Exponents and Powers Class 8 RS Aggarwal Exe-2A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2. We provide step by step Solutions of council prescribe publication for ICSE Board . Our Solutions contain all type Questions of to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Exponents and Powers Class 8 RS Aggarwal Exe-2A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2
| Board | ICSE |
| Publications | Goyal Brothers Prakashan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Topics | Law of Exponents |
| Edition | 2024-2025 |
Law of Exponents
(Exponents and Powers Class 8 RS Aggarwal Exe-2A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2)
- am×an = am+n
- am/an = am-n
- (am)n = amn
- an/bn = (a/b)n
- a0 = 1
- a-m = 1/am
Exercise- 2A
(Exponents and Powers Class 8 RS Aggarwal Exe-2A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2)
Que-1: Simplify :
(i) a^6 x a^8 (ii) x^5 X x^-3 (iii) z^9 x z^3 x z^-6 (iv) a^2b^3 x a^5b^2 (v) 5x^7 X 3x^4 (vi) p^3q^4 x p^5q^-5 (vii) x^7y^-5 X x^-5y^3 (viii) x^-2y^5 X x^0y^-7 (ix) x^6y^4z^-2 X x^-3y^-5z^-1 X x^2z^4
Sol: (i) a^6 × a^8
= a^(6+8) = a^14
(ii) x^5 × x^−3
= x^[5+(−3)]
= x^(5−3) = x^2
(iii) z^9 × z^3 × z^−6
= z^(9+3−6) = z^6
(iv) To simplify the expression a^2 b^3 × a^5 b^2, you add the exponents when the bases are the same. So, for a^2 b^3 × a^5 b^2, you add the exponents of a and b separately:
For a, you have a^(2+5) = a^7, because you add the exponents when multiplying with the same base.
For b, you have b^(3+2) = b^5, following the same rule.
Therefore, the simplified expression is a^7 b^5.
(v) 5x^7 × 3x^4 = 5 × 3 × x^7 × x^4
= 15 × x^(7+4)
= 15x^11
(vi) For the variable p, you add the exponents 3 and 5 since they are being multiplied together:
p^3 × p^5 = p^(3+5) = p^8
For the variable q, you add the exponents 4 and −5 since they are being multiplied together:
q^4 × q^−5 = q^(4−5) = q^−1
So, p^5/q.
(vii) x^7 y^−5 × x^−5 y^3
= x^(7−5) y^(−5+3)
= x^2/y^2.
(viii) x^−2 y^5 × x^0 y^−7
= x^−2 × 1 × y^5 × y^−7
= x^−2 ×y^(5-7)
= x^−2 × y^-2
So, 1/x^2y^2.
(ix) First, multiply the coefficients:
x^(6−3+2) = x^5
y^(4−5) = y^−1
z^(−2−1+4) = z^1 = z
So, x^5 z/y.
Que-2: Simplify :
(i) x^12/x^7 (ii) z^6/z^-3 (iii) m^5n^3/m^2n^-4 (iv) 18x^9 ÷ 6x^7 (v) 7a^12 ÷ 56a^15 (vi) a^13b^7 ÷ a^5b^-3 (vii) 7x^14 ÷ 21x^-10 (viii) p^11 ÷ p^11 (ix) a^7b^5c^4/a^-2b^3c^ 6
Sol: (i) x^12/x^7
= x^(12-7) = x^5
(ii) z^6/z^-3
= z^(6-(-3))
= z^(6+3) = z^9
(iii) Divide the variables with the same base: Since both terms have m as the base, we can subtract the exponents: m^5/m^2 = m^(5−2) = m^3.
Divide the variables with the same base: Since both terms have n as the base, we can subtract the exponents: n^3/n^−4 = n^[3−(−4)] = n^(3+4) = n^7.
So, m^3 n^7.
(iv) First, divide the coefficients: 18/6 = 3
Then, divide the variables: x^9 ÷ x^7 = x^(9−7) = x^2
Putting it all together: 18x^9/6x^7 = 3x^2
(v) Since 7 and 56 share a common factor of 7, we can simplify the numerator:
= 7a^12/56a^15 = a^12/8a^15
Now, we subtract the exponents in the denominator from the exponents in the numerator:
a^(12−15) = a^−3
So, the final simplified expression is:
1/8 a^3.
(vi) When dividing like bases, subtract the exponents:
a^(13−5) and b^[7−(−3)]
a^8 and b^(7+3)
a^8 and b^10
So, the simplified expression is a^8 b^10.
(vii) Combine the constants: 7/21 = 1/3.
Combine the variables with the same base by subtracting their exponents:
x^14 ÷ x^−10 = x^[14−(−10)] = x^(14+10) = x^24.
So, the simplified expression is x^24/3.
(viii) p^11 ÷ p^11 = p^(11-11) = p^0
Any number raised to the power of 0 is equal to 1. So,
p^0 = 1.
(ix) First, let’s deal with the variables:
For a, we subtract the exponents: a^[7−(−2)] = a^9.
For b, we subtract the exponents: b^(5−3) = b^2.
For c, we subtract the exponents: c^(4−6) = c^−2.
So, Now (a^9 b^2)/c^2.
Que-3: Simplify :
(i) (a^3)^2 (ii) (2x^2y)^4 (iii) (m^2n^-3)^4 (iv) (x^2y^-3)^-2 (v)(3x^3y^-3)^2 (vi) (5m^4n^-3)^3
Sol: (i) We can apply the power of a power rule, which states that (x^m)^n = x^(m×n).
So, applying this rule to (a^3)^2, we get:
(a^3)^2 = a^(3×2) = a^6.
(ii) (2x^2 y)^4 = 2^4 ⋅ (x^2)^4 ⋅ y^4
= 16 ⋅ x^(2×4) ⋅ y^4
= 16 x^8 y^4.
(iii) So, we multiply the exponents 2 and −3 by 4:
(m^2 n^−3)^4 = m^(2×4) ⋅ n^(−3×4) = m^8.n^−12
So, Now m^8/n^12.
(iv) Recall that when you raise a power to another power, you multiply the exponents. So, (x^2 y^−3)^−2 is equivalent to x^(2×−2) y^(−3×−2).
This simplifies to:
x^−4 y^6
So, Now y^6/x^4.
(v) So, we have:
(3x^3 y^−3)^2 = 3^2⋅(x^3)^2⋅(y^−3)^2
This simplifies to:
9⋅x^(3×2)⋅y^(−3×2)
Which further simplifies to:
9 x^6 y^−6
Or, if you prefer positive exponents:
9 x^6/y^6.
(vi) (5 m^4 n^−3)^3
= (5^3) (m^4)^3 (n^−3)^3
= 125 m^(4×3) n^(−3×3)
= 125 m^12 n^−9
= 125 m^12⋅1/n^9
= 125 m^12/n^9.
Que-4: Evaluate :
(i) (36)^1/2 (ii) (27)^2/3 (iii) (16)^-3/4 (iv) (64)^-1/3 (v) (81)^-1/4 (vi) (32)^-4/5
Sol: (i) The square root of 36 is 6, because 6×6 = 36.
So, (36)^1/2 = 6.
(ii) (27)^2/3 = 3√(27)^2
3√(27)^2 = 3√729
= 3√729 = 9.
(iii) (16)^−3/4 = 1/4√(16)^3
Next, simplify the inside of the root:
16^3 = (2^4)^3 = 2^(4×3) = 2^12
So, (16)^−3/4 = 1/4√(2)^12.
4√(2)^12 = 2^(12/4) = 2^3 = 8.
Therefore, (16)^−3/4 = 1/8.
(iv) (64)^−1/3 = 1/[(64)^1/3].
Now, we need to calculate (64)^1/3.
64 = 4^3, because 4×4×4=64.
So, (64)^1/3 = (4^3)^1/3.
we can simplify this to 4^(3×1/3) = 4^1 = 4.
(64)^−1/3 = 1/4.
(v) So, (81)^−1/4 is equal to 1/(81)^1/4.
Since 3^4 = 81, the fourth root of 81 is 3.
Therefore, (81)^−1/4 = 1/3.
(vi) First, recognize that 32 can be expressed as 2^5. So, (32)^−4/5 becomes (2^5)^−4/5.
we can simplify this as 2^5×(−4/5).
Now, we multiply the exponents: 5×(−4/5) = −4.
So, (32)^−4/5 becomes 2^−4.
Finally, 2^−4 is the same as 1/2^4, which equals 1/16.
Que-5: Simplify :
(i) (25a^2)^1/2 (ii) (27x^-3)^1/3 (iii) (64m^-6n^3)^2/3 (iv) (81a^4b^8c^-4)^1/4 (v) (3x^-3y^3)^-2 (vi) (6ab^2c^-3)^-1 (vii) (-3a^3/4 b^-1/4)^4 (viii) (32a^10 b^-5)^1/5 (ix) ∛x^18 y^-12 z^3
Sol: (i) (25a^2)^1/2 = 25^1/2⋅(a^2)^1/2
25^1/2 = 5
(a^2)^1/2 = a
(25a^2)^1/2 = 5⋅a = 5a
(ii) we can rewrite (27x^−3)^1/3 as (3^3 x^−3)^1/3.
we can split this into (3^3)^1/3 × (x^−3)^1/3.
For the first part, (3^3)^1/3, we have 3^(3×1/3) = 3^1 = 3.
For the second part, (x^−3)^1/3, we have x^(−3×1/3) = x^−1.
So, the simplified expression is 3 x^−1, which can also be written as 3/x.
(iii) First, let’s break down 64 into its prime factors: 64 = 2^6.
So, (64 m^−6 n^3)^2/3 becomes (2^6 m^−6 n^3)^2/3.
(2^6 m^−6 n^3)^2/3 = 2^(2/3×6) m^(−6×2/3) n^(3×2/3).
This simplifies to:
2^4 m^−4 n^2.
Finally, 2^4 is 16, so the simplified expression is:
16 m^−4 n^2.
(iv) So, you can rewrite the expression as:
(81)^1/4 × (a^4)^1/4 × (b^8)^1/4 × (c^−4)^1/4.
Now, let’s simplify each part separately:
(81)^1/4 is the fourth root of 81, which is 3 because 3^4 = 81.
(a^4)^1/4 is the fourth root of a^4, which is a because (a^4)^1/4 = a^(4×1/4) = a^1 = a.
(b^8)^1/4 is the fourth root of b^8, which is b^2 because (b^8)^1/4 = b^(8×1/4) = b^2.
(c^−4)^1/4 is the fourth root of c^−4, which is c^−1 because (c^−4)^1/4 = c^(−4×1/4) = c^−1.
Putting it all together:
(81 a^4 b^8 c^−4)^1/4 = 3 a b^2 c^−1 = (3 a b^2)/c.
(v) So, you can rewrite the expression as:
(3^−2⋅x^(−3⋅−2)⋅y^(3⋅−2)
Now, let’s simplify each part separately:
3^−2 is the reciprocal of 3^2, which is 1/3^2 = 1/9.
x^(−3⋅−2) is x^6 because x^(−3⋅−2) = x^6.
y^(3⋅−2) is y^−6 because y^(3⋅−2) = y^−6.
Putting it all together:
(3 x^−3 y^3)^−2 = 1/9 x^6 y^−6 = x^6/(9 y^6).
(vi) So, you can rewrite the expression as:
1/(6 a b^2 c^−3)
Now, let’s simplify the expression:
1/(6 a b^2 c^−3) = 1/6 × 1/a × 1/b^2 × c^3
So, (6 a b^2 c^−3)^−1 = 1/(6 a b^2 c^−3) = c^3/(6 a b^2).
(vii) (−3)^4 × (a^3/4)^4 × (b^−1/4)^4
(−3)^4 = 81 because any negative number raised to an even power becomes positive.
(a^3/4)^4 = a^(3/4×4) = a^3 because when you raise a power to another power, you multiply the exponents.
(b^−1/4)^4 = b^(−1/4×4) = b^−1 because b^(−1/4×4) = b^−1.
Putting it all together:
(−3 a^3/4 b^−1/4)^4 = 81 a^3 b^−1 = 81 a^3/b.
(viii) So, you can rewrite the expression as:
(32)^1/5 × (a^10)^1/5 × (b^−5)^1/5.
Now, let’s simplify each part separately:
(32)^1/5 is the fifth root of 32, which is 2 because 2^5 = 32.
(a^10)^1/5 is the fifth root of a^10, which is a^2 because (a^10)^1/5 = a^(10×1/5) = a^2.
(b^−5)^1/5 is the fifth root of b^−5, which is b^−1 because (b^−5)^1/5 = b^(−5×1/5) = b^−1.
Putting it all together:
(32 a^10 b^−5)^1/5 = 2 a^2 b^−1 = 2 a^2/b.
(ix) the property states that n√m = a^(m/n).
So, applying this property to each term:
3√x^18 = x^(18÷3) = x^6
3√y^−12 = y^(−12÷3) = y^−4
3√z^3 = z^(3÷3) = z^1 = z
Putting it all together:
∛x^18 y^-12 z^3 = x^6 y^−4 z = x^6 z/y^4.
Que-6: Show that :
(i) [x^(m+n) X x^(n+l) X x^(l+m)]/(x^m X x^n X x^l)^2 = 1 (ii) √x^(p-q) X √x^(q-r) X √x^(r-p) = 1
Sol: (i) First, we’ll simplify the numerator:
x^(m+n)⋅x^(n+l)⋅x^(l+m) = x^(m+n+n+l+l+m) = x^(2m+2n+2l)
Then, we’ll simplify the denominator:
(x^m⋅x^n⋅x^l)^2 = (x^(m+n+l))^2 = x^2(m+n+l)
Now, we’ll substitute these simplified forms back into the original expression:
[x^(2m+2n+2l)/[x^2(m+n+l)]
Next, we’ll use the rule of subtracting exponents when dividing with the same base:
x^(2m+2n+2l)−[2(m+n+l)] = x^[2m+2n+2l−2m−2n−2l] = x^0
x^0 = 1 is equal to RHS.
(ii) We’ll utilize the properties of exponents and roots.
Let’s start by multiplying the terms together: √x^(p−q) × √x^(q−r) × √x^(r−p)
We know that √a×√b = √ab, so let’s apply this property to the expression:
= √[x^(p−q) × x^(q−r) × x(r−p)]
Now, let’s multiply the exponents together:
= √[x^(p−q+q−r+r−p)]
The exponents simplify: = √x^0
√1 = 1 equals to RHS.
Que-7: Show that : [x^p/x^q]^r X [x^q/x^r]^p X [x^r/x^p]^q = 1
Sol: Let’s simplify each term separately and then multiply them together:
Simplify (x^p x^q)^r : (x^p x^q)^r = (x^(p−q))^r = x^r(p−q)
Simplify (x^q x^r)^p : (x^q x^r)^p = (x^(q−r))^p = x^p(q−r)
Simplify (x^r x^p)^q : (x^r x^p)^q = (x^(r−p))^q = x^q(r−p)
Now, let’s multiply all these simplified terms together:
x^[r(p−q)] × x^[p(q−r)] × x^[q(r−p)]
Using the property of exponents that x^a × x^b = x^(a+b), we can sum the exponents:
x^[r(p−q)+p(q−r)+q(r−p)]
Expanding this expression:
x^[rp−rq+pq−pr+qr−pq]
x^[rp−rq−pr+qr]
Now, notice that rp−rq−pr+qr = 0, because each term cancels out:
rp−rq−pr+qr = 0
Thus, x^(rp−rq−pr+qr) = x^0
Any non-zero number raised to the power of 0 is 1. Therefore:
x^(rp−rq−pr+qr) = 1.
So,
(x^p/x^q)^r × (x^q/x^r)^p × (x^r/x^p)^q = 1. Ans
Que-8: Show that :
(i) (x^a+b)^a-b X (x^b+c)^b-c X (x^c+a)^c-a = 1.
(ii) [x^a/x^-b]^a-b X [x^b/x^-c]^b-c X [x^c/x^-a]^c-a = 1.
(iii) [x^a+b/x^c]^a-b X [x^b+c/x^a]^b-c X [x^c+a/x^b]^c-a = 1.
(iv) [x^a²/x^b²]^1/(a+b) X [x^b²/x^c²]^1/(b+c) X [x^c²/x^a²]^1/(c+a) = 1.
Sol: (i) [x^(a+b)]^a-b X [x^(b+c)]^b-c [x^(c+a)]^c-a
x^(a^2−b^2) X x^(b^2−c^2) X x^(c^2−a^2)
= x^(a^2−b^2+b^2−c^2+c^2−a^2
= x^0 = 1 = RHS.
(ii) [x^a/x^-b]^a-b X [x^b/x^-c]^b-c X [x^c/x^-a]^c-a = 1
[x^(a+b)^(a-b)] [x^(b+c)^(b+c)] [x^(c+a)^(c-a)]
x^(a²−b²) x^(b²−c²) x^(c²−a²)
= x^(a²−b²+b²−c²+c²−a²
= x^0 = 1 = RHS.
(iii) [x^a+b/x^c]^a-b X [x^b+c/x^a]^b-c X [x^c+a/x^b]^c-a
= x^(a+b−c)^(a−b) . x^(b+c−a)^(b−c) . x^(c+a−b)^(c−a)
= x^(a−b)(a+b−c) . x^(b−c)(b+c−a) . x^(c−a (c+a−b)
= x^(a²−b²−ac+bc) . x^(b²−c²−ab+ac) . x^(c²−a²−bc+ab)
= x^(a²−b²−ac+bc+b²−c²−ab+ac+c²−a²−bc+ab)
= x^0 = 1 = RHS.
(iv) [x^a²/x^b²]^1/(a+b) X [x^b²/x^c²]^1/(b+c) X [x^c²/x^a²]^1/(c+a)
= (x^(a²−b²))^1/(a+b) × (x^(b²−c²))^1/(c+b) × (x^(c²−a²))^1/a+c
= x^(a+b)(a−b)/(a+b) × x^(c+b)(b−c)/(c+b) × x^(a+c)(c−a)/(a+c)
= x^(a−b) × x^(b−c) × x^(c−a)
= x^(a−b+b−c+c−a)
= x^0 = 1 = RHS.
Que-9: Show that : [x^a/x^b]^(a²+ab+b²) X [x^b/x^c]^(b²+bc+c²) X [x^c/x^a]^(c²+ca+a²) = 1.
Sol: [x^a/x^b]^(a²+ab+b²) X [x^b/x^c]^(b²+bc+c²) X [x^c/x^a]^(c²+ca+a²) = 1.
LHS = [x^a/x^b]^(a²+ab+b²) X [x^b/x^c]^(b²+bc+c²) X [x^c/x^a]^(c²+ca+a²)
= [x^(a−b)]^(a^2+ab+b^2) X [x^(b−c)]^(b^2+bc+c^2) X [x^(c−a)]^(c^2+ca+a^2)
= x^(a−b)(a^2+ab+b^2) X x^(b−c)(b^2+bc+c^2) X x^(c−a)(c^2+ca+a^2)
= x^(a3−b3) . x^(b3−c3) . x^(c3−a3)
= x^(a3−b3+b3−c3+c3−a3)
= x^0 = 1 = RHS
Que-10: Evaluate :
(i) [x^a/x^b]^1/ab X [x^b/x^c]^1/bc X [x^c/x^a]^1/ca
(ii) (1/1+x^a-b) + (1/1+x^b-a)
Sol: (i) [x^(a-b)]^1/ab × [x^(b-c)]^1/bc × [x^(c-a)]^1/ac
= x^(a-b/ab) × x^(b-c/bc) × x^(c-a/ca)
= x^[a/ab – b/ab] × x^[b/bc – c/bc] × x^[c/ac – a/ac]
= x^[1/b – 1/a] × x^[1/c – 1/b] × x^[1/a – 1/c]
= x^[1/b – 1/a + 1/c – 1/b + 1/a – 1/c]
= x^0 = 1 Ans.
(ii) [1/1+x^(a-b)] + [1/1+x^(b-a)]
= [1/1+(x^a/x^b)] + [1/1+x^b/x^a]
= [1/(x^b+x^a)/x^b] + [1/(x^a+x^b)/x^a]
= x^b/(x^b+x^a) + x^a/(x^a+x^b)
= (x^b+x^a)/(x^b+x^a)
= 1 Ans.
Que-11: Simplify : (ab)^y-z . (bc)^z-x . (ca)^x-y
Sol: Let’s simplify each term separately:
Simplify (ab)^(y−z) : (ab)^(y−z) = a^(y−z)⋅b^(y−z)
Simplify (bc)^(z−x) : (bc)^(z−x) = b^(z−x)⋅c^(z−x)
Simplify (ca)^(x−y) : (ca)^(x−y) = c^(x−y)⋅a^(x−y)
Now, let’s multiply all these simplified terms together:
a^(y−z) ⋅ b^(y−z) ⋅ b^(z−x) ⋅ c^(z−x) ⋅ c^(x−y) ⋅ a^(x−y)
Using the properties of exponents, we can combine like terms:
a^(y−z+x−y) ⋅ b^(y−z+z−x) ⋅ c^(z−x+x−y)
a^(x−z)⋅b^(y−x)⋅c^(z−y) Ans.
Que-12: Simplify : [x^(2n+3) . x^(2n+1)(n+2)]/[(x^3)^2n+1 . (x^n)^(2n+1)]
Sol: [x^(2n+3) . x^(2n+1)(n+2)]/[(x^3)^2n+1 . (x^n)^(2n+1)]
= x^[2n+3+(2n+1)(n+2)]/ x^[6n+3+n(2n+1)]
= x^[2n+3+2n^2+5n+3]/ x^[6n+3+2n^2+n]
= x^[2n^2+7n+5]/ x^[2n^2+7n+3]
= x^[2n^2+7n+5-2n^2-7n-3]
= x² Ans.
Que-13: Simplify : [a^(7n+2) . (a²)^3n+2]/[(a^4)^2n+3].
Sol: [a^(7+2n) . (a^2)^3n+2]/ (a^4)^2n+3
= [a^(7+2n) . a^(6n+4)]/a^(8n+12)
= a^(8n+11)/ a^(8n+12)
= a^−1 = 1/a Ans.
Que-14: Evaluate :
(i) [16/625]^1/4 (ii) [81/16]^-1/4 (iii) (64)^2/3 + ∛125 + 3^0 + 1/2^-5 + (27)^-2/3 X (25/9)^-1/2 (iv) (81)^-1 X 3^-5 X 3^9 X (64)^5/6 X (∛3)^6 (v) (√y³/x) X (√y/x)
Sol: (i) (16/625)^1/4
= √16/√625
= √(2x2x2x2)/√(5x5x5x5)
= 2/5. Ans.
(ii) (81/16)^−(1/4)
= (16/81)^1/4
= 4√(16)/ 4√(81)
= 4√(2×2×2×2)/ 4√(3×3×3×3)
= 2/3 Ans.
(iii) (64)^2/3 + ∛125 + 3^0 + 1/2^-5 + (27)^-2/3 X (25/9)^-1/2
= (4×4×4)^2/3 + 3√(5)^3 + 1 + 2^5 + [1/(3×3×3)^2/3] × [3^2/5^2]^1/2 [∵ x^0 = 1]
= (4)^2 + 5 + 1 + 32 + [1/(3)^2] × (3/5)
= 16 + 5 + 1 + 32 + 1/15
= 54 + 1/15
= 811/15 = 54*(1/15) Ans.
(iv) (81)^-1 X 3^-5 X 3^9 X (64)^5/6 X (∛3)^6
⇒ 1/81 × (3)^9/(3)^5 × [(2)^6] × (3)^(8/3)
⇒ 1/(3)^4 × (3)^(9−5) × (2)^5 × (3)^2
⇒ [(2)^5 × (3)^4 × (3)^2]/ (3)^4
= 9 × (2)^5 = 9×32
⇒ 288 Ans.
(v) (√y³/x) X (√y/x)
= √(y³×y)/√(x×x)
= √y^4/ √x²
= y²/x
Que-15: Find the value of x when :
(i) [-3/11]^x+5 ÷ [-3/11]^-2x+3 = [-3/11]^2x-5 X [(-3/11)¯²]^(x+4)
(ii) [{(2/5)^2}^4]^x+2 = [{(2/5)^-2}^(x-1)]^-3
Sol: (i) [-3/11]^x+5 ÷ [-3/11]^-2x+3 = [-3/11]^2x-5 X [(-3/11)¯²]^(x+4)
Using properties of exponent we get
= (-3/11)^[x+5-(-2x+3)] = (-3/11)^[2x-5+(-2x-8)]
= (-3/11)^3x+2 = (-3/11)^-13
On comparing both sides powers we get
= 3x+2 = -13
= 3x = -13-2
= 3x = -15
= x = -5 Ans.
(ii) [{(2/5)^2}^4]^x+2 = [{(2/5)^-2}^(x-1)]^-3
= (2/5)^8.(x+2) = (2/5)^-2.(x-1).-3
= (2/5)^(8x+16) = (2/5)^(6x-6)
On comparing both sides powers we get
8x+16 = 6x-6
8x-6x = -6-16
2x = -22
x = -11 Ans.
Que-16: Simplify :
(i) [{2p¯¹q²r}³]¯² (ii) [(3p²qr¯²)/(2p¯¹q³)]² ÷ (2p³r)¯¹
Sol: (i) [{2p¯¹q²r}³]¯²
= [2p¯³q^6 r³]¯²
= 2p^6 q^-12 r^-6
= (2p^6)/(q^12 r^6) Ans.
(ii) [(3p²qr¯²)/(2p¯¹q³)]² ÷ (2p³r)¯¹
= [(9p^4 q^2 r^-4)/(4p^-2 q^6)] x 1/(2p³r)¯¹
= 9/4 [p^4+2 q^2-6 r^-4] x (2p³r)¹
= 9/2 [p^6+3 q^-4 r^-4+1]
= 9/2 p^9 q^-4 r^-3. Ans.
— : End of Exponents and Powers Class 8 RS Aggarwal Exe-2A Goyal Brothers Prakashan ICSE Foundation Maths Solutions :–
Return to :- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions
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