Exponent and Power
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1
Introduction
Exponent
Example = 2×2×2×2×2=25.
Here 2 is the baseand five is the Exponent or power.
And the complete number is called exponential form
For example:-
Mass of the sun
2000000000000000000000000000000
= 2×1030.
We read 2×1030 as2 × 10 raised to the power 30
Types of Exponents
Exponents can be divided into four types based on the number in the power. They are:
Positive exponent
Negative exponent
Zero exponent
Rational exponent or fractional exponent
Positive exponent
For any non-zero rational number a and a positive integer na raised to the power n represents a positive exponent
Example =
{1}(8)² {8≠0} {3>O} {8>0}
{2}(-5)8= -5≠0 and 8>0
Negative exponent
For any non-zero rational number 'a' and a negative integer 'm' am represents a negative exponent.
Example :- (5)-3 = 5≠0 , -3<0
(-4) = -4≠0 , -2<0
Important point to note
Difference between these 2 numbers
(-4)²
-4²
-4×-4
-4×4
16
-16
Fractional exponent
Fractional exponent is of the form an, where a>0 and n is a real number.
For example =5 ⅛= 5>0
We can write 5 ⅛ as
8√5
2.3√2=2½
Note
If a<0 , n is any real number
(-2)⅓ ✖ wrong
⅓True
✓✓✓✓
(0)² = 0×0=0
00= this is known as indeterminate form
Zero exponent
Any non zero rational number to the power 0 is equal to 1.
Example :-20= 1
30. = 1
(-4)0=1
a0=1
An= 1 ( if n=0)
Except when a=1 & a= -1
Case 1 :- when a=1
(1)¹ = (1)-3 = (1)-3= 1
This is true for infinity many 'a'
Case 2 :- When a= -1
a = (-1)² = (-1)⁴ = (-1)6 = 1
(-1)² = (a)p (p should be even)
This is true for every even integer.
DECIMAL NUMBER AND ITS EXPANDED FORM
Lets us understand this topic with a example
1.2346.78 = 2×1000 + 3×100 + 4×10 + 6×1 +7/10 + 8× 1/100
Or we can also expand it using power .let us see how
We can write 2×1000 as 2×10³
We can also write 3×100 as 3×10²
We can write 4×10 as 4×10¹
We can also write 6×1 as 100
We can also ⅞ as 7×10-1
We can write 8× 1/100 as 8× 10-2
So the expanded form will be 2×10³ + 3×10² + 4×10¹ + 6×100+ 7×10-1 + 8×10-2
8
Let us see a other question to make the concept clear .
4286.28 = 4×10³ + 2×10² + 8×10¹ + 6×100+ 2×10-1 + 8×10-2
Questions :-
Write the decimal form of 5×1000 + 3×100 + 2×10 + 91/10 + 7×1000
Solution :-
5×1000 + 3×100 + 2×10 + 91/10 +
7×1000
= 5000+300+20+9/10 +7/1000
9
=5320.907.
Ruf work :-
5000.000
300.000
20.000
.9
.007
5320.907
10
If m=3000+200+10+ 9/10 + 5/100 and n= 2000+100+9+ 5/10 + 4/100 then find the value of M-N ?..
Solution :- M=3210.95 & N= 2109.54
So , M-N =
3210.95
2109.54
1101.41
11
M-N = 1101.41
Questions related to your books let's see
Write the multiplicative inverse of
2-3 = 2³
1/72 = 7-2
-4³ = -4-3
85 = 8-5
12
(2)write the expanded form of the following numbers
2378.44 :- 2×1000+3×100+7×10+8×1+4×1/10 + 4× 1/100
79846.13 = 7×10000 + 9×1000 + 8×100 + 4×10 + 6×1 + 1× 1/10 + 3×1/100
Laws of exponent
1 . If a is any non zero rational number and m and n are two integers, then am +an= am+n
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In case of
multiplication when bases are same then the power are added
For example :- (1) ap+ aq+ar = ap+q+r
(2)write the simplified form of 3² × 81 Solution =3² × 81
=3²×3⁴
2+4
= 3
= 3
6
(2 law of exponent) if a is any non zero rational number and m,n are integers then am/an =
14
In case of division when bases are same then the powers are subtracted
For example :-
Q1. Express 420/1024 in its simplest form?
Answer :- Consider 420/1024
= 420/45
15
420-5
= 415
(22)15 =(2)30
16
3.law of exponent (am)n= amn = (an)m
Example:-
Write (16)² in simplest form Answer:- (16)²
=(2⁴)²
=(2)⁴×2
= (2)8
17
Remarks :-
(A ± b) ≠ A ± b
(± this is + and - sign)
mmm.
For better understanding let's take a example based on this
remark
(A+B)3≠ A³ + B³
³√a±b ≠ ³√a ± ³√b
18
4 law of exponent. (ab)m= am × an
Example :- Express 6³ and (72)⁴ as the product of the power of the prime factors ?
Solution :- (i) 6³ = (2×3)³ = 2³×3³
(72)⁴. = (2³×3²)⁴ = (2³)⁴ × (3²)⁴
=(2)3×4 × (3)2×4
= (2)² × (3)8
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(LCM of 6 = 2×3 and LCM of 72 is 2×2×2×3×3)
5thlaw of exponent :-(a/b)m= am/bm
Question based on this law
Simplify
(⅔)³ = (⅔)³
2³/3³
2×2×2/3×3×3
8/27
20
6thlaw of exponent:- (a/b)-m = (b/a)m
Simplify :- (⅘)-3
Solution :- Consider(⅘)-3
= (5/4)³
5³/4³
5×5×5/4×4×4
125/64
21
7th law of exponent :- a¹= a and a0=1
ap× ao= ap+o = ap
ap × ao/ap= ap/ap.
ao= 1
Question based on this law .
Find the value of 20+30-4¹+1¹ Solution : 20+30-4¹+1¹
1+1-4+1
3-4
-1
22
Questions based on law of exponents:
1. Express the following in its simplest forms:
(i) ³√2⁴
Solution: ³√2⁴
∴ (a) = a
: (2⁴)⅓.
m n
mn
:2⁴×⅓
:(2)4/3
23
Express the following as a rational number with a positive exponent ?
(I)(-¾)-2
Solution :- (-¾)-2
(-4/3)²
(-4)²/(3)²
16/9
(-⅓)-4 × (-⅓)-2 Solution : (-⅓)-4+(-2)
:(In case of multiplication when bases are same powers are added (
(-⅓)-4-2
(-⅓)-6
(-3)6 (∴a-m= 1/am)
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(3).Find the value of p?
Solution: (3)2p-1= 1/(27)p-3
(3)2p-1. = 1/(3³)p-3
(3)2p-1 = (⅓)³(p-3)
: (3)
= (3)
.
(
a = 1/a)
2p-1.
-3(p-3).
-m
m
∴
: 2p-1 = -3(p-3)
2p-1 = -3+9
: 5p
= 10
: p
= 5/10
: p
= 2
25
(⅗)³ x (⅗)6 = (⅗)²p-1
Solution:- (⅗)
= (⅗)
Bases are same
3+6
2p-1.
: 3+6
= 2p-1
: 9
= 2p-1
: 2p
= 10
: p
= 2/10
: p
= 5
(3) qp×35×27³/3²×81⁴ = 27
Solution:-(3²)p× (3¹)5× (3³)³/ 3²×(3⁴)⁴ = (3)³
:- 3 ×3×3/3²×3
= (3)³
{ (a)= a
2p 5
9
16
m n
mn
:- 3² /3
= (3)³
p+14 18
26
:- 32p+14-18.
:- 32p-4
:- 2p-4
:- 2p.
:- p.
= (3)³
= (3)³
= 3
= 7
= 7/5
Uses of exponent
Examples
(1). 7800000000
= 7·8 × 10
-3
8
(2) 0.003478
= 3·478 × 10
27
Left to right →→→→→→Power Decreases
Right to left ←←←←←←Power increases
Use of exponent to express small numbers in standard form
Step 1. Obtain the number and see whether the number is between 1 and 10 or it is less than 1
Step 2. If the number is between 1 and 10, then write it as the product of the number itself and 100
Step 3. if the number is less than one, then remove the decimal point to the right so that there is just one digit on the left side of the decimal
point.Write the given number as the product of the number so obtained and 10-n , where n is the number of places the decimal has been moved to the right.
thenumber so obtained as the standard form of the given number
Questions :-
28
1.Express the following numbers in standard form
0·000786= 0·0007·86 = 7·86 ×10-4
725683. × 10-5= 7·25683 × 105×10-5
7·25683 × 105+(-5)
7.25683 × 100
7·25683
(III) 3894·562 = 3·894562 ×10³
2 . EXPRESS THE FOLLOWING NUMBERS IN USUAL FORM :
7
(I) 3·028 × 10
7
Solution: 3·028 × 10
( 4+3=7)
: 3·028 × 10³ × 10.
7
: 3028 × 104
: 30280000
29
(II)5·346 × 10-5
Solution : 5·346 × 1/105 (a-m= 1/am) :5·346/100000
:0·00005346
if diameter of a small spherical ball is 0.03746m, then express the radius in standard form
Solution: Diameter of the ball :- 0.03746m
: Radius of the ball :- d/2 = 0.01873m
: Standard form = 1·873 × 10-3
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