Exponents and Powers
1 Introduction
Do
you know what the mass of earth is?
It
is 5,970,000,000,000,000,000,000,000 kg!
Can
you read this number?
Mass
of Uranus is 86,800,000,000,000,000,000,000,000 kg. Which has greater mass,
Earth or Uranus?
Distance
between Sun and Saturn is 1,433,500,000,000 m and distance between Saturn and
Uranus is 1,439,000,000,000 m. Can you read these numbers? Which distance is
less?
These
very large numbers are difficult to read, understand and compare. To make these
numbers easy to read, understand and compare, we use exponents. In this
Chapter, we shall learn about exponents and also learn how to use them
2 Exponents
We can write large numbers in a
shorter form using exponents.
Observe 10, 000 = 10 × 10 × 10 × 10 = 104
The short notation 104
stands for the product 10×10×10×10. Here ‘10’ is called the base and ‘4’ the
exponent. The number 104 is read as 10 raised to the power of 4 or
simply as fourth power of 10.
104
is called the exponential form of 10,000.
We can similarly express 1,000 as a
power of 10. Note that
1000 = 10 × 10 × 10 = 103
Here again, 103 is the
exponential form of 1,000.
Similarly, 1, 00, 000 = 10 × 10 × 10
× 10 × 10 = 105
105 is the exponential
form of 1, 00,000
In both these examples, the base is
10; in case of 103, the exponent is 3 and in case of 105
the exponent is 5.
In all the above given examples, we
have seen numbers whose base is 10. However the base can be any other number
also.
For
example: 81 = 3 × 3 × 3 × 3 can be written as 81 = 34, here 3 is the
base and 4 is the exponent.
Some powers have special names.
For
example,
102, which is 10 raised
to the power 2, also read as ‘10 squared’ and 103, which is 10
raised to the power 3, also read as ‘10 cubed’.
You can also extend this way of
writing when the base is a negative integer.
What does (–2)3 mean?
It is (–2)3 = (–2) × (–2)
× (–2) = – 8
Is (–2)4 = 16? Check it
Instead of taking a fixed number let
us take any integer a as the base, and write the numbers as,
a × a = a2 (read
as ‘a squared’ or ‘a raised to the power 2’)
a
× a × a = a3 (read as ‘a cubed’ or ‘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
soon
a × a × a × b × b can be
expressed as a3 b2 (read as a cubed b squared)
Example 1: Express 256 as a power
2.
Solution
We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2
× 2.
So we can say that 256 = 28
Example 2: Which one is greater 23
or 32 ?
Solution
We have, 23 = 2 × 2 × 2 = 8
and 32 = 3 × 3 = 9.
Since 9 > 8, so, 32 is
greater than 2
Example 3: Expand a3 b2, a2 b3, b2 a3, b3 a2. Are they all same?
Solution
a3 b2
= a3 × b2
= (a × a × a) × (b × b)
= a × a × a × b × b
a2 b3
= a2 × b3
= a × a × b × b × b
b2 a3
= b2 × a3
= b × b × a × a × a
b3 a2
= b3 × a2
= b × b × b × a × a
Note
that in the case of terms a3 b2 and a2 b3
the powers of a and b are different. Thus a3 b2 and a2
b3 are different.
On
the other hand, a3 b2 and b2 a3 are
the same, since the powers of a and b in these two terms are the same. The
order of factors does not matter.
Thus, a3
b2 = a3 × b2 = b2 × a3 =
b2 a3. Similarly, a2 b3 and b3
a2 are the same.
Example 4: Express the following
numbers as a product of powers of prime factors:
(i) 72 (ii) 432 (iii) 1000 (iv)
16000
Solution
(i)
72 = 2 × 36 = 2 × 2 × 18
= 2 × 2 × 2 × 9
= 2 × 2 × 2
× 3 × 3
= 23 × 32
Thus, 72 = 23
× 32 (required prime factor product form).
(ii)
432 = 2 × 216 = 2 × 2 × 108 = 2
× 2 × 2 × 54
= 2 × 2 × 2 × 2 × 27 =
2 × 2 × 2 × 2 × 3 × 9
= 2 × 2 × 2
× 2 × 3 × 3 × 3
or
432 = 24 × 33
(iii)
1000 = 2 × 500 = 2 × 2 × 250 =
2 × 2 × 2 × 125
= 2 × 2
× 2 × 5 × 25 = 2 × 2 × 2 × 5 × 5 × 5
Or
1000 = 23 × 53
(iv)
16,000 = 16 × 1000 = (2 × 2 × 2
× 2) ×1000 = 24 ×103 (as 16 = 2 × 2 × 2 × 2)
= (2 × 2 × 2 × 2) × (2 × 2 × 2 × 5
× 5 × 5)
= 24 × 23 × 53 (Since 1000 = 2 × 2 × 2
× 5 × 5 × 5)
=
(2 × 2 × 2 × 2 × 2 × 2 × 2 ) × (5 × 5 × 5)
Or
16,000 = 27 ×
53
Example 5 :Work out (1)5 ,
(–1)3 , (–1)4 , (–10)3 , (–5)4 .
Solution
(i)
We have (1)5 = 1 × 1
× 1 × 1 × 1 = 1
In fact, you will
realise that 1 raised to any power is 1.
(ii)
(–1)3 = (–1) × (–1)
× (–1) = 1 × (–1) = –1
(iii)
(–1)4 = (–1) × (–1)
× (–1) × (–1) = 1 ×1 = 1
You may check that (–1)
raised to any odd power is (–1), and (–1) raised to any even power is (+1).
(iv)
(–10)3 = (–10) ×
(–10) × (–10) = 100 × (–10) = – 1000.
(v)
(–5)4 = (–5) × (–5)
× (–5) × (–5) = 25 × 25 = 625
3 Laws Of
Exponents
3.1 Multiplying Powers with the Same Base
(i)
Let us calculate 22
× 23
22
× 23 = (2 × 2) × (2 × 2 × 2)
= 2 × 2 × 2
× 2 × 2 = 25 = 22+3
Note
that the base in 22 and 23 is sa2me and the sum of the
exponents, i.e., 2 and 3 is 5
(ii)(–3)4
× (–3)3 = [(–3) × (–3) × (–3)× (–3)] × [(–3) × (–3) × (–3)]
= (–3) × (–3) × (–3) × (–3) × (–3) ×
(–3) × (–3) = (–3)7 = (–3)4+3 .
Again,
note that the base is same and the sum of exponents, i.e., 4 and 3, is 7
(iii)
a2 × a4 =
(a × a) × (a × a × a × a)
= a × a × a × a × a × a = a6
(Note:
the base is the same and the sum of the exponents is 2 + 4 = 6)
Similarly,
verify: 42 × 42 =
42+2,
32
× 33 = 32+3
Caution!
Consider
23 × 32 Can you add the exponents? No! Do you see ‘why’?
The base of 23 is 2 and base of 32 is 3. The bases are
not same.so,we can’t add
exponents.
3.2 Dividing
Powers with the Same Base
Let
us simplify 37 ÷ 34 ?
Thus,
(Note,
in 37 and 34 the base is same and 37 ÷ 34
becomes 37-4)
Let
a be a non-zero integer, then,
Thus,
In
general, for any non-zero integer a,
am
÷ an = am-n
Where
m and n are whole numbers and m > n.
3.3 Taking Power
of a Power
Consider
the following
Simplify
Now,
means 23 is multiplied two times
with itself.
(Since am × an
= am+n)
Thus
Similarly
From
this we can generalise for any non-zero integer ‘a’, where ‘m’ and ‘n’ are
whole numbers,
3.4 Multiplying
Powers with the Same Exponents
Can
you simplify 23 × 33 ? Notice that here the two terms 23
and 33 have different bases, but the same exponents.
Now,
23 × 33
= (2 × 2 × 2) × (3 × 3 × 3)
= (2 × 3) × (2 × 3) × (2 × 3)
= 6 × 6 × 6 = 63
(Observe 6 is the product of bases 2 and 3)
Consider 44 × 34 = (4 × 4 × 4 × 4) × (3 × 3 × 3 × 3)
= (4 × 3) ×
(4 × 3) × (4 × 3) × (4 × 3)
= 12 × 12 ×
12 × 12
= 124
Consider,
also, 32 × a2 = (3 × 3) × (a × a)
= (3 × a) ×
(3 × a)
= (3 × a)2
= (3a)2 (Note: 3×a = 3a)
Similarly,
a4 × b4 = (a × a × a × a) × (b × b × b × b)
= (a × b) ×
(a × b) × (a × b) × (a × b)
= (a × b)4
= (ab)4 (Note a × b = ab)
In
general, for any non-zero integer a
am × bm = (ab)m
Example 1: Express the following terms in
the exponential form:
(i) (2 × 3)5 (ii) (2a)4 (iii) (– 4m)3
Solution
(i)
(2 × 3)5 = (2 × 3) ×
(2 × 3) × (2 × 3) × (2 × 3) × (2 × 3)
= (2 × 2 × 2 × 2 × 2) × (3 × 3× 3 × 3 ×
3)
= 25 × 35
(ii)
(2a)4 = 2a × 2a × 2a
× 2a
= (2 × 2 × 2 × 2) × (a × a × a × a)
= 24 × a4
(iii)
(– 4m)3 = (– 4 × m)3
= (– 4 × m) × (– 4 × m) × (– 4 × m)
= (– 4) × (– 4) × (– 4) × (m × m × m)
= (– 4)3 × (m)3
3.5 Dividing
Powers with the Same Exponents
Observe
the following simplifications:
(i)
(ii)
From
these examples we may generalise
Where
a and b are any non-zero integers and m is a whole number.
Example 1: Expand: (i) (ii)
Solution
(i)
(ii)
v
Numbers with exponent zero
Can
you tell what
equals to?
By
using laws of exponents
So,
Similarly
And
Thus
a0
= 1 (for any non-zero integer a)
So,
we can say that any number (except 0) raised to the power (or exponent) 0 is 1.
Example 1: Write
exponential form for 8 × 8 × 8 × 8 taking base as 2.
Solution
We
have, 8 × 8 × 8 × 8 = 84
But
we know that
8 = 2 × 2 × 2 = 23
Therefore
84 = (23
)4
= 23
× 23 × 23 × 23
Example 1: Simplify
and write the answer in the exponential form.
(i) (ii) 23
× 22 × 55 (iii)
(62 × 64 ) ÷ 63 (iv) [(22 )3 × 36 ] × 56
(v) 82 ÷ 23
Solution
(i)
= (37-2) ×35 = 35
×35
= 35+5 = 310
(ii)
23 × 22 ×
55
= 23+2 × 55 = 25 ×
55
= (2 × 5)5 = 105
(iii)
(62 × 64
) ÷ 63
66-3 = 63
(iv)
[(22 )3 ×
36 ] × 56
= [26 × 36] × 56
= (2 × 3)6 × 56
= (2×3 ×5)6
= 306
(v)
82 ÷ 23
82=
2 × 2 × 2 = 23
Therefore 82
÷ 23 = (23)2 ÷ 23 = 26 ÷
23 = 26-3 = 23
Example 2: Simplify:
(i) (ii) 23 × a3 × 5a4
Solution
(i)
We have
= 210-9
× 310-6 = 21 ×34
= 2 × 81 = 162
(ii)
23 × a3 ×
5a4
= 23 × a3 × 5 × a4
= 23 × 5 × a3 × a4
= 8 × 5 × a3+4
= 40 a7
5 Decimal Number
System
Let
us look at the expansion of 47561, which we already know:
47561 = 4 × 10000 + 7 × 1000 + 5 ×
100 + 6 × 10 + 1
We
can express it using powers of 10 in the exponent form:
Therefore,
47561 = 4 × 104
+ 7 × 103 + 5 × 102 + 6 × 101 + 1 × 100
(Note 10,000 = 104,
1000 = 103, 100 = 102, 10 = 101 and 1 = 100
Let
us expand another number:
104278 = 1 × 100,000 + 0
× 10,000 + 4 × 1000 + 2 × 100 + 7 × 10 + 8 × 1
= 1 × 105
+ 0 × 104 + 4 × 103 + 2 × 102 + 7 × 101
+ 8 × 100
= 1 × 105
+ 4 × 103 + 2 × 102 + 7 × 101 + 8 × 100
Notice
how the exponents of 10 start from a maximum value of 5 and go on decreasing by
1 at a step from the left to the right upto 0.
6 Expressing
Large Numbers in the Standard Form
Let us
now go back to the beginning of the chapter. We said that large numbers can be
conveniently expressed using exponents. We have not as yet shown this. We shall
do so now.
1.
Sun is located
300,000,000,000,000,000,000 m from the centre of our Milky Way Galaxy.
2.
Number of stars in our Galaxy
is 100,000,000,000.
3.
Mass of the Earth is
5,976,000,000,000,000,000,000,000 kg.
These
numbers are not convenient to write and read. To make it convenient we use
powers.
Observe
the following:
59 = 5.9 × 10 = 5.9 × 101
590 = 5.9 × 100 = 5.9 × 102
5900 = 5.9 × 1000 = 5.9 × 103
59000 = 5.9 × 10000 = 5.9 × 104
and so on.
We have
expressed all these numbers in the standard form. Any number can be expressed
as a decimal number between 1.0 and 10.0 including 1.0 multiplied by a power of
10. Such a form of a number is called its standard form. Thus,
5,985 = 5.985 × 1,000 = 5.985 × 103
is the standard form of 5,985.
Note,
5,985 can also be expressed as 59.85 × 100 or 59.85 × 102. But these
are not the standard forms, of 5,985. Similarly, 5,985 = 0.5985 × 10,000 =
0.5985 × 104 is also not the standard form of 5,985.
We are
now ready to express the large numbers we came across at the beginning of the
chapter in this form.
The, distance of Sun from the centre
of our Galaxy i.e.,
300,000,000,000,000,000,000
m can be written as
3.0
× 100,000,000,000,000,000,000 =
3.0 × 1020 m
Example 1
Express the following numbers in the standard form:
(i) 5985.3 (ii) 65,950 (iii) 3,430,000 (iv) 70,040,000,000
Solution
(i)
5985.3 = 5.9853 × 1000 = 5.9853
× 103
(ii)
65,950 = 6.595 × 10,000 = 6.595
× 104
(iii)
3,430,000 = 3.43 × 1,000,000 =
3.43 × 106
(iv)
70,040,000,000 = 7.004 ×
10,000,000,000 = 7.004 × 1010
A point
to remember is that one less than the digit count (number of digits) to the
left of the decimal point in a given number is the exponent of 10 in the
standard form. Thus, in 70,040,000,000 there is no decimal point shown; we
assume it to be at the (right) end. From there, the count of the places (digits)
to the left is 11. The exponent of 10 in the standard form is 11 – 1 = 10. In
5985.3 there are 4 digits to the left of the decimal point and hence the
exponent of 10 in the standard form is 4 – 1 = 3.
Summary
Numbers
in exponential form obey certain laws, which are: For any non-zero integers a and b and whole numbers m and n,
a)
am ÷ an = am-n
b)
c)
am × bm = (ab)m
d)
e)
a0
= 1
f)
(–1)even
number = 1
(–1) odd number = – 1