Logarithms Class 9 OP Malhotra Exe-7B ICSE Maths Solutions

Logarithms Class 9 OP Malhotra Exe-7B ICSE Maths Solutions Ch-7. We Provide Step by Step Solutions / Answer of Questions on Law of Logarithms of OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Logarithms Class 9 OP Malhotra Exe-7B ICSE Maths Solutions Ch-7

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-7	Indices Exponents
Writer	OP Malhotra
Exe-7B	Logarithms Problems Questions with Answer
Edition	2025-2026

Exercise- 7B Logarithms Problems / Questions Answer

 Class 9 OP Malhotra ICSE Maths Solutions Ch-7

Que-1:  Express each sum or difference as a single logarithm :
(i) log 6 + log 5
(ii) log 12 – log 2
(iii) log₃ 5 + log₃ 2 + log₃ 4
(iv) log₂ 12 – log₂ 2 + log₂ 5

Sol:  (i) log 6 + log 5
= log (6×5)

(ii) log 12 – log 2
= log (12/2)

(iii) log₃ 5 + log₃ 2 + log₃ 4
= log3 (5×2×4)

(iv) log₂ 12 – log₂ 2 + log₂ 5
= log (12×5)/2

Que-2: Simplify without using tables :
(i) {(log40) 1000}/{(log40) 100}
(ii) (log32)/(log4)
(iii) log2 8

Sol:  (i) {(log40) 1000}/{(log40) 100}
= {(log40) (10)³}/{(log40) (10²)}
= {3 (log40) 10}/{2 (log40) 10}
= 3/2.

(ii) (log32)/(log4)
= {log (2^5)}/{log 2²}
= {5 log 2}/{2 log 2}
= 5/2

(iii) log2 8
= log2 (2)³
= 3 (log2) 2
= 3 × 1
= 3.

Que-3: Simplify:
(i) log (m²) – log m
(ii) log y² ÷ log y
(iii) log 24 – log 3
(iv) log 32 + log 4 – log 16
(v) log 256 – log 1024
(vi) log 256 – log 1024

Sol:  (i) log (m²) – log m
= 2 log m – log m
= log m

(ii) log y² ÷ log y
= 2 log y ÷ log y
= 2 log y / log y
= 2.

(iii) log 24 – log 3
= log (24/3)
= log 8
= log 2³
= 3 log 2.

(iv) log 32 + log 4 – log 16
= log {(32×4)/16}
= log 8
= log 2³
= 3 log 2.

(v) log 256 – log 1024
= log (356/1024)
= log (1/4)
= log (1/2)²
= log 2¯²
= -2 log²

(vi) log 256 ÷ log 1024
= log (2^8) ÷ log (2^10)
= 8 log 2 ÷ 10 log 2
= (8 log 2)/(10 log 2)
= 8/10
= 4/5

Que-4: (i) log 7 + log (1/7) = 0
(ii) log 72 = 3 log 2 + 2 log 3
(iii) log 448 = 6 log 2 + log 7
(iv) log (4/7) + log (33/18) – log (22/21) = 0
(v) log √{6*(2/9)} = (1/3)(log 2 – 2 log 3) + log 2
(vi) (log a)² – (log b)² = log (ab) log (a/b)

Sol: (i) log 7 + log (1/7) = 0
L.H.S. = log 7 + log (1/7)
= log 7 x (1/7) (∵ log mn = log m + log n)
= log 1
= 0 = R.H.S. (∵ log 1 = 0)

(ii) log 72 = 3 log 2 + 2 log 3
L.H.S. = log 72 = log (2³ x 3²)
= log 2³ + log 3²
(∵ log mn = log n + log n and log mⁿ = n log m)
72 = 2³ x 3²
= 3 log 2 + 2 log 3 = R.H.S

(iii) log 448 = 6 log 2 + log 7
L.H.S. = log 448 = log (2⁶ x 7)
= log 2⁶ + log 7
(∵ log mn = log n + log n and log mⁿ = n log m)
6 log 2 + log 7 = R.H.S

(iv) log (4/7) + log (33/18) – log (22/21) = 0
LHS =  log (4/7) + log (33/18) – log (22/21)
= log {(4/7)×(33/18)} – log (22/21)
= log (66/63) – log (22/21)
= log {(66/63)÷(22/21)}
= log {(66/63)×(21/22)}
= log 1 = 0 RHS.

(v) log √{6*(2/9)} = (1/3)(log 2 – 2 log 3) + log 2
LHS = log √{6*(2/9)}
= log (56/9)^(1/3)
= (1/3) log (56/9)
= (1/3) (log56 – log9)
= (1/3) (log 2³×7-3²)
= (1/3) (log 2³ – log 7 – 2 log 3)
= (1/3) (3 log 2 – log 7 – 2 log 3)
= (1/3)× 3 log 2 + (1/3) (log 7 – 2 log 3)
= log + (1/3) (log 7 – 2 log 3)
= (1/3) (log 7 – 2 log 3) + (log 2)
= RHS Proved.

(vi) (log a)² – (log b)² = log (ab) log (a/b)
LHS = (log a)² – (log b)²
= (log a + log b) (log a – log b)
= log (a×b) log (a/b)
= log (ab) log (a/b)
= RHS.

Que-5: Solve:
(i) log₁₀n + log₁₀5 = 1
(ii) log₃ n – log₃ 4 = 2
(iii) log₆ n – log₆ (n – 1) = log₆ 3
(iv) 2 log x = 4 log 3

Sol: (i) log₁₀ n + log₁₀ 5 = 1
⇒ log₁₀ (n x 5) = log₁₀ 10
{∵ log m + log n = log mn and log_a a = 1}
⇒ log₁₀ (5n) = log₁₀ 10
Comparing, we get
5z = 10 ⇒ n = 10/5 = 2
n = 2

(ii) log3 n – log3 4 = 2 = 2 x 1
⇒ log3 (n/4) = 2 log3 3 {∵ loga a = 1}
⇒ log3 (n/4) = log3 3² = log3 9
Comparing, we get
(n/4) = 9
⇒ n = 9 x 4 = 36
∴ n = 36

(iii) log₆ n – log₆ (n – 1) = log₆ 3
= log6 {n/(n-1)} = log6 3
Comparing we get,
n/(n-1) = 3
n = 3n – 3
2n = 3
n = 3/2

(iv) 2 log x = 4 log 3
⇒ log x = 2 log 3 (Dividing by 2)
⇒ log x = log 3² = log 9
Comparing, we get
x = 9

Que-6: Simplify : (Do not use tables)
(i) log₁₀ 5 + log₁₀ 2
(ii) log₁₀4 + log₁₀ 5 – log₁₀ 2
(iii) 2 log₁₀ 5 + log₁₀ 8 – (1/2) log₁₀ 4

Sol: (i) log10(4) +log10(5)-log10(2)
= log10(4×5) – log10(2) (Rule: logA + logB = logA×B)
= log10(20) – log10(2)
= log10(20÷2) (Rule: logA – logB = logA÷B)
= log10(10) ( logX(X) = 1 for all x )
= 1

(ii) (log10)4 + (log10)5 – (log10)2
= log10 {(4×5)/2}
= (log10) 10
= 1.

(iii) 2 log10 5 + log10 8 – (1/2) log10 4
= (log10) 5² + (log10) 8 – (1/2) (log10) 4^(1/2)
= (log10) 25 + (log10) 8 – (log10) 2
= log 10 {(25×8)/2}
= (log10) 100
= 2.

— : End of Logarithms Class 9 OP Malhotra Exe-7B ICSE Maths Solutions Ch-7 Step by Step Solutions :–

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

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