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 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 ax = by 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) $$\frac{\log a}{\log b}=\frac{y}{x}$$
ax = by
Taking log both sides,
logax = bby ⇒ xloga = ylogb
⇒ $$\frac{\log a}{\log b}=\frac{y}{x}$$

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
= 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 logx 4 = 0.4, then the value of x is
(a) 1
(b) 4
(c) 16
(d) 32
Solution:
logx 4 = 0.4
⇒ 4 = x0.4
⇒ 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 logπ [log2 (log7 x)] = 0 is
(a) 2
(b) π²
(c) 72
(d) None of these
Solution:
logπ [log2 (log7 x)] = 0
⇒ log2 (log7 x) = π° = 1
⇒ log7x = 21 = 2
⇒ x = 7²
∴ x = 7²