Logarithm Aptitude Questions and Answer – Quantitative Aptitude

Logarithm Aptitude Questions and Answer – Formulas & Tricks

General Rule: The Logarithm of any positive real number y, other than 1 when ax = y, then component x is called the logarithm of y to the base a, then x = logay.

Logarithm Properties:

  • loga(xy) = logax + logay
  • loga(x/y) = logax – logay
  • logxx = 1
  • loga1 = 0
  • loga(xn) = n(logax)
  • logax = (1/logxa)
  • logax = (logbx / logba) = (log x / log a)

Important Terms:

  • The logarithm of one(1) to any base is zero = Loga1 = 0
  • The logarithm of zero (0) to any base greater than unity is Loga0 = -∞
  • The logarithm of any number (a) to the same base is always unity, logaa = 1
  • Let logbx=p, then x=bp, or x=blogbx
  • The logarithm of product: Loga(m * n) = logam * logan
  • The logarithm of the fraction: loga(m / n) = logam – logan
  • Power Formula: Logamn = nlogam
  • Logba = logca / logcb
  • Logbc = 1 / logcb
  • Logba = logca * logbc
  • Logxn(ym) = mlogay / nlogax

Common Logarithm: logarithm to the base 10 are known as a common logarithm. Therefore, Log1010 = 1

Mantissa:

Every logarithm has two parts

  1. Characteristics Part
  2. Mantissa Part

The integer part is called Characteristics Part

The decimal part of the logarithm called as Mantissa.

Characteristics Rules:

  1. To find the Characteristics of a number greater than one.
  2. To find the Characteristics of a number less than one.

Logarithm Aptitude Questions and Answer With Detailed ExplanationQuantitative Aptitude

Logarithm

Question 1
If log2 = 0.3010 and log3 = 0.4771, the value of log5 512 is:
A
2.870
B
2.967
C
3.876
D
3.912
Question 1 Explanation: 
Answer: Option C
Explanation:
Question 2
If log27 = 1.431, then the value of log9 is:
A
0.934
B
0.945
C
0.954
D
0.958
Question 2 Explanation: 
Answer: Option C
Explanation:
log27 = 1.431
=> log(33) = 1.431
=> 3log3 = 1.431
=> log3 = 0.477
Therefore, log9 = log(32) = 2 log 3 = ( 2 * 0.4777)
log9 = 0.954
Question 3
If ax = by,then:
A
B
C
D
None of these
Question 3 Explanation: 
Answer: Option C
Explanation:
Question 4
If logx y = 100 and log2x = 10, then the value of y is:
A
210
B
2100
C
21000
D
210000
Question 4 Explanation: 
Answer: Option C
Explanation:
log2x = 10 => x=210
Therefore, logxy = 100
=> y = x100
=> y = (210) [Put value of x]
=> y = 21000
Question 5
The value of log2 16 is:
A
2
B
4
C
8
D
16
Question 5 Explanation: 
Answer: Option B
Explanation:
Let log2 16 = n
Then, 2n = 16 =24
=> n =4
Therefore, log216 = 4
Question 6
If log102 = 0.3010, the value of log1080 is:
A
1.6020
B
1.9030
C
3.9030
D
4.1003
Question 6 Explanation: 
Answer: Option B
Explanation:
log1080 = log10 (8 * 10)
=log108 + log1010
=log10(23) + 1
=3log10 2 + 1
=(3 * 0.3010) + 1
=1.9030.
Question 7
If log105 + log10(5x + 1) = log10(x + 5) + 1, then x is equal to
A
1
B
3
C
5
D
10
Question 7 Explanation: 
Answer: Option B
Explanation:
log105 + log10(5x + 1) = log10(x + 5) + 1
=> log105 + log10(5x + 1) = log10(x + 5) + log1010
=>log10[5 (5x + 1)] = log10[10(x + 5)]
=>5(5x + 1) = 10(x + 5)
=>5x + 1 = 2x + 10
=>3x = 9
=> x = 3.
Question 8
A
B
C
D
Question 8 Explanation: 
Answer: Option C
Explanation:
Question 9
A
a + b = 1
B
a - b =1
C
a = b
D
a2 - b2 = 1
Question 9 Explanation: 
Answer: Option A
Explanation:
Question 10
A
B
C
D
Question 10 Explanation: 
Answer: Option A
Explanation:
There are 10 questions to complete.