can be written as 243 = 35, where 3 is base and 5 is exponent.
Some powers have special names.
For example : 52 which is 5 raised to the power 2 is also read as ‘5 square’ and 53 which is 5 raised to the power 3 is also read as ‘5 cube’.
Can you tell what 43 (4 cube) means ?
43 means 4 is to be multiplied by itself three times i.e., 43 = 4 * 4 * 4 = 64
So, we can say 64 is the third power of 4.
What is the exponent and base in 43 ?
Similarly, 25 = 2 * 2 * 2 * 2 * 2 = 32 which is the fifth power of 2.
In 25, 2 is known as base and 5 as exponent.
In the same way,
81 = 3 * 3 * 3 * 3 = 34
64 = 2 * 2 * 2 * 2 * 2 * 2 = 26
3125 = 5 * 5 * 5 * 5 * 5 = 55
Instead of taking a fixed number, let us take any integer a as the base,
a * a = a2 (commonly read as ‘a square’ rather than ‘a raised to the power 2’)
a * a * a = a3 (read as ‘a cube’ rather than ‘a raised to the power 3’)
a * a * a * a = a4 (read as a raised to the power 4 or the 4th power of a)...
a * a * a * a * a * a * a = a7 (read as a raised to the power 7 or the 7th power of a) and so on.
a * a * a * b * b can be expressed as a3b2 (read as a cube 6 square)
a * a * b * b * b * b can be expressed as a2b4 (read as a square 6 raised to the power of 4)
Activity
Let us find the value of 20, 21, 22 ... practically.
Method :
1. Cut the coloured chart paper into four (or more) equal rectangular pieces, with the help of these four pieces we will be able to find the value of 20, 21, 22 and 23.
2. In this activity the number of folds are made according to the number at the base of a number in the exponential notation i.e., in 2x base is 2 so 2 parts folds are followed as so on.
Observation :
1. In fig.l, the rectangular piece represents the base 2.
2. From the four rectangular pieces. Take one rectangular piece.
Do not fold it (or we can say it has zero folds). This represents 1 piece of paper so fig.2, represents 20 = 1.
(2) number of 2 parts fold = number of parts into which a rectangular piece has been divided.
(2)0 = 1. This is representation of above step in the exponential form.
3. Take the second rectangular piece and fold it once along its width (or length) called as one 2 parts fold and then unfold it as shown in fig.3,
(2) number of 2 parts fold = number of parts into which a rectangular piece has been divided.
(2)1 = 2.
4. Take the third rectangular piece and fold it into two equal parts along its width (or length). Again fold this paper along the length (or width) to divide it further into 2 equal parts called as second 2 parts fold and then unfold it as shown in fig-4,
(2) number of 2 parts fold = number of parts into which a rectangular piece has been divided.
(2)2 = 4. This is representation of above step in the exponential form.
5. Take the fourth rectangular piece and proceed exactly in the same way as in step 4, then again fold the paper along the width (or length) to divide it further into two equal parts and then unfold it as shown in fig.5,
(2) number of 2 parts fold = number of parts
into which a rectangular piece has been divided.
(2)3 = 8. This is representation of above step in the exponential form.
6. In the same manner by increasing number of 2 parts fold we can obtain values of 24 = 16, 25 = 32 etc.
* Let us find the value of 30, 31, 32 ... practically.
Method :
1. Cut the coloured chart paper into four (or more) equal rectangular pieces, with the help of these four pieces we will be able to find the value of 30, 31, 32 and 33.
2. In this activity the number of folds are made according to the number at the base of a number in exponential notation i.e., in 3x; base is 3 so three parts fold are followed and so on.
Observation :
1. In fig. 6, the rectangular piece represents the base 3.
2. From the four rectangular pieces, take one rectangular piece. Do not fold it (or we can say it has zero folds). This represents 1 piece of paper. So fig. 7 represents 30 = 1.
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)0 = 1. This is representation of above step in the exponential form.
3. Take the second rectangular piece and fold it into three equal parts along its width (or length) called as one 3 parts folds and then unfold it as shown in fig. 8,
(2) number of 2 parts fold = number of parts into which a rectangular piece has been divided.
(2)1 = 2.
4. Take the third rectangular piece and fold it into two equal parts along its width (or length). Again fold this paper along the length (or width) to divide it further into 2 equal parts called as second 2 parts fold and then unfold it as shown in fig-4,
(2) number of 2 parts fold = number of parts into which a rectangular piece has been divided.
(2)2 = 4. This is representation of above step in the exponential form.
5. Take the fourth rectangular piece and proceed exactly in the same way as in step 4, then again fold the paper along the width (or length) to divide it further into two equal parts and then unfold it as shown in fig.5,
(2) number of 2 parts fold = number of parts
into which a rectangular piece has been divided.
(2)3 = 8. This is representation of above step in the exponential form.
6. In the same manner by increasing number of 2 parts fold we can obtain values of 24 = 16, 25 = 32 etc.
* Let us find the value of 30, 31, 32 ... practically.
Method :
1. Cut the coloured chart paper into four (or more) equal rectangular pieces, with the help of these four pieces we will be able to find the value of 30, 31, 32 and 33.
2. In this activity the number of folds are made according to the number at the base of a number in exponential notation i.e., in 3x; base is 3 so three parts fold are followed and so on.
Observation :
1. In fig. 6, the rectangular piece represents the base 3.
2. From the four rectangular pieces, take one rectangular piece. Do not fold it (or we can say it has zero folds). This represents 1 piece of paper. So fig. 7 represents 30 = 1.
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)0 = 1. This is representation of above step in the exponential form.
3. Take the second rectangular piece and fold it into three equal parts along its width (or length) called as one 3 parts folds and then unfold it as shown in fig. 8,
4. Take the third rectangular piece and fold it into three equal parts along its width (or length) called as one 3 parts fold. Again fold this paper along the length (or width) to divide it further into 3 equal parts called as second 3 parts fold and then unfold it as shown in fig.9,
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)2 = 9. This is representation of above step in the exponential form.
5. Take the fourth rectangular piece and proceed exactly in the same way as in step 4, then again fold the paper along the width (or length) to divide it further into three equal parts and then unfold it as shown in fig. 10,
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)3 = 27. This is representation of above step in the exponential form.
6. In the same manner by increasing number of 3 parts fold we can obtain values of 34 = 81, 35 = 243 etc.
Illustrative Examples
Example 2. Find the value of :
(i) 26 (ii) 54 (iii) 112
Solution. (i) 26 = 2 * 2 * 2 * 2 * 2 * 2 = 64
(ii) 54 = 5 * 5 * 5 * 5 = 625
(iii) 112 = 11 * 11 = 121
Example 3. Express the following in exponential form :
(i) 6 * 6 * 6 * 6
(ii) 5 * 5 * 7 * 7 * 7
(iii) 2 * 2 * a * a
Solution. (i) We have 6 * 6 * 6 * 6 = 64
(ii) We have 5 * 5 * 7 * 7 * 7 = 52 * 73
(iii) We have 2 * 2 * a * a = 22 * a2
Example 4. Identify the greater number, wherever possible, in each of the following :
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100
Solution. (i) We have 43 = 4 * 4 * 4 = 64 and 34 = 3 * 3 * 3 * 3 = 81
Clearly, 81 > 64
Hence, 34 is greater
(ii) We have 28 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256
and 82 = 8 x 8 = 64
Clearly, 256 > 64
Hence, 28 is greater.
(iii) We have 1002 = 100 x 100 = 10000
and 2100 = (210)10 = ( 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)10
= (1024)10 = [(1024)2]5 = (1024 x 1024)5 = (1048576)5
Clearly, 2100 is greater.
Example 5. Express the following numbers as a product of powers of prime factors :
(i) 648
(ii) 540
(iii) 3600
Solution. (i) 648 = 2 * 324 = 2 * 2 * 162
= 2 * 2 * 2 * 8 1
= 2 * 2 * 2 * 3 * 2 7
= 2 * 2 * 2 * 3 * 3 * 9
= 2 * 2 * 2 * 3 * 3 * 3 * 3
= 23 * 34
Thus, 648 = 23 * 34
(ii) 540 = 2 * 270 = 2 * 2 * 135
= 2 * 2 * 3 * 45
= 2 * 2 * 3 * 3 * 15
= 2 * 2 * 3 * 3 * 3 * 5
= 22 * 33 * 5
Thus, 540 = 22 * 33 * 5
(iii) 3600 = 2 * 1800
= 2 * 2 * 900
= 2 * 2 * 2 * 450
= 2 * 2 * 2 * 2 * 225
= 2 * 2 * 2 * 2 * 3 * 75
= 2 * 2 * 2 * 2 * 3 * 3 * 25
= 2 * 2 * 2 * 2 * 3 * 3 * 5 * 5
= 24 * 32 * 52
Thus, 3600 = 24 * 32 * 52
Example 6. Simplify :
(i) 2 * 103
(ii) 52 * 33
(iii)(-4)3
(iv) (-2)3 * (-10)3
Solution. (i) 2 * 1000 = 2000
(ii) 52 x 33 = 5 * 5 * 3 * 3 * 3
= 25 * 27
= 675
(iii) (-4)3 = (-4) * (-4) * (-4)
= -64
(iv) (-2)3 * (-10)3 = (-2 * -10)3
= (20)3
= 20 * 20 * 20
= 8000
Example 7. Compare 2.7 * 1012 and 1.5 * 108.
Solution. We have
2.7 * 1012 = 2.7 * 10 * 1011
= 27 * 1011, contains 13 digits
and 1.5 * 108 = 1.5 * 10 * 107
= 15 * 107, contains 9 digits
Clearly, 2.7 * 1012 > 1.5 * 108.
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)2 = 9. This is representation of above step in the exponential form.
5. Take the fourth rectangular piece and proceed exactly in the same way as in step 4, then again fold the paper along the width (or length) to divide it further into three equal parts and then unfold it as shown in fig. 10,
(3) number of 3 parts fold = number of parts into which a rectangular piece has been divided.
(3)3 = 27. This is representation of above step in the exponential form.
6. In the same manner by increasing number of 3 parts fold we can obtain values of 34 = 81, 35 = 243 etc.
Illustrative Examples
Example 2. Find the value of :
(i) 26 (ii) 54 (iii) 112
Solution. (i) 26 = 2 * 2 * 2 * 2 * 2 * 2 = 64
(ii) 54 = 5 * 5 * 5 * 5 = 625
(iii) 112 = 11 * 11 = 121
Example 3. Express the following in exponential form :
(i) 6 * 6 * 6 * 6
(ii) 5 * 5 * 7 * 7 * 7
(iii) 2 * 2 * a * a
Solution. (i) We have 6 * 6 * 6 * 6 = 64
(ii) We have 5 * 5 * 7 * 7 * 7 = 52 * 73
(iii) We have 2 * 2 * a * a = 22 * a2
Example 4. Identify the greater number, wherever possible, in each of the following :
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100
Solution. (i) We have 43 = 4 * 4 * 4 = 64 and 34 = 3 * 3 * 3 * 3 = 81
Clearly, 81 > 64
Hence, 34 is greater
(ii) We have 28 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256
and 82 = 8 x 8 = 64
Clearly, 256 > 64
Hence, 28 is greater.
(iii) We have 1002 = 100 x 100 = 10000
and 2100 = (210)10 = ( 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)10
= (1024)10 = [(1024)2]5 = (1024 x 1024)5 = (1048576)5
Clearly, 2100 is greater.
Example 5. Express the following numbers as a product of powers of prime factors :
(i) 648
(ii) 540
(iii) 3600
Solution. (i) 648 = 2 * 324 = 2 * 2 * 162
= 2 * 2 * 2 * 8 1
= 2 * 2 * 2 * 3 * 2 7
= 2 * 2 * 2 * 3 * 3 * 9
= 2 * 2 * 2 * 3 * 3 * 3 * 3
= 23 * 34
Thus, 648 = 23 * 34
(ii) 540 = 2 * 270 = 2 * 2 * 135
= 2 * 2 * 3 * 45
= 2 * 2 * 3 * 3 * 15
= 2 * 2 * 3 * 3 * 3 * 5
= 22 * 33 * 5
Thus, 540 = 22 * 33 * 5
(iii) 3600 = 2 * 1800
= 2 * 2 * 900
= 2 * 2 * 2 * 450
= 2 * 2 * 2 * 2 * 225
= 2 * 2 * 2 * 2 * 3 * 75
= 2 * 2 * 2 * 2 * 3 * 3 * 25
= 2 * 2 * 2 * 2 * 3 * 3 * 5 * 5
= 24 * 32 * 52
Thus, 3600 = 24 * 32 * 52
Example 6. Simplify :
(i) 2 * 103
(ii) 52 * 33
(iii)(-4)3
(iv) (-2)3 * (-10)3
Solution. (i) 2 * 1000 = 2000
(ii) 52 x 33 = 5 * 5 * 3 * 3 * 3
= 25 * 27
= 675
(iii) (-4)3 = (-4) * (-4) * (-4)
= -64
(iv) (-2)3 * (-10)3 = (-2 * -10)3
= (20)3
= 20 * 20 * 20
= 8000
Example 7. Compare 2.7 * 1012 and 1.5 * 108.
Solution. We have
2.7 * 1012 = 2.7 * 10 * 1011
= 27 * 1011, contains 13 digits
and 1.5 * 108 = 1.5 * 10 * 107
= 15 * 107, contains 9 digits
Clearly, 2.7 * 1012 > 1.5 * 108.
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