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

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

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-7	Indices Exponents
Writer	OP Malhotra
Exe-7A	Introduction to Logarithms
Edition	2025-2026

Introduction to Logarithms

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

Que-1: Given an equivalent exponential form for each statement :

(i) log₂ 8 = 3
(ii) log₃ 81 = 4
(iii) log₂ (1/2) = -1
(iv) log₅ (1/25) = -2
(v) (1/2) = log₄ 2
(vi) (1/3) = log₂₇ 3

Sol: (i) log2 8 = 3
⇒ 2³ = 8

(ii) log3 81 = 4
⇒ 3^4 = 81

(iii) log2 (1/2) = – 1
⇒ 2^-1 = (1/2)

(iv) log2 (1/25) = – 2
⇒ 5^-2 = 1/25

(v) (1/2) = log4 2
⇒ 4^(1/2) = 2

(vi) 1/3 = log27 3
⇒ 27^(1/3) = 3

Que-2: Give an equivalent logarithmic form for each statement :
(i) 16 = 24
(ii) 25 = 5²
(iii) 81 = 3^4
(iv) 6° = 1
(v) 8^(1/3) = 2
(vi) 1/9 = 3^-2
(vii) 1/32 = 2^-5
(viii) 10^1.4969 = 31.4

Sol: (i) 16 = 24
⇒ log2 16 = 4

(ii) 25 = 5²
⇒ log5 25 = 2

(iii) 81 = 34
⇒ log3 81 = 4

(iv) 6° = 1
⇒ log6 1 = 0

(v) 8^(1/3)
⇒ log8 2 = 1/3

(vi) 1/9 = 3^-2
⇒ log3 (1/9) = – 2

(vii) 1/32 = 2^-5
⇒ log3 (1/32) = – 5

(viii) 10^1.4969 = 31.4
⇒ log10 31.4 = 1.4969

Que-3: Find the value of each logarithm given below:
(i) log10 1000
(ii) log2 8
(iii) log3 81
(iv) log10 0.1
(v) log10 0.01
(vi) log10 0.0001
(vii) log2 (1/4)
(viii) log3 (1/27)
(ix) log3 1
(x) log(1/2) (1/4)
(xi) log27 9
(xii) log(1/125) 125
(xiii) log(5/6) 1
(xiv) log(1/3) 9
(xv) log10 10

Sol: (i) log₁₀ 1000

Let log₁₀ 1000 = x, then 10^x = 1000
⇒ 10^x = (10)³
Comparing, we get
x = 3
∴ log₁₀ 1000 = 3

(ii) log₂ 8

Let log₂ 8 = x, then 2^x = 8
⇒ 2^x = 23
Comparing, we get
x = 3
∴ log₂ 8 = 3

(iii) log₃ 81

Let log₃ 81 = x, then 3^x = 81 = 3⁴
∴ 3^x = 3⁴
Comparing we get,
x = 4
∴ log₃ 81 = 4

(iv) log₁₀ 0.1

Let log₁₀ 0.1 = , then 10^x = 0.1 = 1/10 = 10^-1
Comparing we get,
x = – 1
∴ log₁₀ 0.1 = – 1

(v) log10 0.01

Let log10 0.01 = x then 10x = 0.01 = 1/100
= 1/10²
∴ 10^x = 10^-2
Comparing we get,
x = – 2
∴ log10 0.01 = – 2

(vi) log10 0.0001

Let log10 0.0001 = x, then 10x = 0.0001
= 1/10000 = 1/10^4
⇒ 10^x = 10^-4
Comparing we get, x = – 4
∴ log10  0.0001 = – 4

(vii) log2 (1/4)

Let log2 (1/4) = x, then 2x = (1/4) = 1/2² = 2^-2
Comparing we get,
x = – 2
∴ log2 (1/4) = – 2

(viii) log3 = (1/27)

Let log3 (1/27) = x, then 3x = (1/27) = 1/3³
⇒ 3^x = 3^-3
Comparing, we get
x = – 3
∴ log3 (1/27) = – 3

(ix) log3 1

Let log3 1 = x, then 3x = 1 = 3° (∵ 3° = 1)
Comparing, we get
x = 0
∴ log3 1 = 0

(x) log(1/2) (1/4)

Let log(1/2) (1/4) = x, then (1/2)^x = 1/4 = (1/2)²
Comparing, we get
x = 2
∴ log(1/2) (1/4) = 2

(xi) log27 9

Let log27 9 = x, then 27x = 9
[(3)³]x = (3)² ⇒ 3^3x = 3²
Comparing, we get
3x = 2 ⇒ x = 2/3
∴ log27 9 = 2/3

(xii) log(1/125) 125

Let log(1/125) 125 = x, then (1/5)^x = 1/125 = (1/5)³
Comparing, we get
x = 3
∴ log(1/125) 125 = 3

(xiii) log(5/6) 1

Let,  log(5/6) 1 = x then,
(5/6)^x = 1 = (5/6)^0
(5/6)^x = (5/6)^0
Comparing, we get,
x = 0
log(5/6) 1 = 0

(xiv) log(1/3) 9

Let log(1/3) 9, then (1/3)^x = 9 = 3²
⇒ (1/3)^x = (1/3)¯²
Comparing, we get
x = – 2
∴ log(1/3) 9 = – 2

(xv) log10 10

Let log10 10 = x, then 10^x = 10 = (10)¹
Comparing, we get x = 1
∴ log10 10 = 1

Que-4: Find the value of x
(i) log_x 216 = 3
(ii) log₄ x = – 4
(iii) log₃ x = 0
(iv) log₈ x = 2/3
(v) log₁₀ 100 = x
(vi) log₂ 0.5 = x

Sol: (i) log_x 216 = 3
⇒ x³ = 216 = (6)³
Comparing, we get
x = 6

(ii) log4 x = – 4
⇒ (4)^-4 = x
⇒ 1/4^4 = x
⇒ x = 1/256
∴ x = 1/256

(iii) log3 x = 0
⇒ 3° = x
⇒ x = 1
∴ x = 1

(iv) log8 x = 2/3
⇒ (8)^(2/3)
⇒ x = (2³)^(2/3) = 2^(3×(2/3)) = 2²
⇒ x = 2 x 2 = 4
∴ x = 4

(v) log10 100 = x ⇒ 10x = 100
⇒ 10x = (10)²
Comparing, we get
x = 2

(vi) log₂ 0.5 = x ⇒ 2^x = 0.5 = 1/2 = (2)^-1
Comparing both sides
x = – 1

Que-5: Answer true or false : If log₁₀x = a, then 10^a = x.

Sol: ∵ log₁₀ x = a ⇒ 10^a = x
Which is true.

— : End of Logarithms Class 9 OP Malhotra Exe-7A ICSE Maths Solutions Ch-7 Solutions / :–

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