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

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

Question 1.
Given an equivalent exponential form for each statement :
(i) log₂ 8 = 3
(ii) log₃ 81 = 4
(iii) log₂ \(\frac { 1 }{ 2 }\) = – 1
(iv) log₅ \(\frac { 1 }{ 25 }\)= – 2
(v) \(\frac { 1 }{ 2 }\) = log₄ 2
(vi) \(\frac { 1 }{ 3 }\) = log₂₇ 3
Solution:
(i) log₂ 8 = 3 ⇒ 2³ = 8
(ii) log₃ 81 = 4 ⇒ 3⁴ = 81
(iii) log₂ \(\frac { 1 }{ 2 }\) = – 1 ⇒ 2^-1 = \(\frac { 1 }{ 2 }\)
(iv) log₂ \(\frac { 1 }{ 25 }\) = – 2 ⇒ 5^-2 = \(\frac { 1 }{ 25 }\)
(v) \(\frac { 1 }{ 2 }\) = log₄ 2 ⇒ 4^{\(\frac { 1 }{ 2 }\)} = 2
(vi) \(\frac { 1 }{ 3 }\) = log₂₇ 3 ⇒ 27^{\(\frac { 1 }{ 3 }\)} = 3

Question 2.
Give an equivalent logarithmic form for each statement :
(i) 16 = 2⁴
(ii) 25 = 5²
(iii) 81 = 3⁴
(iv) 6° = 1
(v) 8^{\(\frac { 1 }{ 3 }\)} = 2
(vi) \(\frac { 1 }{ 9 }\) = 3^-2
(vii) \(\frac { 1 }{ 32 }\) = 2^-5
(viii) 10^1.4969 = 31.4
Solution:
(i) 16 = 2⁴ ⇒ log₂ 16 = 4
(ii) 25 = 5² ⇒ log₅ 25 = 2
(iii) 81 = 3⁴ ⇒ log₃ 81 = 4
(iv) 6° = 1 ⇒ log₆ 1 = 0
(v) 8^{\(\frac { 1 }{ 3 }\)} ⇒ log₈ 2 = \(\frac { 1 }{ 3 }\)
(vi) \(\frac { 1 }{ 9 }\) = 3^-2 ⇒ log₃ \(\frac { 1 }{ 9 }\) = – 2
(vii) \(\frac { 1 }{ 32 }\) = 2^-5 ⇒ log₃ \(\frac { 1 }{ 32 }\) = – 5
(viii) 10^1.4969 = 31.4 ⇒ log₁₀ 31.4 = 1.4969

Question 3.
Find the value of each logarithm given below:
(i) log₁₀ 1000
(ii) log₂ 8
(iii) log₃ 81
(iv) log₁₀ 0.1
(v) log₁₀ 0.01
(vi) log₁₀ 0.0001
(vii) log₂ \(\frac { 1 }{ 4 }\)
(viii) log₃ \(\frac { 1 }{ 27 }\)
(ix) log₃ 1
(x) log_{\(\frac { 1 }{ 2 }\)} \(\frac { 1 }{ 4 }\)
(xi) log₂₇ 9
(xii) log_{\(\frac { 1 }{ 125 }\)} 125
(xiii) log_{\(\frac { 5 }{ 6 }\)} 1
(xiv) log_{\(\frac { 1 }{ 3 }\)} 9
(xv) log₁₀ 10
Solution:
(i) log₁₀ 1000
Let log₁₀ 1000 = x, then 10^x = 1000
⇒ 10^x = (10)³
Comparing, we get
x = 3
∴ log₁₀ 1000 = 3

(ii) log₂ 8
Let log₂ 8 = x, then 2^x = 8
⇒ 2^x = 23
Comparing, we get
x = 3
∴ log₂ 8 = 3

(iii) log₃ 81
Let log₃ 81 = x, then 3^x = 81 = 3⁴
∴ 3^x = 3⁴
Comparing we get,
x = 4
∴ log₃ 81 = 4

(iv) log₁₀ 0.1
Let log₁₀ 0.1 = , then 10^x = 0.1 = \(\frac { 1 }{ 10 }\) = 10^-1
Comparing we get,
x = – 1
∴ log₁₀ 0.1 = – 1

(v) log₁₀ 0.01
Let log₁₀ 0.01 = x then 10^x = 0.01 = \(\frac { 1 }{ 100 }\)
= \(\frac { 1 }{ 10² }\)
∴ 10^x = 10^-2
Comparing we get,
x = – 2
∴ log₁₀0.01 = – 2

(vi) log₁₀ 0.0001
Let log₁₀ 0.0001 = x, then 10^x = 0.0001
= \(\frac{1}{10000}=\frac{1}{10^4}\)
⇒ 10^x = 10^-4
Comparing we get, x = – 4
∴ log₁₀ 0.0001 = – 4

(vii) log₂ \(\frac { 1 }{ 4 }\)
Let log₂ \(\frac { 1 }{ 4 }\) = x, then 2^x = \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2² }\) = 2^-2
Comparing we get,
x = – 2
∴ log₂ \(\frac { 1 }{ 4 }\) = – 2

(viii) log₃ = \(\frac { 1 }{ 27 }\)
Let log₃ \(\frac { 1 }{ 27 }\) = x, then 3^x = \(\frac{1}{27}=\frac{1}{3^3}\)
⇒ 3^x = 3^-3
Comparing, we get
x = – 3
∴ log₃ \(\frac { 1 }{ 27 }\) = – 3

(ix) log₃ 1
Let log₃ 1 = x, then 3^x = 1 = 3° (∵ 3° = 1)
Comparing, we get
x = 0
∴ log₃ 1 = 0

(x) \(\log _{\frac{1}{2}} \frac{1}{4}\)
Let \(\log _{\frac{1}{2}} \frac{1}{4}=x \text {, then }\left(\frac{1}{2}\right)^x=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)
Comparing, we get
x = 2
∴ \(\log _{\frac{1}{2}} \frac{1}{4}\) = 2

(xi) log₂₇ 9
Let log₂₇ 9 = x, then 27^x = 9
[(3)³]^x = (3)² ⇒ 3^3x = 3²
Comparing, we get
3x = 2 ⇒ x = \(\frac { 2 }{ 3 }\)
∴ log₂₇ 9 = \(\frac { 2 }{ 3 }\)

(xii) log_{\(\frac { 1 }{ 125 }\)} 125
Let log_{\(\frac { 1 }{ 125 }\)} 125 = x, then \(\left(\frac{1}{5}\right)^x=\frac{1}{125}=\left(\frac{1}{5}\right)^3\)
Comparing, we get
x = 3
∴ log_{\(\frac { 1 }{ 125 }\)} 125 = 3

(xiii) log_{\(\frac { 5 }{ 6 }\)} 1

(xiv) log_{\(\frac { 1 }{ 3 }\)} 9
Let log_{\(\frac { 1 }{ 3 }\)} 9, then (\(\frac { 1 }{ 3 }\))^x = 9 = 3²
⇒ \(\left(\frac{1}{3}\right)^x=\left(\frac{1}{3}\right)^{-2}\)
Comparing, we get
x = – 2
∴ log_{\(\frac { 1 }{ 3 }\)} 9 = – 2

(xv) log₁₀ 10
Let log₁₀ 10 = x, then 10^x = 10 = (10)¹
Comparing, we get x = 1
∴ log₁₀ 10 = 1

Question 4.
Find the value of x
(i) log_x 216 = 3
(ii) log₄ x = – 4
(iii) log₃ x = 0
(iv) log₈ x = \(\frac { 2 }{ 3 }\)
(v) log₁₀ 100 = x
(vi) log₂ 0.5 = x
Solution:
(i) log_x 216 = 3
⇒ x³ = 216 = (6)³
Comparing, we get
x = 6

(ii) log₄ x = – 4
⇒ (4)^-4 = x ⇒ \(\frac{1}{4^4}\) = x
⇒ x = \(\frac { 1 }{ 256 }\)
∴ x = \(\frac { 1 }{ 256 }\)

(iii) log₃ x = 0 ⇒ 3° = x
⇒ x = 1
∴ x = 1

(iv) log₈ x = \(\frac { 2 }{ 3 }\) ⇒ \((8)^{\frac{2}{3}}\)
⇒ x = \(\left(2^3\right)^{\frac{2}{3}}=2^{3 \times \frac{2}{3}}\) = 2²
⇒ x = 2 x 2 = 4
∴ x = 4

(v) log₁₀ 100 = x ⇒ 10^x = 100
⇒ 10^x = (10)²
Comparing, we get
x = 2

(vi) log₂ 0.5 = x ⇒ 2^x = 0.5 = \(\frac { 1 }{ 2 }\)
⇒ 2^x = (2)^-1
Comparing both sides
x = – 1

Question 5.
Answer true or false : If log₁₀x = a, then 10^a = x.
Solution:
∵ log₁₀ x = a ⇒ 10^a = x
Which is true.