OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Chapter Test

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S Chand Class 9 ICSE Maths Solutions Chapter 7 Logarithms Chapter Test

Question 1.
The value of log₂ 16 is
(a) \(\frac { 1 }{ 8 }\)
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
log₂16 = log₂2⁴ = 4log₂² = 4 x 1 (∵ log₂a = 1)
= 4

Question 2.
If a^x = b^y then
(a) log\(\frac{a}{b}=\frac{x}{y}\)
(b) \(\frac{\log a}{\log b}=\frac{x}{y}\)
(c) \(\frac{\log a}{\log b}=\frac{y}{x}\)
(d) None of these
Solution:
(c) \(\frac{\log a}{\log b}=\frac{y}{x}\)
a^x = b^y
Taking log both sides,
log_ax = b_by ⇒ xlog_a = ylog_b
⇒ \(\frac{\log a}{\log b}=\frac{y}{x}\)

Question 3.
If log 3 = 0.477 and (1000)^x = 3, then x equals
(a) 0.0159
(b) 0.0477
(c) 0.159
(d) 10
Solution:
(c) 0.159
log 3 = 0.477
(1000)^x = 3
Taking log of both sides
xlog 1000 = log3
⇒ x × 3 = log3 (∵ log 1000 = 3)
⇒ 3x = 0.477 ⇒ x = \(\frac { 0.477 }{ 3 }\)
x = 0.159

Question 4.
If log₁₀2 = 0.3010, the value of log₁₀5 is
(a) 0.3241
(b) 0.6911
(c) 0.6990
(d) 0.7525
Solution:
log₁₀2 = 0.3010
log₁₀5 = log₁₀ (\(\frac { 10 }{ 2 }\)) = log₁₀ 10 – log₁₀ 2
= 1 – 0.3010 = 0.6990 (c)

Question 5.
If log₁₀2 = 0.3010, the value of log₁₀80 is
(a) 1.6020
(b) 1.9030
(c) 3.9030
(d) None of these
Solution:
(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

Question 6.
If log₁₀7 = a, then log₁₀ (\(\frac { 1 }{ 70 }\)) is equal to
(a) – (1 + a)
(b) (1 + a)^-1
(c) \(\frac { a }{ 10 }\)
(d) \(\frac { 1 }{ 10a }\)
Solution:
(a) – (1 + a)
log₁₀7 = a
log₁₀(\(\frac { 1 }{ 70 }\)) = log₁₀ 1 – log₁₀ 70
= 0 – log₁₀ (7 x 10)
= 0 – log₁₀ 7 – log₁₀ 10
= 0 – o – 1 = – (1 + a)

Question 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
Solution:
(c) 0.954
log 27 = 1.431 ⇒ log 3³ = 1.431
⇒ 31og 3 = 1.431
⇒ log 3 = \(\frac { 1.431 }{ 3 }\) = 0.477
log 9 = log 3² = 2 log 2
= 2 x 0.477 = 0.954

Question 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
Solution:
(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 = \(\frac { 45 }{ 15 }\) = 3
x = 3

Question 9.
If log_x 4 = 0.4, then the value of x is
(a) 1
(b) 4
(c) 16
(d) 32
Solution:
log_x 4 = 0.4
⇒ 4 = x^0.4
⇒ x = \(4^{\frac{1}{0.4}}=\left(2^2\right)^{\frac{1}{0.4}}\)
= \(2^{2 \times \frac{1}{0.4}}=2^{\frac{1}{0.2}}\)
= \(2^{\frac{1}{5}}=2^{1 \times \frac{5}{1}}=2^5\)
= 2 x 2 x 2 x 2 x 2 = 32
∴ x = 32

Question 10.
The solution of log_π [log₂ (log₇ x)] = 0 is
(a) 2
(b) π²
(c) 72
(d) None of these
Solution:
log_π [log₂ (log₇ x)] = 0
⇒ log₂ (log₇ x) = π° = 1
⇒ log₇x = 2¹ = 2
⇒ x = 7²
∴ x = 7²