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

Logarithms Class 9 OP Malhotra Ch-Test ICSE Maths Solutions Ch-7

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-7	Logarithms
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Logarithms

Logarithms Class 9 OP Malhotra ICSE Maths Solutions Ch-7

Que-1: The value of log2 16 is

(a) 1/8
(b) 4
(c) 8
(d) 16

Sol: (b) 4
log2 16 = log2 (2^4)
= 4log2² = 4 x 1 (∵ log2(a) = 1)
= 4

Que-2: If a^x = b^y then

(a) log (a/b) = x/y
(b) loga / logb = x/y
(c) loga / logb = y/x
(d) None of these

Sol: (c) loga / logb = y/x
a^x = b^y
Taking log both sides,
log_ax = b_by ⇒ xlog_a = ylog_b
⇒ loga / logb = y/x

Que-3: If log 3 = 0.477 and (1000)^x = 3, then x equals

(a) 0.0159
(b) 0.0477
(c) 0.159
(d) 10

Sol: (c) 0.159
log 3 = 0.477
(1000)^x = 3
Taking log of both sides
x log 1000 = log3
⇒ x × 3 = log3 (∵ log 1000 = 3)
⇒ 3x = 0.477
⇒ x = 0.477/3
x = 0.159

Que-4: If log₁₀2 = 0.3010, the value of log₁₀5 is

(a) 0.3241
(b) 0.6911
(c) 0.6990
(d) 0.7525

Sol: (c) 0.6990
log₁₀2 = 0.3010
log₁₀5 = log₁₀ (10/2)
log₁₀ 10 – log₁₀ 2
= 1 – 0.3010 = 0.6990

Que-5: If log₁₀2 = 0.3010, the value of log₁₀80 is

(a) 1.6020
(b) 1.9030
(c) 3.9030
(d) None of these

Sol: (b) 1.9030
log₁₀ 2 = 0.3010
log₁₀ 80 = log₁₀ 10 x 2³
= log₁₀ 10 + log₁₀ 2³
= log₁₀ 10 + 3log₁₀ 2
= 1 + 3 x 0.3010
= 1 + 0.9030 = 1.9030

Que-6: If log₁₀7 = a, then log₁₀ (1/70) is equal to

(a) – (1 + a)
(b) (1 + a)^-1
(c) a/10
(d) 1/10a

Sol: (a) – (1 + a)
log₁₀7 = a
log₁₀(1/70) = log₁₀ 1 – log₁₀ 70
= 0 – log₁₀ (7 x 10)
= 0 – log₁₀ 7 – log₁₀ 10
= 0 – o – 1 = – (1 + a)

Que-7: If log 27 = 1.431, then the value of log 9 is

(a) 0.934
(b) 0.945
(c) 0.954
(d) 0.958

Sol: (c) 0.954
log 27 = 1.431 ⇒ log 3³ = 1.431
⇒ 31og 3 = 1.431
⇒ log 3 = 1.431/3 = 0.477
log 9 = log 3² = 2 log 2
= 2 x 0.477 = 0.954

Que-8: If log₁₀ 5 + log₁₀(5x + 1) = log ₁₀(x + 5) + 1, then x is equal to

(a) 1
(b) 3
(c) 5
(d) 10

Sol: (b) 3
log₁₀5 + log₁₀(5x + 1) = log₁₀(x + 5) + 1
log₁₀5 (5x + 1) = log₁₀ (x + 5) + log₁₀10
log₁₀(5x + 1) = log₁₀10 (x + 5)
Comparing, we get
⇒ 5(5x + 1) = 10(x + 5)
⇒ 25x + 5 = 10x + 50
⇒ 25x – 10x = 50 – 5
⇒ 15x = 45
⇒ x = 45/15 = 3
x = 3

Que-9: If log_x 4 = 0.4, then the value of x is

(a) 1
(b) 4
(c) 16
(d) 32

Sol: log_x 4 = 0.4
⇒ 4 = x^0.4
⇒ x = 4^(1/0.4) = (2²)^(1/0.4)
= 2^{2×(1/0.4)} = 2^(1/0.2)
= 2^(1/5) = 2^{1×(5/1)} = 2^5
= 2 x 2 x 2 x 2 x 2 = 32
∴ x = 32

Que-10: The solution of log_π [log₂ (log₇ x)] = 0 is

(a) 2
(b) π²
(c) 72
(d) None of these

Sol: log_π [log₂ (log₇ x)] = 0
⇒ log₂ (log₇ x) = π° = 1
⇒ log₇x = 2¹ = 2
⇒ x = 7²
∴ x = 7²

— : End of Logarithms 9 OP Malhotra Ch-Test ICSE Maths Step by step Solutions :–

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

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