Parents can use OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(C) to provide additional support to their children.

## S Chand Class 9 ICSE Maths Solutions Chapter 7 Logarithms Ex 7(C)

If log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, log 7 = 0.8451 and log 11 = find the value of the following :

Question 1.

log 6

Solution:

log 6 = log (2 x 3) = log 2 + log 3

= 0.3010 + 0.4771 = 0.7781

Question 2.

log 12

Solution:

log 12 = log (2 x 2 x 3) = log (2² x 3)

= log 2² + log 3 = 2 log 2 + log 3

= 2 (0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791

Question 3.

log 15

Solution:

log 15 = log (3 x 5) = log 3 + log 5

= 0.4771 + 0.6990 = 1.1761

Question 4.

log 200

Solution:

log 200 = log (2³ x 5²) = log 2³ + log 5²

= 3 log 2 + 2 log 5

= 3 (0.3010) + 0.6990 x 2

= 0.9030 + 1.3980 = 2.3010

Question 5.

log 36

Solution:

sog 36 – log (4 x 9) = log (2² x 3²)

= log 2² + log 3² = 2 log 2 + 2 log 3

= 2 (0.3010) + 2 (0.4771)

= 0.6020 + 0.9542

= 1.5562

Question 6.

log 80

Solution:

log 80 = log (2^{4} x 5)

= log 2^{4} + log 5

= 4 log 2 + log 5 = 4 (0.3010) + 0.6990

= 1.2040 + 0.6990 = 1.9030

Question 7.

log 2\(\frac { 1 }{ 3 }\)

Solution:

log (2\(\frac { 1 }{ 3 }\)) = log(\(\frac { 7 }{ 3 }\)) = log 7 – log 3

= 0.8451 – 0.4771

= 0.3680

Question 8.

log 11³

Solution:

log (11)³ = 3 log 11

= 3 x 1.0414 = 3.1242

Question 9.

log (2\(\frac { 1 }{ 3 }\))^{5}

Solution:

log (2\(\frac { 1 }{ 3 }\))^{5} = log (\(\frac { 7 }{ 3 }\))^{5} = 5 log (\(\frac { 7 }{ 3 }\))

= 5 [log 7 – log 3] = 5 [0.8451 – 0.4771]

= 5 (0.3680) = 1.8400

Question 10.

If log 6 = 0.7782, find the value of log 36.

Solution:

log 6 = 0.7782

log 36 = log (6)² = 2 log 6

= 2 (0.7782) = 1.5564

Question 11.

Given log_{10} 25 = x, log_{10} 75 = y, evaluate without using logarithmic tables, in terms of x and y.

(i) log_{10} 3

(ii) log_{10}2

Solution:

Question 12.

Given log 31.87 = x, write down in terms of x.

(i) log (31.87)²

(ii) log_{10} 0.03187

(iii) log_{10} \(\sqrt{31870}\)

Solution:

log 31.87 = x

(i) log (31.87)² = 2 log (31.87)

= 2x

(ii) log_{10} 0.03187 = log_{10}\(\frac{31.87}{1000}\)

= log_{10} 31.87 – log_{10} 1000

= x – 3 (log 1000 = 3)

(iii) log_{10} \(\sqrt{31870}\) = log_{10} (31870)^{1/2}

= \(\frac { 1 }{ 2 }\) log_{10} 31870 = \(\frac { 1 }{ 2 }\) log (31.87 x 1000)

= \(\frac { 1 }{ 2 }\) [log 31.87 + log 1000]

= \(\frac { 1 }{ 2 }\) (x + 3) = \(\frac { x+3 }{ 2 }\)

Question 13.

Solve the equation

(i) log_{10} (x + 1) + log_{10} (A – 1) = log_{10} 11 + 2 log_{10}3

(ii) log (10x + 5) – log (x – 4) = 2

Solution:

(i) log_{10} (x + 1) + log_{10} (A – 1) = log_{10} 11 + 2 log_{10}3

⇒ log (x + 1) (x – 1) = log (11 x 3²) (∵ 2 log_{10} 3 = log_{10} 3²)

⇒ log (x² – 1) = log (11 x 9) ⇒ log (x² – 1) = log 99

Comparing we get.

x² – 1 = 99 ⇒ x² = 99 + 1 = 100 = (10)²

⇒ x = 10

∴ x = 10

(ii) log (10x + 5) – log (x – 4) = 2

⇒ log \(\frac{10 x+5}{x-4}\) = log 100 (∵ log 100 = 2)

Comparing both sides,

\(\frac{10 x+5}{x-4}=\frac{100}{1}\)

100x – 400 = 10x + 5

⇒ 100x – 10x = 5 + 400 ⇒ 90x = 405

⇒ x = \(\frac { 405 }{ 90 }\) = \(\frac { 45 }{ 10 }\) = 4.5

∴ x = 4.5

Question 14.

(a) Given 2 log_{10}x + \(\frac { 1 }{ 2 }\) log_{10}y = 1, express y in terms of x.

(b) Express as a single logarithm :

2 log 3 – \(\frac { 1 }{ 2 }\) log 16 + log 12

Solution:

Question 15.

Given that log_{10} y + 2log_{10} x = 2, express y in terms of x.

Solution:

Question 16.

If a = 1 + log_{10} 2 – log_{10} 5, b = 2 log10 3 and c = log_{10} m – log_{10} 5, find the value of m if a + b = 2c (Do not use log tables).

Solution:

a = 1 + log_{10} 2 – log_{10} 5 = log_{10} 10 + log_{10} 2 – log_{10} 5

= log_{10} \(\frac { 10×2 }{ 5 }\)– = log_{10} 4

b = 2 log_{10} 3 = log_{10} 3² = log_{10} 9

c = log_{10} m – log_{10} 5 = log_{10} \(\frac { m }{ 5 }\)

∵ a + b = 2c

∴ log_{10} 4 + log_{10} 9 = 2 log \(\frac { m }{ 5 }\)

⇒ log (4 × 9) = log (\(\frac { m }{ 5 }\))² = log\(\frac { m² }{ 25 }\)

Comparing both sides,

4 x 9 = \(\frac { m² }{ 25 }\) ⇒ m² = 4 x 9 x 25 = 900 = (30)²

∴ m = 30

Question 17.

Express as a single logarithm

2 + \(\frac { 1 }{ 2 }\) log_{10} 9 – 2 log_{10} 5

Solution:

Question 18.

If a = log 12, b – log 6 and c = 2 log \(\sqrt{2}\), find

(i) a – b – c

(ii) 9^{a-b-c}

Solution:

a = log 12, b = log 6, c = 2 log \(\sqrt{2}\) = log

\((\sqrt{2})^2\) = log 2

(i) a – b – c = log 12 – log 6 – log 2

= log \(\frac { 12 }{ 6×2 }\) = log 1 = 0 (∵ log 1 = 0)

(ii) 9^{a-b-c} = 9° [∵ a – b – c = 0 proved in (i)]

= 1 (∵ x° = 1)

Question 19.

If x = log_{10} 12, y = log_{4} 2 x log_{10} 9 and z = log_{10} (0.4) then find

(i) x – y – z

(ii) prove that 6^{x – y- z} = 6

Solution:

x = log_{10} 12, y = log_{4} 2 x log_{10} 9, z = log_{10} (0.4)

(i) x = log_{10} 12 = log_{10} (2² x 3) = log 2² + log 3

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

y = log_{4} 2 x log_{10}9 = log_{4}2 x log_{10} 3²

= log_{4} 2 x 2 log_{10} 3 = log_{4} (4)^{\(\frac { 1 }{ 2 }\)} x 2 log_{10} 3 (∴ log_{a} = 1)

= log_{10}3

z = log_{10} 0.4 = log_{10} \(\frac { 4 }{ 10 }\) = log_{10} 4 – log_{10} 10

= log_{10} 2² – 1 = 2 log_{10} 2 – 1

x – y – z = 2 log_{10} 2 + log_{10} 3 – log_{10} 3 – 2 log_{10}2 + 1 = 1

(ii) 6^{x-y-z} = 6^{1} = 6 [∵ x – y – z = 1 (proved in (i)]

Hence proved.

Question 20.

If p = log_{10} 20 and q = log_{10} 25, find x such that 2 log_{10} (x + 1) = 2p – q.

Solution:

p = log_{10} 20 = log_{10} (22 x 5) = log_{10} 2² + log_{10} 5

= 2 log 2 + log 5

q = log_{10} 25 = log_{10} (5²) = 2 log_{10} 5

Now,

2p – q = 2 [2 log_{10} 2 + log_{10} 5] – 2 log_{10} 5

= 4 log_{10} 2 + 2 log_{10} 5 – 2 log_{10} 5

= 4 log_{10} 2 = 2 log_{10} 2²

= 2 log_{10} 4

2 log_{10}(x + 1) = 2 log_{10} 4

Comparing, we get

x + 1 = 4 ⇒ x = 4 – 1 = 3

∴ x = 3

Question 21.

Without using logarithm tables, evaluate :

3 + log_{10 }(10^{-2})

Solution:

3 + log_{10} (10^{-2}) = 3 + (- 2 log_{10} 10)

= 3 – 2 log_{10} 10 = 3 – 2 x 1

= 3 – 2 = 1 (∵ log_{a} a = 1)

= 1

Question 22.

Given log_{10} x = a, log_{10} y = b

(i) Write down 10^{n-1} in terms of x

(ii) Write down 10^{2b} in terms of y

(iii) If log_{10} P = 2a – b, express P in terms of x and y.

Solution:

(i) log_{10} x = a, log_{10}y = b

∵ log_{10} x = a and log_{10} y = b

∴ 10^{n} = x … (i)

∴ 10^{b} = y … (ii)

Now,

(i) 10^{n-1} = \(\frac{10^a}{10^1}=\frac{x}{10}\) [From (i)]

(ii) 10^{2b} = (10^{b})² = y² [From (ii)]

(iii) log_{10} P = 2a – b

⇒ log_{10}P = 2 log_{10}x – log_{10}y

⇒ log_{10} P = log x² – log_{10} y

⇒ log_{10} P = log \(\frac { x² }{ y }\)

Comparing, we get

p = \(\frac { x² }{ y }\)

Question 23.

Simplify without using tables :

\(2 \log _{10} 5+\log _{10} 8-\frac{1}{2} \log _{10} 4\)

Solution:

Question 24.

Given that log_{10}2 = x, log_{10} 3 = y, find

(i) log_{10} 60

(ii) log_{10}1.2 in terms of x and y

Solution:

(i) log_{10} 60 = log_{10} (2 x 3 x 10)

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

= x + y + 1 (∵ log_{10} 10 = 1)

(ii) log_{10}1.2 = log_{10} \(\frac { 12 }{ 10 }\) = log_{10} 12 – log_{10} 10

= log (2² x 3) – log_{10} 10

= 2 log 2 + log 3 – log_{10} 10 (∵ log_{10} 10 = 1)

= 2x + y – 1

Question 25.

Given 2 log_{10} x + 1 = log_{10} 250, find

(i) x

(ii) log_{10}2x

Solution:

(i) 2 log_{10} x + 1 = log_{10} 250

⇒ log_{10}x² + 1 = log_{10} 250

⇒ log_{10} x² + log_{10} 10 = log_{10} 250 (∵ log_{a} a = 1)

⇒ log (x² x 10) = log_{10} 250

Comparing both side,

10x² = 250 ⇒ x² = 25 = (5)²

∴ x = 5

(ii) log_{10} 2x = log_{10} 2 x 5 = log_{10} 10 = 1

Question 26.

(a) Given that log x = m + n and log y = m – n, express the value of log_{10} \(\frac { 10x }{ y² }\) in terms of in and n. (b) Solve for x : log_{10}x = – 2.

Solution:

(a) log x = m + n, log y = m – n

log_{10} \(\frac{10 x}{y^2}=\log _{10} 10 x-\log _{10} y^2\)

= log_{10} 10 + log_{10} x – 2 log_{10} y

= 1 + m + n – 2 (m – n)

= 1 + m + n – 2m + 2n

= 1 – m + 3n

(b) log_{10} x = – 2 = log_{10}\(\frac { 1 }{ 100 }\)

Comparing we get,

x = \(\frac { 1 }{ 100 }\)

Question 27.

If log \(\left(\frac{p+q}{3}\right)=\frac{1}{2}\) = (log p + log q) prove that p² + q² = 7pq.

Solution:

Log \(\left(\frac{p+q}{3}\right)=\frac{1}{2}\) = (log p + log q)

⇒ \(\log \frac{p+q}{3}=\frac{1}{2}(\log p \times q)=\log (p q)^{\frac{1}{2}}\)

Comparing we get,

\(\frac{p+q}{3}=(p q)^{\frac{1}{2}}\)

Squaring both sides,

\(\left(\frac{p+q}{3}\right)^2=p q \Rightarrow \frac{p^2+q^2+2 p q}{9}\) = pq

⇒ P² + q² + 2 pq

⇒ p² + q² + 2pq = 9pq

⇒ p² + q² = 9pq – 2pq ⇒ 7pq

∴ p² + q² = 2pq

Hence proved.

Question 28.

If x² + y² = 51xy, prove that log \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) (log x + log y).

Solution:

x² + y² = 51xy

⇒ x² + y² – 2xy = 51 xy – 2xy

(Subtracting 2xy)

⇒ (x – y)² = 49xy