Indices Exponents Class 9 OP Malhotra Exe-6D ICSE Maths Solutions

Indices Exponents Class 9 OP Malhotra Exe-6D ICSE Maths Solutions Ch-6. We Provide Step by Step Solutions / Answer of Questions on Indices Exponents OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Indices Exponents Class 9 OP Malhotra Exe-6D ICSE Maths Solutions Ch-6

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-6	Indices Exponents
Writer	OP Malhotra
Exe-6D	More complex Problems on Indices Exponents
Edition	2025-2026

Exercise- 6D

Indices Exponents Class 9 OP Malhotra ICSE Maths Solutions Ch-6

Que-1: 2^x+1 = 4^x-3

Sol: 2^x+1 = 4^x-3 ⇒ 2^{x +1} = (2²)^x-3
⇒ 2^x+1 = 2^2x-6
Comparing both sides, we get
x + 1 = 2x – 6 ⇒ 2x – x = 1 + 6
⇒ x = 7
∴ x = 7

Que-2: x^(−3/4) = 1/8.

Sol: x^(−3/4) = 1/8 = 1/2³ = 2¯³
⇒ x^(1/4)×(-3) = 2¯³
⇒ (x^(1/4))¯³ = (2)¯³
Comparing,
⇒ x^(1/4) = 2
⇒ x = 2^4
⇒ x = 16
∴x = 16

Que-3: (x – 1)^(2/3) = 25

Sol: (x – 1)^(2/3) = 25 = 5²
(x – 1)^(2/3) = (5³)^(2/3)
∴ Comparing, we get
x – 1 = 5³
⇒ x – 1 = 125
⇒ x = 125 + 1 = 126
∴ x = 126

Que-4: 2^5x+3 = 8^{x + 3}

Sol: 2^5x+3 = 8^{x + 3} = (2³)^x+3
⇒ 2^5x+3 = 8^{3x + 9}
Comparing, we get
5x + 3 = 3x + 9
⇒ 5x – 3x = 9 – 3
⇒ 2x = 6
⇒ x = 6/2 = 3
x = 3.

Que-5: (√(5/7))^(x−1) = (125/343)^−1

Sol: (√(5/7))^(x−1) = (125/343)^−1
⇒ (5/7)^(x−1)/2 = (5³/7³)¯¹ = (5/7)¯³
Comparing both sides,
(x−1)/2 = – 3
⇒ x – 1 = – 6
⇒ x = – 6 + 1
⇒ x = – 5
∴ x = – 5

Que-6: 11^(3-4x) = (√(1/121))^−2

Sol: 11^(3-4x) = (√(1/121))^−2 = (√(1/11)²)^−2
⇒ 11^(3-4x) = 11{((−2)×(−2))/2}
⇒ 11^(3−4x) = 11²
Comparing, we get.
3 – 4x = 2
⇒ – 4x = 2 – 3 = – 1
⇒ – 4x = – 1
⇒ x = + 1/4
∴ x = 1/4

Que-7: Solve for x, (√4)^{2x+(1/2)} = 1/32.

Sol:  (√4)^{2x+(1/2)} = 1/32
⇒ (√(2)²)^{2x+(1/2)} = 1/2^5
⇒ (2^(2/3))^{2x+(1/2)} = 2^-5
⇒ 2^(2/3){2x+(1/2)} = 2^-5
Comparing, we get
(2/3){(2x+(1/2))} = – 5
(4x/3) + (1/3) = – 5
⇒ 4x + 1 = – 15
⇒ 4x = – 15 – 1 = – 16
⇒ x = −16/4 = – 4
∴ x = – 4

Que-8: Find the value of x if √(p/q) = (q/p)^(1−2x)

Sol: √(p/q) = (q/p)^(1−2x)
= (p/q)^(1/2) = (p/q)^-(1-2x)
Comparing,
-(1-2x) = 1/2
2x-1 = 1/2
4x-2 = 1
4x = 3
x = 3/4.

Que-9: Solve for x, 2³ (5° + 3^2x) = 8*(8/27)

Sol:  2³ (5° + 3^2x) = 8*(8/27)
= 8(1+3^2x) = 224/7
= (1+3^2x) = (224/27) × (1/8) = 28/27
= 3^2x = (28/27) – 1
= 3^2x = (28-27)/27 = 1/27
= 3^2x = 1/3³ = 3¯³
Comparing we get,
2x = -3
x = -3/2

Que-10: Solve for x, √{(8^0)+(2/3)} = (0.6)^(2-3x).

Sol: √{(8^0)+(2/3)} = (0.6)^(2-3x)
= √{1 + (2/3)} = (6/10)^(2-3x)
= √(5/3)  = (3/5)^(2-3x)
= (5/3)^(1/2) = (5/3)^-(2-3x)
Comparing we get.
-2+3x = 1/2
6x-4 = 1
6x = 5
x = 5/6

Solve the following equations for x

Que-11: 3^{2x+ 4} + 1 = 2.3^{x+ 2}.

Sol: 3^{2x+ 4} + 1 = 2.3^{x+ 2}
⇒ 3^2x.3⁴ + 1 = 2.3^3x.3²
⇒ 81.3^2x + 1 = 18.3^x
⇒ 81.3^2x – 18.3^x + 1 = 0
⇒ Let 3^x = a, then 3^2x = a²
∴ 81a² – 18a + 1 = 0
⇒ (9a)² – 2 x 9a + (1)² = 0
⇒ (9a – 1)² = 0
⇒ 9a – 1 = 0 ⇒ 9a = 1
⇒ 9.3^x = 1 ⇒ 3^x = 1/9 = 3^-2
Comparing both sides,
∴ x = – 2

Que-12: 5^2x+1 = 6.5^x – 1.

Sol: 5^2x+1 = 6.5^x – 1.
⇒ 5^2x.5¹ – 6.5^x + 1 = 0
⇒ 5.5^2x – 6.5^x + 1 = 0
Let 5^x = a, then 5^2x = a²
∴ 5a² – 6a + 1 = 0
⇒ 5a² – 5a – a + 1 = 0
⇒ 5a (a – 1) – 1 (a – 1) = 0
⇒ (a – 1) (5a – 1) = 0
Either a – 1 = 0, then a = 1
or 5a – 1 = 0 then 5a = 1 ⇒ a = 1/5
(i) If a = 1, then 5^x = 1 = 5° (∵ 5° = 1)
∴ x = 0
(ii) If a = (1/5) then 5^x = (1/5) = 5^-1
∴ x = – 1
Hence x = 0 or x = – 1

Que-13: 2^2x – 2^x+3 = – 2⁴

Sol: 2^2x – 2^x+3 = – 2⁴
⇒ 2^2x – 2^x.2³ + 2⁴ = 0
⇒ 2^2x – 8.2^x + 16 = 0
Let 2^2x = a, then 2^2x = a²
a² – 8a + 16 = 0
⇒ (a)² – 2 x a x 4 + (4)² = 0
⇒ (a – 4)² = 0
⇒ a – 4 = 0 ⇒ a = 4
∴ 2^x = 4 = 2²
Comparing we get, x = 2

Solve for x and y :

Que-14: 9^x = 3^y-2, 81^y = 3² x (27)^x

Sol: 9^x = 3^y-2, 81^y = 3² x (27)^x
[(3)²]x = (3)^y-2 ⇒ 3^2x = 3^y-2
Comparing,
2x = y – 2
⇒ y = 2x + 2 … (i)
81^y = 3² x (27)^x ⇒ (3⁴)^y = 3² x (3³)^x
⇒ (3⁴)^y = 3² x 3^3x ⇒ 3^4y = 3^3x+2
∴ 4y = 3x + 2 … (ii)
From (i)
4 (2x + 2) = 3x + 2
8x + 8 = 3x + 2 ⇒ 8x – 3x = 2 – 8
⇒ 5x = – 6 ⇒ x = (-6/5)
∴ y = 2x + 2 = 2 x (−6/5) + 2
= −12/5 + 2 = (−12+10)/5 = −2/5
Hence x = −6/5 and y = −2/5

Que-15: 2^{1−(x/2)} = 4^y, (7^(1+x)) × (49)^−2y = 1

Sol: 2^{1−(x/2)} = 4^y, (7^(1+x)) × (49)^−2y = 1
∴ 1 – (x/2) = 2y
⇒ 2 – x = 4y
⇒ 4y + x = 2
⇒ x = 2 – 4y … (i)
and 71 + x x (49)-2y = 1
71+x x (7²)-2y = 7° (∵ 7° = 1)
⇒ 71 + x.7-4y = 7° ⇒ 71 + x-4y = 7°
∴ 1 + x – 4y = 0
⇒ 1 + (2 – 4y) – 4y = 0
From (i) x = 2 – 4y
⇒ 1 + 2 – 4y – 4y = 0
⇒ 3 – 8y = 0
⇒ 8y = 3
⇒ y = 3/8
∴ x = 2 – 4y = 2 – 4 x (3/8) = 2 – (3/2)
= (4−3)/2 = 1/2
Hence x = 1/2, y = 3/8

Que-16: 2^x = 16 x 2^y, (27)^x = 9 x 3^2y

Sol: 2^x = 16 x 2^y ⇒ 2^x = 2⁴ x 2^y
⇒ 2^x = 2^4+y
Comparing, we get
∴ x = 4 + y … (i)
and (27)^x = 9 x 3^2y ⇒ (3³)^x = 3² x 3^2y
⇒ 3^3x = 3^2y+2
∴ 3x = 2y + 2
⇒ 3 (4 + y) = 2y + 2 [From (i)]
12 + 3y = 2v + 2 ⇒ 3y – 2y = 2 – 12
⇒ y = – 10
∴ x = 4 + 7 = 4 – 10 = – 6
Hence x = – 6, y = – 10

— : End of Indices Exponents Class 9 OP Malhotra Exe-6D ICSE Maths Solutions Ch-6 :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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