Logarithms Questions and Answers – Expanding and Condensing Logarithm – Set 2

This set of Aptitude Questions and Answers (MCQs) focuses on “Expanding and Condensing Logarithm – Set 2”.

1. Which of the following options is the correct simplified form of the expression log (75 / 16) – 2log (5 / 9) + log (32 / 343)?
a) log (2 / 3)
b) log (5 / 3)
c) log 2
d) 3 / 19
View Answer

Answer: c
Explanation: Given,
➩ log (75 / 16) – 2log (5 / 9) + log (32 / 343)
➩ log (25 * 3 / 4 * 4) – log (25 / 81) + log ((16 * 2) / (81 * 3))
➩ log (25 * 3) – log (4 * 4) – log (25) + log (81) + log (16 * 2) – log (81 * 3)
➩ log 25 + log 3 – log 16 – log 25 + log 81 + log 16 + log 2 – log 81 – log 3
➩ on simplifying gives log 2.

2. What is the value of log101.5?
a) 0.1751
b) 0.2468
c) 0.8423
d) 0.3752
View Answer

Answer: a
Explanation: Given,
log101.5
➩ log10 (15 / 10)
➩ log 15 – log 10
➩ log (5 * 3) – log 10
➩ log 5 + log 3 – log 10
➩ 0.698 + 0.4771 – 1
➩ 1.1751 – 1
➩ 0.1751

3. Which of the following is the correct solution of the expression log (x2 / yz) + log (y2 / xz) + log (z2 / xy)?
a) x2y2z2
b) 0
c) 1
d) xyz
View Answer

Answer: b
Explanation: Given,
log (x2 / yz) + log (y2 / xz) + log (z2 / xy)
➩ log ((x2 * y2 * z2) / (yz * xz * xy))
➩ log ((x2 * y2 * z2) / (x2 * y2 * z2))
➩ log 1
➩ 0
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4. Which of the following is a correct alternate representation of the expression logc z + logc (1 + z) = 0?
a) z2 + z – 1 = 0
b) z2 + 2z + 2 = 0
c) z3 + z2 + z = 0
d) z + 5 = 0
View Answer

Answer: a
Explanation: Given,
logc z + logc (1 + z) = 0
➩ logc z (1 + z) = 0
➩ logc z (1 + z) = logc1
➩ z (1 + z) = 1
➩ z + z2 = 1
➩ z2 + z – 1 = 0

5. Which values of m would satisfy the equation 2log2m = 1 + log2((5m / 2) – 3)?
a) 4, 1
b) 5, 3
c) 3, 2
d) 2, 9
View Answer

Answer: c
Explanation: Given,
2log2m = 1 + log2((5m / 2) – 3)
➩ log2m2 = log22 + log2((5m / 2) – 3)
➩ m2 = 2 ((5m / 2) – 3)
➩ m2 = 5m – 6
➩ m2 – 5m + 6 = 0
➩ m2 – 3m – 2m + 6 = 0
➩ m (m – 3) – 2 (m – 3) = 0
➩ (m – 3) (m – 2) = 0
➩ m = 3, 2

6. Which of the following options represent the right value of variable z if, (10000)z = 4?
a) 0.5692
b) 0.2558
c) 0.3010
d) 0.1505
View Answer

Answer: d
Explanation: Given,
(10000)z = 4
➩ z log 104 = log 4
➩ 4z = log 4
➩ z = log 4 / 4
➩ z = log 22 / 4
➩ z = 2 log 2 / 4
➩ z = log 2 / 2
➩ z = 0.3010 / 2
➩ z = 0.1505

7. Which of the following values of a satisfies the expression 55-a = 2-(5-a)?
a) 5
b) 2
c) 4
d) Can’t be determined
View Answer

Answer: a
Explanation: Given,
55-a = 2-(5-a)
➩ (5 – a) log 5 = – (5 – a) log 2
➩ (5 – a) log 5 + (5 – a) log 2 = 0
➩ (5 – a) (log 5 + log 2) = 0
➩ (5 – a) (log 10 / 2 + log 2) = 0
➩ (5 – a) (log 10 – log 2 + log 2) = 0
➩ (5 – a) (log 10) = 0
➩ 5 – a = 0
➩ a = 5
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8. What will be the value of log0.12532?
a) -3.54
b) 3.54
c) -1.67
d) 1.67
View Answer

Answer: c
Explanation: Given,
log0.12532
➩ 0.125 can be written as 2-3.
➩ 32 can be written as 25.
Thus, the expression can be written as:
➩ log2-3 25
➩ (5 / (- 3)) log22
➩ (5 / (- 3)) * 1
➩ (5 / (- 3))
➩ -1.67

9. What will be the value of log1010 + log101000 + log10100000 + log1010000000?
a) 16 log10100
b) 16
c) log1610
d) 132
View Answer

Answer: b
Explanation: Given,
log1010 + log101000 + log10100000 + log1010000000
➩ log10101 + log10103 + log10105 + log10107
➩ 1 log1010 + 3 log1010 + 5 log1010 + 7 log1010
➩ 1 + 3 + 5 + 7
➩ 16

10. What will be the value of b if log10a + log10b = c?
a) 10c / a
b) a / c
c) a2c2 / 2
d) 10a / c
View Answer

Answer: a
Explanation: Given,
log10a + log10b = c
➩ log10 (ab) = c
➩ ab = 10c
➩ b = 10c / a
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To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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