Exponents ICSE Class-8th Concise Selina Power Solutions Chapter-2. We provide step by step Solutions of Exercise / lesson-2 Exponents ICSE Class-8th Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-2 A and Exe-2 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8.
Exponents ICSE Class-8th Concise Selina Power Solutions Chapter-2
–: Select Topics :–
Laws of Exponents / Power
| Law | Example |
|---|---|
| x1 = x | 61 = 6 |
| x0 = 1 | 70 = 1 |
| x-1 = 1/x | 4-1 = 1/4 |
| xmxn = xm+n | x2x3 = x2+3 = x5 |
| xm/xn = xm-n | x6/x2 = x6-2 = x4 |
| (xm)n = xmn | (x2)3 = x2×3 = x6 |
| (xy)n = xnyn | (xy)3 = x3y3 |
| (x/y)n = xn/yn | (x/y)2 = x2 / y2 |
| x-n = 1/xn | x-3 = 1/x3 |
| And the law about Fractional Exponents: | |
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Exercise – 2 A Exponents ICSE Class-8th Concise Selina Power Solutions
Question 1 :-
Evaluate:

Answer :-





Question 2 :-
If 1125 = 3m x 5n; find m and n.
Answer 2:-
1125 = 32 x 53
The factor of 1125 are 3 x 3 x 5 x 5 x 5
| 3 | 1125 |
| 3 | 375 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
∴ 1125 = 3 x 3 x 5 x 5 x 5
Now comparing, 32 x 53 = 3m x 5n
∴ m = 2, n = 3
Question 3 :-
Find x, if 9 × 3x = (27)2x-3
Answer 3 :-
9 x 3x = (27) 2x-3

Since, bases are same, compare them,
x + 2 = 6x – 9
6x – x = 9 + 2
⇒ 5x = 11
⇒ x = 11/2
⇒ x = 21⁄5
Exe -2 B Exponents / Power ICSE Class-8th Concise selina
Question 1 :-
Compute:


Answer 1 :-








Question 2 :-
Simplify:
(i) 
(ii) [(64)-2]-3 ÷ [{(-8)2}3]2
(iii) (2-3 – 2-4) (2-3 + 2-4)
Answer 2 :-

(ii)

(iii)

Question 3 :-
Evaluate:
(i) (−5)0
(ii) 80+40+20
(iii) (8+4+2)0
(iv) 4X0
(v) (4x)0
(vi) [(103)0)0]5
(vii) (7X0)2
(viii) 90+9-1−9-2+ 91/2 − 91/2
Answer 3:-
(i) (−5)0
(−5)0 =1
( a0 =1)
(ii)

(iii)
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(iv)
(v)
![]()
(vi)

(vii)
![]()
(viii)
90+9-1−9-2+ 91/2 − 91/2
Question 4 :-
Simplify:
(i) 
(ii) ![]()
(iii) ![]()
(iv) ![]()
(v) ![]()
(vi) ![]()
(vii) 
(viii)
Answer 4 :-
(i) 
(ii) 
(iii) 
(iv) 
(v)
(vi) 
(vii) 
(viii) 
Question 5 :-
Simplify:
(xa+b)a-b.(xb+c)b-c.(xc+a)c-a
Answer 5 :-
(xa+b)a-b.(xb+c)b-c.(xc+a)c-a
(xa+b)(a-b).(xb+c)(b-c).(xc+a)(c-a)
=xa²-b². xb²-c².. xc²-a².
= xa²–b²+b²- c²+c²–a²
=x0
=1
Question 6 :-
Simplify:
(i) 
(ii) 
Answer 6 :-
(i) 
(ii)

Question 7 :-
(i) (a-2)-2. (ab)-3
(ii) (xny-m)4 × (x3y-2)-n
(iii) 
(iv)
(v) (a-2b)1/2 × (ab-3)1/3
(vi) (xy)m-n . (yz)n-1 . (zx)1-m
Answer 7 :-
(i) 
(ii) 
(iii) 

(iv) 
(v) 
(vi) 
Question 8 :-
Show that:

Answer 8:-

Question 9 :-
Evaluate:

Answer 9 :-

Question 10 :-
Evaluate:

Answer 10 :-

Question 11 :-
(m + n)-1 (m-1 + n-1) = (mn)-1
Answer 11:-

Hence Proved.
Question 12 :-
Prove that:
(i) 
(ii) 
Answer :-
(i) 
(ii) 
Question 13 :-
Find the values of n, when:
(i) 12-5 × 122n+1 = 1213 ÷ 127
(ii) 
Answer :-
(i) 
(ii) 
Question 14 :-
Simplify:
(i) 
(ii) 
Answer 14 :-
(i) 
(ii) 
— End of Exponents ICSE Class-8th Concise Solutions :–
Return to – Concise Selina Maths Solutions for ICSE Class -8
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