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Natural Log Formula

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  • Last Updated : 15 May, 2022

The logarithm is an important topic for mathematics. The logarithm is used in various formulae in mathematics. This article is about different important natural log formulas which are very useful in various fields. The knowledge of the natural log formula helps to solve various problems. The base of the natural log is the mathematical constant ‘e‘, whose value is e = 2.718. 

Natural Log Formulae

A natural log is a power to the base ‘e’, which is raised to find the value of natural log of a given number. The base of the natural log is mathematical constant ‘e’. It is the logarithm of a number to the base of ‘e’. Natural log is written as logex or ln x. (or log x with e as an implicit base). Representation of natural log, the natural log can be represented in two forms: ln x or logex. As natural log has its base as ‘e’ so it is written as logex.

  • ln of 1: The natural log of 1 is 0.
  • ln of e: The natural log of e is 1.
  • ln of a negative number: The natural log of a negative number is not defined.
  • ln of ∞: The natural log of ∞ is.
  • ln in the form of e: As ln has the base of e, it can be represented in the powers of e. ln x = y <=> ey = x 
  • e to the power ln x: The result of e to the power ln x is x for x>0. eln x  = x, x > 0
  • ln of e to the power x: Since the base of ln is e, then ln of e to the power x results in x for all x belongs to the real number. ln(ex) = x, x ∈ R
  • Product Rule: When we have a natural log of the product of two numbers, then it can be represented as the addition of the natural log of the first number and the natural log of the second number. ln(xy) = ln x + ln y
  • Quotient Rule: When we have a natural log of a fraction of two numbers, then it can be represented as the subtraction of natural log of the first number and the natural log of the second number.   ln(x/y) = ln x – ln y
  • Power Rule: When we have a natural log of x to power r, then it can be represented as r times ln x. ln(xr) = r.ln x
  • Reciprocal Rule: When we have a natural log of reciprocal of x, it can be represented as minus of the natural log of x. ln(1/x) = -ln x

Natural Log Formulae

Representation of natural log in both forms      loge x = ln x 
ln of 1      ln 1 = 0
ln of e      ln e = 1
ln of a negative number      Not defined
ln of infinity      ln ∞ = ∞ 
The conversion of ln in the form of e

      ln x = y  <=>   ey = x 

      (e is the base of natural log)         

e to the power ln x      eln x = x ,  x>0
ln of e to the power x      ln(ex) = x , x ∈ R
Product Rule      ln(xy) = ln x + ln y
Quotient Rule      ln(x/y) = ln x – ln y
Power Rule      ln(xr) = r.ln x
Reciprocal Rule      ln(1/x) = -ln x
Base change Rule      logba = (ln a)/(ln b)
Equality of ln      ln x = ln y   <=>   x = y

Sample Problems

Question 1: Solve: 

  1. ex = 10  
  2. ln x = 2  
  3. eln 15  
  4. ln(e29)  
  5. ln(39)  
  6. ln(15/4) 
  7. ln(39
  8. log57

Solution: 

  1. ex = 10 ⇒ x = ln 10 ⇒ x = 2.303
  2. ln x = 2 ⇒ x = e2  ⇒ x = 7.389
  3. eln 15 = 15 
  4. ln(e29) = 29
  5. ln(39) = ln(13×3) = ln 13 + ln 3 = 2.565 + 1.099 = 3.664
  6. ln(15/4) = ln 15 – ln 4 = 2.708 – 1.386 = 1.322
  7. ln(39) = 9×ln 3 = 9×1.099 = 9.891
  8. log57 = (ln 7)/(ln 5) = 1.946/1.609 = 1.209

Question 2: Solve: 

  1. ln(15x – 3) = 2  
  2. e2x-2 = 4  
  3. 5e4x + 3 = 13

Solution: 

1. ln(15x – 3) = 2 ⇒ 15x – 3 = e2 ⇒ 15x -3 = 7.389 ⇒ 15x = 10.389 ⇒ x = 10.389/15 ⇒ x = 0.6926

2. e2x-2 = 4 (Applying ln both sides)

⇒ ln(e2x-2) = ln 4

⇒ 2x-2 = 1.386

⇒ 2x = 1.386 + 2

⇒ 2x = 3.386

⇒ x = 3.386/2

x = 1.693

3. 5e4x + 3 = 13

⇒ 5e4x = 13 – 3 ⇒ 5e4x = 10

⇒ e4x = 10/5 ⇒ e4x = 2 [Applying ln both the sides]

⇒ ln(e4x) = ln 2   

⇒ 4x = ln 2 ⇒ 4x = 0.693 ⇒ x = 0.693/4 

x = 0.173  

Question 3: If 8exy + 2 = 98  and  2ez + 3 = 79, then find the value of x + y,  where z = x2 + y2

Solution: 

8exy + 2 = 98

⇒ 8exy = 98-2 = 96

⇒ exy = 96/8 = 12 [Applying ln on both sides] 

⇒ ln(exy) = ln 12 [Since ln(ex )= x]

⇒ xy = 2.4849 ⇢ Equation 1

2ez + 3 = 79

⇒ 2ez = 79-3 = 76

⇒ ez = 76/2 = 38 [Applying ln on both sides] 

⇒ ln(ez) = ln 38 [Since ln(ex )= x]

⇒ z = x2 + y2 = 3.6375 ⇢ Equation 2

Now, (x + y)2 = x2 + y2 + 2xy [Putting value of Equation 1 and Equation 2]

(x + y)2 = 3.6375 + 2 × 2.4849

(x + y)2 = 3.6375 + 4.9698

(x + y)2 = 8.6073

(x + y) = √8.6073 = 2.933

Question 4: Evaluate: y = ln 25 – ln 15

Solution: 

y = ln 25 – ln 15 = ln(5 × 5) – ln(5 × 3) 

= ln 5 + ln 5 – [ln 5 + ln3]  

= ln 5 + ln 5 – ln 5 – ln3  

= ln 5 – ln 3

= ln (5/3)

y = 0.511 

Question 5: Solve: ln(e15) + e 2+x = 16

Solution: 

ln(e15) + e 2+x = 16

⇒ 15 + e2+x = 16

⇒ e2+x = 16 – 15

⇒ e2+x = 1 [Applying ln both the sides]

⇒ ln( e2+x )= ln 1

⇒ 2 + x = 0

x = -2

Question 6: Evaluate: p = log35 – log36 + log310

Solution : 

p = (ln 5/ ln 3) – (ln 6/ ln 3) + (ln 10/ ln 3)

= [ln 5 -(ln 6 + ln 10)] / ln 3 

= [ln 5 – ln (6 × 10)]/ ln 3 

= [ln 5 – ln 60]/ ln 3 

= [ln(5/60)] / ln 3

= [ln(1/12)] / ln 3

= [ln (12)-1] / ln 3

= [-1×ln 12] / ln 3

= -ln 12 / ln 3

p = -2.262

Question 7: If ln 6 = a , ln 8 = b , ln 16 = c , ln 12 = d  then write d in terms of a, b and c.

Solution:

(a + c) – b = (ln 6 + ln 16) – ln 8 [ln a + ln b = ln (ab)]

⇒ a + c – b = ln 96 – ln 8 [ln a – ln b = ln (a/b)]

⇒ a + c – b = ln 96/8

⇒ a + c – b = ln 12

⇒ a + c – b =  d 

d = a + c – b

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