# 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 2.87 B 2.967 C 3.876 D 3.912
Question 1 Explanation:
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:
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:
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:
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:
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 1.602 B 1.903 C 3.903 D 4.1003
Question 6 Explanation:
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:
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:
Explanation: Question 9 A a + b = 1 B a - b =1 C a = b D a2 - b2 = 1
Question 9 Explanation:  A  B  C  D   