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In general, for any non-zero integer a, a = 1.
0
Let us summarize all the laws of exponents discussed above.

Let a, b be the two non-zero integers and m, n be the whole numbers.

Then the laws of exponents are given as follows:
a m
n
(i) a × a = a m + n (ii) a ÷ a = = a m – n create and solve
n
m
m
a n
m
(iii) (a ) = a mn (iv) a × b = (ab) m Frame any three problems
m n
m
based on laws of exponents
a m  a  m with the answer 1.
m
m
(v) a÷ b= =   (vi) a = 1
0
b m  b 
Miscellaneous Examples Using the Laws of Exponents
Example 15: Express 9 × 9 × 9 × 9 × 9 taking base as 3.
Solution: We have, 9 × 9 × 9 × 9 × 9 = 9 5
We know that, 9 = 3 × 3 = 3 2

5
2 5
\ 9 = (3 )
10
mn
m n
= 3 2 × 5 = 3 (Using III law of exponents: (a ) = a ]
Example 16: Simplify the following in exponential form:
(
4
3
 ×() 3 (c) ( ) ×
(a) 2 × ) ÷ 2 2 (b)  − ( ) 3 5  3  5 2 3 5 4  ÷ 5 2
2

2
3
 − ( )   


3
m
4
n
2
Solution: (a) (2 × 2 ) ÷ 2 = (2 4 + 3 ) ÷ 2 2 [Q a × a = a m + n ]
7
= 2 ÷ 2 = 2 7 – 2 = 2 5 [Q a ÷ a = a m – n ]
m
n
2
 − ( ) 3 5  52 3

−
−3
m
n
(b)   ×− ( ) 3 3 = ( )   ×− ( ) 3 [Q a ÷ a = a m – n ]

2
3
  − ( )  
n
m
= (–3) 3 + 3 = (–3) 6 [Q a × a = a m + n ]
mn
m n
4
2 3
4
2
(c) ((5 ) × 5 ) ÷ 5 = (5 2 × 3 × 5 ) ÷ 5 2 [Q (a ) = a ]
4
6
= (5 × 5 ) ÷ 5 2
n
m
= 5 6 + 4 ÷ 5 2 [Q a × a = a m + n ]
10
n
m
= 5 ÷ 5 2 [Q a ÷ a = a m – n ]
= 5 10 – 2 = 5 8
Example 17: Evaluate the following:
2
3
3
18 × 9 × 16 2 2 × a 3 × 5a 4
(a) (b)
3
12 × 6 4 10 a 2
)
23
3
2
22
×
Solution: (a) 18 × 9 × 16 2 = ( 23 ) × 3 () × 2 ( 42
12 × 6 4 ( 2 × ) 3 23) 4
2
3 ×(
×
3
2
2
[Q 18 = 2 × 3 × 3 = 2 × 3 and 12 = 2 × 2 × 3 = 2 × 3]
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