Students can track their progress and improvement through regular use of OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Chapter Test.

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

Question 1.

The value of log_{2} 16 is

(a) \(\frac { 1 }{ 8 }\)

(b) 4

(c) 8

(d) 16

Solution:

(b) 4

log_{2}16 = log_{2}2^{4} = 4log_{2}² = 4 x 1 (∵ log_{2}a = 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_{a}x = b_{b}y ⇒ 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 log_{10}2 = 0.3010, the value of log_{10}5 is

(a) 0.3241

(b) 0.6911

(c) 0.6990

(d) 0.7525

Solution:

log_{10}2 = 0.3010

log_{10}5 = log_{10} (\(\frac { 10 }{ 2 }\)) = log_{10} 10 – log_{10} 2

= 1 – 0.3010 = 0.6990 (c)

Question 5.

If log_{10}2 = 0.3010, the value of log_{10}80 is

(a) 1.6020

(b) 1.9030

(c) 3.9030

(d) None of these

Solution:

(b) 1.9030

log_{10} 2 = 0.3010

log_{10} 80 = log_{10} 10 x 2³

= log_{10} 10 + log_{10} 2³

= log_{10} 10 + 3log_{10} 2

= 1 + 3 x 0.3010

= 1 + 0.9030 = 1.9030

Question 6.

If log_{10}7 = a, then log_{10} (\(\frac { 1 }{ 70 }\)) is equal to

(a) – (1 + a)

(b) (1 + a)^{-1}

(c) \(\frac { a }{ 10 }\)

(d) \(\frac { 1 }{ 10a }\)

Solution:

(a) – (1 + a)

log_{10}7 = a

log_{10}(\(\frac { 1 }{ 70 }\)) = log_{10} 1 – log_{10} 70

= 0 – log_{10} (7 x 10)

= 0 – log_{10} 7 – log_{10} 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 log_{10} 5 + log_{10}(5x + 1) = log _{10}(x + 5) + 1, then x is equal to

(a) 1

(b) 3

(c) 5

(d) 10

Solution:

(b) 3

log_{10}5 + log_{10}(5x + 1) = log_{10}(x + 5) + 1

log_{10}5 (5x + 1) = log_{10} (x + 5) + log_{10}10

log_{10}(5x + 1) = log_{10}10 (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} [log_{2} (log_{7} x)] = 0 is

(a) 2

(b) π²

(c) 72

(d) None of these

Solution:

log_{π} [log_{2} (log_{7} x)] = 0

⇒ log_{2} (log_{7} x) = π° = 1

⇒ log_{7}x = 2^{1} = 2

⇒ x = 7²

∴ x = 7²