Natural Log

Natural log is the log of a number with base "e" where 'e' is Euler number and its value is 2.718 (approximately). The natural log is defined by the symbol 'ln'. The natural log formula is given as, suppose, e^x = a then log_e = a, and vice versa.

Here log_e is called a natural log i.e., log with base e. The natural log is always represented by the symbol "ln". Thus, ln x = log_e x.
The natural log of a positive number is represented as 'ln x'. The natural log of numbers is a way of representing an exponent.

If e^x = y for any number "x"
then ln (x) = y

Natural Log Formulas

Various natural log formulas are,

• ln (1) = 0
• ln (e) = 1
• ln (-x) = Not Defined {log of negative number is Not-Defined}
• ln (∞) = ∞
• ln(e^x) = x, x ∈ R

Change of Base for Natural Log

Base of log can be easily changed using the formula,

log_e a = (log_c a)/(log_c e)

Natural Logarithms Table

Natural log of any number is the log with base e. The natural log of various number are added in the table below,

Please refer Difference Between Log and Ln for more details on how natural differs from log.

Examples Using Natural Log Formula

Example 1: Solve,

• e^x = 10  
• ln x = 2  
• e^{ln 15}  
• ln(e²⁹)  

Solution:

• e^x = 10
  x = ln 10
  x = 2.303

• ln x = 2
  x = e²
  x = 7.389

• e^{ln 15} = 15

• ln(e²⁹) = 29

Example 2: Solve, ln(15x - 3) = 2
Solution: 

ln(15x - 3) = 2

15x - 3 = e²
15x -3 = 7.389
15x = 10.389

x = 10.389/15 ⇒ x = 0.6926

Example 3: If 8e^xy + 2 = 98  and  2e^z + 3 = 79, then find the value of x + y,  where z = x² + y²
Solution: 

8e^xy + 2 = 98

8e^xy = 98-2 = 96
e^xy = 96/8 = 12
ln(e^xy) = ln 12
xy = 2.4849...(i)

2e^z + 3 = 79
2e^z = 79-3 = 76
e^z = 76/2 = 38

ln(e^z) = ln 38

z = 3.6375...(ii)

z = x² + y²
Now, (x + y)² = x² + y² + 2xy

From eq(i) and eq(ii)

(x + y)² = 3.6375 + 2 × 2.4849
(x + y)² = 3.6375 + 4.9698
(x + y)² = 8.6073
(x + y) = √8.6073

x + y = 2.933

Example 4: Simplify y = ln 25 - ln 15
Solution: 

y = ln 25 - ln 15
y = ln(5 × 5) - ln(5 × 3) 
y = ln 5 + ln 5 - [ln 5 + ln3]  
y = ln 5 + ln 5 - ln 5 - ln3  
y = ln 5 - ln 3
y = ln (5/3)

y = 0.511 

Example 5: Solve: ln(e¹⁵) + e ^2+x = 16
Solution: 

ln(e¹⁵) + e ^2+x = 16

⇒ 15 + e^2+x = 16
⇒ e^2+x = 16 - 15
⇒ e^2+x = 1
⇒ ln( e^2+x )= ln 1
⇒ 2 + x = 0

⇒ x = -2