The big picture
The natural logarithm ln β‘ \ln ln inverse of the exponential. It turns a product into a sum, making it essential for every model that multiplies (growth, decibels, pHβ¦).
1 2 3 4 5 6 7 8 -3 -2 -1 1 2 3 (1, 0) (e, 1) f(x) = ln(x) Definition
ln β‘ : R + β β R \ln : \mathbb{R}_+^* \to \mathbb{R} ln : R + β β β R
The inverse of exp β‘ \exp exp
ln β‘ ( e x ) = x \ln(e^x) = x ln ( e x ) = x x x x e ln β‘ x = x e^{\ln x} = x e l n x = x x > 0 x > 0 x > 0
Key values: ln β‘ ( 1 ) = 0 \ln(1) = 0 ln ( 1 ) = 0 ln β‘ ( e ) = 1 \ln(e) = 1 ln ( e ) = 1
The 5 algebraic properties
These mirror the exponential rules. Memorize them.
Property Formula Product ln β‘ ( a b ) = ln β‘ a + ln β‘ b \ln(ab) = \ln a + \ln b ln ( ab ) = ln a + ln b Quotient ln β‘ ( a b ) = ln β‘ a β ln β‘ b \ln\left(\dfrac{a}{b}\right) = \ln a - \ln b ln ( b a β ) = ln a β ln b Reciprocal ln β‘ ( 1 a ) = β ln β‘ a \ln\left(\dfrac{1}{a}\right) = -\ln a ln ( a 1 β ) = β ln a Power ln β‘ ( a n ) = n ln β‘ a \ln(a^n) = n \ln a ln ( a n ) = n ln a Root ln β‘ ( a ) = 1 2 ln β‘ a \ln(\sqrt{a}) = \dfrac{1}{2} \ln a ln ( a β ) = 2 1 β ln a
β οΈ Only valid for a , b > 0 a, b > 0 a , b > 0
Derivative
Fundamental :
( ln β‘ x ) β² = 1 x forΒ x > 0 (\ln x)' = \dfrac{1}{x} \quad \text{for } x > 0 ( ln x ) β² = x 1 β forΒ x > 0
General (chain) :
( ln β‘ u ( x ) ) β² = u β² ( x ) u ( x ) (\ln u(x))' = \dfrac{u'(x)}{u(x)} ( ln u ( x ) ) β² = u ( x ) u β² ( x ) β
Examples :
( ln β‘ ( 2 x + 1 ) ) β² = 2 2 x + 1 (\ln(2x + 1))' = \dfrac{2}{2x + 1} ( ln ( 2 x + 1 ) ) β² = 2 x + 1 2 β ( ln β‘ ( x 2 + 1 ) ) β² = 2 x x 2 + 1 (\ln(x^2 + 1))' = \dfrac{2x}{x^2 + 1} ( ln ( x 2 + 1 ) ) β² = x 2 + 1 2 x β
Monotonicity
On R + β \mathbb{R}_+^* R + β β 1 / x > 0 1/x > 0 1/ x > 0 ln β‘ \ln ln
Bijective from R + β \mathbb{R}_+^* R + β β R \mathbb{R} R
Limits
Limit Value lim β‘ x β + β ln β‘ x \lim_{x \to +\infty} \ln x lim x β + β β ln x + β +\infty + β lim β‘ x β 0 + ln β‘ x \lim_{x \to 0^+} \ln x lim x β 0 + β ln x β β -\infty β β lim β‘ x β + β ln β‘ x x \lim_{x \to +\infty} \dfrac{\ln x}{x} lim x β + β β x ln x β 0 0 0 lim β‘ x β + β ln β‘ x x n \lim_{x \to +\infty} \dfrac{\ln x}{x^n} lim x β + β β x n ln x β 0 0 0 n > 0 n > 0 n > 0 lim β‘ x β 0 + x ln β‘ x \lim_{x \to 0^+} x \ln x lim x β 0 + β x ln x 0 0 0 lim β‘ x β 0 ln β‘ ( 1 + x ) x \lim_{x \to 0} \dfrac{\ln(1 + x)}{x} lim x β 0 β x ln ( 1 + x ) β 1 1 1
Comparative growth : logarithm loses to every power (opposite of exponential).
Equations and inequalities
ln β‘ \ln ln
ln β‘ a = ln β‘ b β
β βΊ β
β a = b \ln a = \ln b \iff a = b ln a = ln b βΊ a = b a , b > 0 a, b > 0 a , b > 0 ln β‘ a < ln β‘ b β
β βΊ β
β a < b \ln a < \ln b \iff a < b ln a < ln b βΊ a < b
Example 1
Solve ln β‘ ( 2 x β 3 ) = ln β‘ ( x + 1 ) \ln(2x - 3) = \ln(x + 1) ln ( 2 x β 3 ) = ln ( x + 1 )
Domain : x > 3 / 2 x > 3/2 x > 3/2 2 x β 3 = x + 1 β
β βΊ β
β x = 4 2x - 3 = x + 1 \iff x = 4 2 x β 3 = x + 1 βΊ x = 4
Example 2
Solve ln β‘ ( x 2 β 1 ) β€ 0 \ln(x^2 - 1) \leq 0 ln ( x 2 β 1 ) β€ 0
ln β‘ y β€ 0 β
β βΊ β
β y β€ 1 \ln y \leq 0 \iff y \leq 1 ln y β€ 0 βΊ y β€ 1 x 2 β 1 β€ 1 x^2 - 1 \leq 1 x 2 β 1 β€ 1 x 2 β€ 2 x^2 \leq 2 x 2 β€ 2
Domain : x 2 β 1 > 0 β x < β 1 x^2 - 1 > 0 \Rightarrow x < -1 x 2 β 1 > 0 β x < β 1 x > 1 x > 1 x > 1
Union: x β [ β 2 , β 1 [ βͺ ] 1 , 2 ] x \in [-\sqrt{2}, -1[ \cup ]1, \sqrt{2}] x β [ β 2 β , β 1 [ βͺ ] 1 , 2 β ]
Example 3
Solve ln β‘ x = 3 \ln x = 3 ln x = 3
x = e 3 β 20.09 x = e^3 \approx 20.09 x = e 3 β 20.09
Full study: f ( x ) = x ln β‘ x β x f(x) = x \ln x - x f ( x ) = x ln x β x
Domain : x > 0 x > 0 x > 0
Derivative : f β² ( x ) = ln β‘ x + 1 β 1 = ln β‘ x f'(x) = \ln x + 1 - 1 = \ln x f β² ( x ) = ln x + 1 β 1 = ln x
Sign : ln β‘ x > 0 β
β βΊ β
β x > 1 \ln x > 0 \iff x > 1 ln x > 0 βΊ x > 1
f β² < 0 f' < 0 f β² < 0 ( 0 , 1 ) (0, 1) ( 0 , 1 ) f β² > 0 f' > 0 f β² > 0 ( 1 , + β ) (1, +\infty) ( 1 , + β )
Minimum at x = 1 x = 1 x = 1 f ( 1 ) = β 1 f(1) = -1 f ( 1 ) = β 1
Limits :
x β 0 + x \to 0^+ x β 0 + f ( x ) β 0 f(x) \to 0 f ( x ) β 0 x β + β x \to +\infty x β + β f ( x ) β + β f(x) \to +\infty f ( x ) β + β
1 2 3 4 5 -1 1 2 min f(x) = xΒ·ln(x) β x Real applications
pH : pH = β log β‘ [ H + ] \text{pH} = -\log[H^+] pH = β log [ H + ] Decibels : L = 10 log β‘ ( I / I 0 ) L = 10 \log(I/I_0) L = 10 log ( I / I 0 β ) Earthquake magnitude : log scaleDoubling time : T = ln β‘ 2 k T = \dfrac{\ln 2}{k} T = k ln 2 β
Common mistakes
ln β‘ ( a + b ) β ln β‘ a + ln β‘ b \ln(a + b) \neq \ln a + \ln b ln ( a + b ) ξ = ln a + ln b products : ln β‘ ( a b ) \ln(ab) ln ( ab ) ln β‘ \ln ln Forgetting domain conditions : always check arguments are positive.ln β‘ \ln ln log β‘ \log log log β‘ \log log ln β‘ \ln ln
Solved exercises
Exercise 1
Simplify ln β‘ 12 β ln β‘ 3 + ln β‘ 4 \ln 12 - \ln 3 + \ln 4 ln 12 β ln 3 + ln 4
Show solution = ln β‘ ( 12 / 3 ) + ln β‘ 4 = ln β‘ 4 + ln β‘ 4 = 2 ln β‘ 4 = ln β‘ 16 = \ln(12/3) + \ln 4 = \ln 4 + \ln 4 = 2\ln 4 = \ln 16 = ln ( 12/3 ) + ln 4 = ln 4 + ln 4 = 2 ln 4 = ln 16
Exercise 2
Solve ln β‘ ( x + 2 ) + ln β‘ ( x β 1 ) = ln β‘ 4 \ln(x + 2) + \ln(x - 1) = \ln 4 ln ( x + 2 ) + ln ( x β 1 ) = ln 4
Show solution Domain: x > 1 x > 1 x > 1 ln β‘ ( ( x + 2 ) ( x β 1 ) ) = ln β‘ 4 \ln((x+2)(x-1)) = \ln 4 ln (( x + 2 ) ( x β 1 )) = ln 4 ( x + 2 ) ( x β 1 ) = 4 (x+2)(x-1) = 4 ( x + 2 ) ( x β 1 ) = 4 x 2 + x β 6 = 0 x^2 + x - 6 = 0 x 2 + x β 6 = 0 2 2 2 β 3 -3 β 3 2 2 2
Exercise 3
Derivative of f ( x ) = x ln β‘ x f(x) = x \ln x f ( x ) = x ln x
Show solution f β² ( x ) = ln β‘ x + x β
1 x = ln β‘ x + 1 f'(x) = \ln x + x \cdot \dfrac{1}{x} = \ln x + 1 f β² ( x ) = ln x + x β
x 1 β = ln x + 1
Takeaways
Domain R + β \mathbb{R}_+^* R + β β R \mathbb{R} R
Inverse of exp β‘ \exp exp ln β‘ ( e x ) = x \ln(e^x) = x ln ( e x ) = x e ln β‘ x = x e^{\ln x} = x e l n x = x
ln β‘ ( 1 ) = 0 \ln(1) = 0 ln ( 1 ) = 0 ln β‘ ( e ) = 1 \ln(e) = 1 ln ( e ) = 1 Product to sum : ln β‘ ( a b ) = ln β‘ a + ln β‘ b \ln(ab) = \ln a + \ln b ln ( ab ) = ln a + ln b Derivative: 1 / x 1/x 1/ x u β² / u u'/u u β² / u
Strictly increasing, bijective.
Loses to any power at + β +\infty + β
Going further
