Mathematics Formula Algebra
In algebra, we have various topics having formulas and identities, Some are listed below:-
- Algebraic Identity
- Exponents and Powers
- Least Common Multiple (LCM) & Highest Common Factor(HCF)
Exponents and Powers
if we factorize 4,9,16 or 25 then can be written as
4 = 2 x 2 x 1
9 = 3 x 3 x 1
16 = 2 x 2 x 2 x 2 x 2 x 1
25 = 5 x 5 x 1
The above expression can be also written in a shorter way using exponents in this way.
4 = 2 x 2 = 22
9 = 3 x 3 = = 32
16 = 2 x 2 x 2 x 2 x 2 = 24
25 = 5 x 5 = 52
In the above expression that represents repeated multiplication of the same factor is called power.
The conventions for reading are such as
The above short notation stands for the product 2 x 2 x 2 x 2. Here '2' is called the base and '4' the exponent. The above number is read as " 2 raised to the power of 4" or simply as the "fourth power of 2". It is also called the exponential form of 16.
Example 1:- Express 64 as the power of 2?
Solution:- By prime factorisation method we have
64 = 2 x 2 x 2 x 2 x 2 x 2 x 1
So we can say 64 = 26
Example 2:- Express 128 as the power of 2?
Solution:- By prime factorisation method we have
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 1
So we can say 128 = 27
Example 3:- Express 81 as the power of 3?
Solution:- By prime factorisation method we have
81 =3 x 3 x 3 x 3 x 1
So we can say 81 = 34
Example 4:- Express 320 as a product of power of prime factors?
Solution:- By prime factorisation method we have
320 = 2 x 2 x 2 x 2 x 2 x 2 x 5
= 26 × 5
= 320
Laws of Exponents and Powers
The followings are important laws of exponents and are very useful to solve problems;
Laws of Exponents - 1
On Multiplying Power with the Same Base
For any non zero real number a , where m and n are any rational numbers then
am × an = am + n
or in generally we can say on multiplication if bases are equals then powers may be added.
it can be easily verified by following illustrative example;
Example 5:- Evaluate 23 × 22 ?
Solution:- By general concepts we have
23 × 22 = (2 x 2 x 2) x (2 x 2)
= 2 x 2 x 2 x 2 x 2
= 25
So we can say 23 × 22 = 23+2
We have to note that the base in 23 and 22 is same and the sum of the exponents,i.e, 3 and 2 is 5.
Laws of exponent - 2
On Dividing Power with the Same Base
For any non zero real number a, where m and n are any rational numbers then
am
÷
an = am - nor in general, we can say on division if bases are equals then powers may be subtracted.
it can be easily verified by following illustrative examples;
Example 6:- Evaluate 23
÷
22 ?Solution:- By general concepts we have
23
÷
22 = [ 2 x 2 x 2 ] ÷
[ 2 x 2 ] = 2
= 21
So we can say 23
÷
22 = 23 - 2We have to note that the base in 23 and 22 is same and the difference of the exponents,i.e, 3 and 2 is 1.
We have consider now one more illustration for verification of above results
Example 7:- Evaluate 10 5
÷
10 2 ?Solution:- By general concepts of mathematics we have,
10 5
÷
10 2 = (10 x 10 x 10 x 10 x 10 ) ÷
(10 x 10) = 10 x 10 x 10
= 103
So we can say 10 5
÷
10 2 = 10 5 - 2We have to note that the base in 105 and 102 is the same and the difference of the exponents,i.e, 5 and 2 is 3.
Laws of Exponents - 3
On Multiplying Power with the Same Exponents
For any non zero real number a and b, where m is any rational number then
am × b m = (a x b)m = (ab)m
or in general, we can say on multiplication if bases are different but powers are equals then bases may be multiplied.
it can be easily verified by following illustrative example;
Example 8:- Evaluate 22 × 5 2?
Solution:- By general logical concepts we have,
22 × 52 = (2 x 2 ) x (5 x 5)
= (2 x 5 ) x (2 x 5)
= 10 x 10
= 102
So we can say 22 × 52 = 102
We have to note that the base in 22 and 52 is different but power is the same 2 and the products of the bases,i.e, 2 and 5 are 10.
Laws of exponent - 4
On Dividing power with the Same Exponents
For any non zero real number a and b, where m is any rational number then
am
÷
b m = (a / b) mor in general, we can say on division if bases are different and powers are equals then bases may be divided.
it can be easily verified by following illustrative examples;
Example 9:- Evaluate 83
÷
23 ?Solution:- By general concepts we can write,
83
÷
23 = [ 8 x 8 x 8 ] ÷
[ 2 x 2 x 2 ] = 4 x 4 x 4
= 43
So we can say 83
÷
23 = ( 8 / 2) 3 = 43Now in the above illustration the bases in 83 and 23 is different but power is same so the bases of the exponents,i.e, 8 and 2 is divided hence result is 4 so required solution is 43.
We may consider now one more illustration for verification of above results
Example 10:- Evaluate 100 5
÷
10 5 ?Solution:- By general concepts we have
100 5
÷
10 5 = (100 x 100 x 100 x 100 x 100 )
÷
(10 x 10 x 10 x 10 x 10)= 10 x 10 x 10 x 10 x 10
= 105
So we can say 100 5
÷
10 5 = ( 100 / 10 ) = 10 5 Now in the above examples 1005 and 105 bases is different but power is same so the bases of the exponents,i.e, 100 and 10 is divided hence result is 10 so required solution is 105.
Laws of exponent - 5
On taking Numbers with Exponents ZERO
For any non zero real number a
a 0 = 1
or in general, we can say that any number ( except 0 ) raised to the power ( or Exponent ) 0 is always ONE ( 1 ).
It can be easily verified by following illustrative examples;
Example 11:- Evaluate 83
÷
83 ?Solution:- By general concepts we have
83
÷
83 = [ 8 x 8 x 8 ] ÷
[ 8 x 8 x 8 ] = 1 x 1 x 1
= 1
By using laws of Exponents am
÷
an = am - n again we can write in other way 83
÷
83= 8 3 - 3
= 8 0
So we have 83
÷
83 = 8 0 = 1 We may consider now one more illustration for verification of above results
Example 12:- Evaluate 1001 5
÷
1001 5 ?Solution:- By general concepts we can write in the following way
1001 5
÷
1001 5 = (1001 x 1001 x 1001 x 1001 x 1001 )
÷
(1001 x 1001 x 1001 x 1001 x 1001 ) = 1 x 1 x 1 x 1 x 1
= 1
By using laws of Exponents am
÷
an = am - n we can write 1001 5
÷
1001 5 = 1001 5 - 5
= 10010
So, we have 1001 5
÷
1001 5 = 10010 = 1 Laws of exponent - 6
On taking Power of Power
For any non zero real number a
( a m ) n = a m n = ( a n ) m
where m and n are rational numbers.
it can be easily verified by following illustrative examples;
Example 13:- Evaluate ( 2 2 ) 3 ?
Solution:- By general concepts we that it means 2 2 is multiplied three times itself
( 2 2 ) 3
= [ 2 2 ] x [ 2 2 ] x [ 2 2 ]
= 4 x 4 x 4
= 64
On using above property this problem can be also solved in the following way;
( 2 2 ) 3
= 2 6
= 2 x 2 x 2 x 2 x 2 x 2
= 64
On the basis of the above laws of Powers and Exponents, We can optimize our time to solve problems not only in algebra but in all branches of pure sciences as well as in the fields of engineering sciences.
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