OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(C)

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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 (2⁴ x 5)

= log 2⁴ + 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 log₁₀ 25 = x, log₁₀ 75 = y, evaluate without using logarithmic tables, in terms of x and y.
(i) log₁₀ 3
(ii) log₁₀2
Solution:

Question 12.
Given log 31.87 = x, write down in terms of x.
(i) log (31.87)²
(ii) log₁₀ 0.03187
(iii) log₁₀ \(\sqrt{31870}\)
Solution:
log 31.87 = x
(i) log (31.87)² = 2 log (31.87)
= 2x

(ii) log₁₀ 0.03187 = log₁₀\(\frac{31.87}{1000}\)
= log₁₀ 31.87 – log₁₀ 1000
= x – 3 (log 1000 = 3)

(iii) log₁₀ \(\sqrt{31870}\) = log₁₀ (31870)^1/2
= \(\frac { 1 }{ 2 }\) log₁₀ 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) log₁₀ (x + 1) + log₁₀ (A – 1) = log₁₀ 11 + 2 log₁₀3
(ii) log (10x + 5) – log (x – 4) = 2
Solution:
(i) log₁₀ (x + 1) + log₁₀ (A – 1) = log₁₀ 11 + 2 log₁₀3
⇒ log (x + 1) (x – 1) = log (11 x 3²) (∵ 2 log₁₀ 3 = log₁₀ 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 log₁₀x + \(\frac { 1 }{ 2 }\) log₁₀y = 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 log₁₀ y + 2log₁₀ x = 2, express y in terms of x.
Solution:

Question 16.
If a = 1 + log₁₀ 2 – log₁₀ 5, b = 2 log10 3 and c = log₁₀ m – log₁₀ 5, find the value of m if a + b = 2c (Do not use log tables).
Solution:
a = 1 + log₁₀ 2 – log₁₀ 5 = log₁₀ 10 + log₁₀ 2 – log₁₀ 5
= log₁₀ \(\frac { 10×2 }{ 5 }\)– = log₁₀ 4
b = 2 log₁₀ 3 = log₁₀ 3² = log₁₀ 9
c = log₁₀ m – log₁₀ 5 = log₁₀ \(\frac { m }{ 5 }\)
∵ a + b = 2c
∴ log₁₀ 4 + log₁₀ 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 + \(\frac { 1 }{ 2 }\) log₁₀ 9 – 2 log₁₀ 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 = log₁₀ 12, y = log₄ 2 x log₁₀ 9 and z = log₁₀ (0.4) then find
(i) x – y – z
(ii) prove that 6^{x – y- z} = 6
Solution:
x = log₁₀ 12, y = log₄ 2 x log₁₀ 9, z = log₁₀ (0.4)
(i) x = log₁₀ 12 = log₁₀ (2² x 3) = log 2² + log 3
= 2 log₁₀ 2 + log₁₀ 3
y = log₄ 2 x log₁₀9 = log₄2 x log₁₀ 3²
= log₄ 2 x 2 log₁₀ 3 = log₄ (4)^{\(\frac { 1 }{ 2 }\)} x 2 log₁₀ 3 (∴ log_a = 1)
= log₁₀3
z = log₁₀ 0.4 = log₁₀ \(\frac { 4 }{ 10 }\) = log₁₀ 4 – log₁₀ 10
= log₁₀ 2² – 1 = 2 log₁₀ 2 – 1
x – y – z = 2 log₁₀ 2 + log₁₀ 3 – log₁₀ 3 – 2 log₁₀2 + 1 = 1

(ii) 6^x-y-z = 6¹ = 6 [∵ x – y – z = 1 (proved in (i)]
Hence proved.

Question 20.
If p = log₁₀ 20 and q = log₁₀ 25, find x such that 2 log₁₀ (x + 1) = 2p – q.
Solution:
p = log₁₀ 20 = log₁₀ (22 x 5) = log₁₀ 2² + log₁₀ 5
= 2 log 2 + log 5
q = log₁₀ 25 = log₁₀ (5²) = 2 log₁₀ 5
Now,
2p – q = 2 [2 log₁₀ 2 + log₁₀ 5] – 2 log₁₀ 5
= 4 log₁₀ 2 + 2 log₁₀ 5 – 2 log₁₀ 5
= 4 log₁₀ 2 = 2 log₁₀ 2²
= 2 log₁₀ 4
2 log₁₀(x + 1) = 2 log₁₀ 4
Comparing, we get
x + 1 = 4 ⇒ x = 4 – 1 = 3
∴ x = 3

Question 21.
Without using logarithm tables, evaluate :
3 + log₁₀ (10^-2)
Solution:
3 + log₁₀ (10^-2) = 3 + (- 2 log₁₀ 10)
= 3 – 2 log₁₀ 10 = 3 – 2 x 1
= 3 – 2 = 1 (∵ log_a a = 1)
= 1

Question 22.
Given log₁₀ x = a, log₁₀ y = b
(i) Write down 10^n-1 in terms of x
(ii) Write down 10^2b in terms of y
(iii) If log₁₀ P = 2a – b, express P in terms of x and y.
Solution:
(i) log₁₀ x = a, log₁₀y = b
∵ log₁₀ x = a and log₁₀ y = b
∴ 10ⁿ = 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) log₁₀ P = 2a – b
⇒ log₁₀P = 2 log₁₀x – log₁₀y
⇒ log₁₀ P = log x² – log₁₀ y
⇒ log₁₀ 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 log₁₀2 = x, log₁₀ 3 = y, find
(i) log₁₀ 60
(ii) log₁₀1.2 in terms of x and y
Solution:
(i) log₁₀ 60 = log₁₀ (2 x 3 x 10)

= log₁₀ 2 + log₁₀ 3 + log₁₀ 10
= x + y + 1 (∵ log₁₀ 10 = 1)

(ii) log₁₀1.2 = log₁₀ \(\frac { 12 }{ 10 }\) = log₁₀ 12 – log₁₀ 10
= log (2² x 3) – log₁₀ 10
= 2 log 2 + log 3 – log₁₀ 10 (∵ log₁₀ 10 = 1)
= 2x + y – 1

Question 25.
Given 2 log₁₀ x + 1 = log₁₀ 250, find
(i) x
(ii) log₁₀2x
Solution:
(i) 2 log₁₀ x + 1 = log₁₀ 250
⇒ log₁₀x² + 1 = log₁₀ 250
⇒ log₁₀ x² + log₁₀ 10 = log₁₀ 250 (∵ log_a a = 1)
⇒ log (x² x 10) = log₁₀ 250
Comparing both side,
10x² = 250 ⇒ x² = 25 = (5)²
∴ x = 5

(ii) log₁₀ 2x = log₁₀ 2 x 5 = log₁₀ 10 = 1

Question 26.
(a) Given that log x = m + n and log y = m – n, express the value of log₁₀ \(\frac { 10x }{ y² }\) in terms of in and n. (b) Solve for x : log₁₀x = – 2.
Solution:
(a) log x = m + n, log y = m – n
log₁₀ \(\frac{10 x}{y^2}=\log _{10} 10 x-\log _{10} y^2\)
= log₁₀ 10 + log₁₀ x – 2 log₁₀ y
= 1 + m + n – 2 (m – n)
= 1 + m + n – 2m + 2n
= 1 – m + 3n

(b) log₁₀ x = – 2 = log₁₀\(\frac { 1 }{ 100 }\)

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