Important Formulas: Logarithms | Quantitative Aptitude for SSC CGL PDF Download

Logarithm

A logarithm is a mathematical function designed to determine the exponent to which a particular base must be elevated to achieve a specified number. It stands as the reciprocal operation of exponentiation.

General Formula of Logarithm

• where b is the base of the logarithm.
• y is is the number for which we want to find the exponent.
• x is the exponent to which b must be raised to get y.

Definition & Logarithmic Formulas

Logarithms represent the exponent to which a number must be raised to attain another specified number.

Types of Logarithms

There are two types:

• Common logarithm
• Natural logarithm

Common Logarithm

Logarithm with a base of 10 is referred to as the common logarithm and is expressed as log10 X. If the base is not specified, it is assumed to be 10.
(a) Natural Logarithm:
(b) Logarithm with a base of 'e' is known as the natural logarithm and is expressed as loge X.

Crucial Note: In the absence of a specified base, always assume the base to be 10

Formulas for Logarithm

• log_x X= 1
• log_a 1= 0
• a log^log_ax =X
• log_a(x ^p) = p(log _ax)
• log_a (xy)= log_aX+ log_aY

Value of log(2 to 10): Remember

• Log 2 = 0.301
• Log 3 = 0.477= 0.48
• Log 4 = 0.60
• Log 5 = 0.698 = 0.7
• Log 6 = 0.778 = 0.78
• Log 7 = 0.845 = 0.85
• Log 8 = 0.90
• Log 9 = 0.954= 0.96
• Log 10 = 1

Logarithm Formulas (Antilog):

• An antilog is the inverse function of a logarithm.
  log(b) x = y means that antilog (b) y = x.
• The best way to understand any problem is by having a look at the Solved Example.
• We are going to do the same here, and we are going to understand the Antilog problem by Solved Example.

Examples

Example 1: If log 27= 1.431, then the value of log 9 is?
(a) 0.945
(b) 0.934
(c) 0.958
(d) 0.954
Ans: (d)
log 27= 1.431
⇒ log(3)³= 1.431
⇒ 3log 3= 1.431
⇒ log3= 0.477
therefore, log9= log 3²= 2 log3= (2×0.477)= 0.954

Example 2: Solve the equation log x= 1- log(x-3)
(a) 2
(b) 1/2
(c) 5
(d) 4
Ans: (c)
By combining both the equation we get
logx + log (x-3)=1
log(x(x-3))= log 10¹
Now convert it into exponential form,
x (x-3)= 10¹
x² – 3x-10= 0
(x-5) (x+2)=0
x= -2, x=5
By solving this equation we get two values for x.
x= -2, x=5
Put the different value of x in different equation and solve them,
x= -2
log(-2) = 1- log (-2-3)
x= 5
log5 = 1-log(5-3)
log5 = 1-log2
Negative value is not considered in logarithm. So, we have a single value of x i.e, x=5.

Example 3: If log ₁₀ 5+ log(5x+1) = log ₁₀ (x+5) +1, Find the value of X?
(a) 3
(b) 1
(c) 10
(d) 5
Ans: (a)
log ₁₀ 5+ log(5x+1) = log ₁₀ (x+5) +1.
log ₁₀ 5+ log(5x+1) = log ₁₀ (x+5) +log ₁₀ 10
log₁₀ [5 (5x+1) ] = log₁₀( 10 (x+5)]
5 (5x+1) = 10 (x+5)
5x+1 = 2x+ 10
3x= 9
x=3.

Example 4: If log   log(a+b), then
(a) a-b=1
(b) a=b
(c) a+b=1
(d) a²-b² = 1
Ans: (c)


so, a+b=1

Example 5: log₉ (3log₂ (1+log₃ (1+2log₂ x)))= 1/2. Find x.
(a) 2
(b) 1/2
(c) 1
(d) 4
Ans: (a)

 
log₂(1+log₃(1+2log₂ x) = 1
1+log₃ (1+2log₂ x)= 2¹
log₃ (1+2log₂ x)=2¹ − 1 
log₃ (1+2log₂ x) = 1
(1+2log₂ x) = 3¹ 
1+ 2log₂ x= 3
2log₂ x = 2
log₂x = 1
x= 2