The natural logarithm of a number is its logarithm to the base of the mathematical constant e, which is an irrational and transcendental number approximately equal to 2.718. The natural logarithm of x is generally written as ln x, loge x, or sometimes, if the base e is implicit, simply log x. Parentheses are sometimes added for clarity, giving ln(x), loge(x), or log(x). This is done particularly when the argument to the logarithm is not a single symbol, so as to prevent ambiguity. The natural logarithm of x is the power to which e would have to be raised to equal x. For example, ln 7.5 is 2.0149..., because e2.0149... = 7.5. The natural logarithm of e itself, ln e, is 1, because e1 = e, while the natural logarithm of 1 is 0, since e0 = 1. The natural logarithm can be defined for any positive real number a as the area under the curve y = 1/x from 1 to a (with the area being negative when 0 < a < 1). The simplicity of this definition, which is matched in many other formulas involving the natural logarithm, leads to the term "natural". The definition of the natural logarithm can then be extended to give logarithm values for negative numbers and for all non-zero complex numbers, although this leads to a multi-valued function: see complex logarithm for more. The natural logarithm function, if considered as a real-valued function of a positive real variable, is the inverse function of the exponential function, leading to the identities:
e ln x = x if x ∈ R + ln e x = x if x ∈ R {\displaystyle {\begin{aligned}e^{\ln x}&=x\qquad {\text{ if }}x\in \mathbb {R} _{+}\\\ln e^{x}&=x\qquad {\text{ if }}x\in \mathbb {R} \end{aligned}}}
Like all logarithms, the natural logarithm maps multiplication of positive numbers into addition:
ln ( x ⋅ y ) = ln x + ln y . {\displaystyle \ln(x\cdot y)=\ln x+\ln y~.}
Logarithms can be defined for any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and can be defined in terms of the latter, log b x = ln x / ln b {\displaystyle \log _{b}x=\ln x/\ln b} . Logarithms are useful for solving equations in which the unknown appears as the exponent of some other quantity. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches of mathematics and scientific disciplines, and are used to solve problems involving compound interest.
History
The concept of the natural logarithm was worked out by Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa before 1649. Their work involved quadrature of the hyperbola with equation xy = 1, by determination of the area of hyperbolic sectors. Their solution generated the requisite "hyperbolic logarithm" function, which had the properties now associated with the natural logarithm. An early mention of the natural logarithm was by Nicholas Mercator in his work Logarithmotechnia, published in 1668, although the mathematics teacher John Speidell had already compiled a table of what in fact were effectively natural logarithms in 1619. It has been said that Speidell's logarithms were to the base e, but this is not entirely true due to complications with the values being expressed as integers.
Definition and properties The natural logarithm can be defined more generally as the inverse function of the exponential function e x {\displaystyle e^{x}} , so that e ln ( x ) = x {\displaystyle e^{\ln(x)}=x} or ln ( e x ) = x {\displaystyle \ln(e^{x})=x} . Because the exponential function e x {\displaystyle e^{x}} is positive and invertible for any real input x {\displaystyle x} , this definition of ln ( x ) {\displaystyle \ln(x)} is well defined for any positive x {\displaystyle x} . That is, the domain of a natural logarithm is x ∈ ( 0 , ∞ ) {\displaystyle x\in (0,\infty )} .
The natural logarithm can be defined as the area under the graph of a rectangular hyperbola with equation y = 1 / x {\displaystyle y=1/x} between x = 1 {\displaystyle x=1} and x = a {\displaystyle x=a} . That is, ln a = ∫ 1 a 1 x d x . {\displaystyle \ln a=\int _{1}^{a}{\frac {1}{x}}\,dx.} If a is in ( 0 , 1 ) {\displaystyle (0,1)} , then the region has negative area, and the logarithm is negative. The natural logarithm obeys the following mathematical properties, which can be used to simplify formulae that combine them with multiplication or exponentiation:
ln 1 = 0 , ln e = 1 , ln ( x y ) = ln x + ln y for x > 0 and y > 0 , ln ( x / y ) = ln x − ln y for x > 0 and y > 0 , ln ( x y ) = y ln x for x > 0 , ln ( x y ) = ( ln x ) / y for x > 0 and y ≠ 0 , ln x < ln y for 0 < x < y . {\displaystyle {\begin{aligned}\ln 1&=0,\\\ln e&=1,\\\ln(xy)&=\ln x+\ln y\quad {\text{for }}\;x>0\;{\text{and }}\;y>0,\\\ln(x/y)&=\ln x-\ln y\quad {\text{for }}\;x>0\;{\text{and }}\;y>0,\\\ln(x^{y})&=y\ln x\quad {\text{for }}\;x>0,\\\ln({\sqrt[{y}]{x}})&=(\ln x)/y\quad {\text{for }}\;x>0\;{\text{and }}\;y\neq 0,\\\ln x&<\ln y\quad {\text{for }}\;0<x<y.\end{aligned}}}
As x {\displaystyle x} approaches 0 {\displaystyle 0} from the right, the natural logarithm approaches negative infinity, − ∞ {\displaystyle -\infty } . As x {\displaystyle x} approaches ∞ {\displaystyle \infty } , the natural logarithm approaches ∞ {\displaystyle \infty } . By the concept of limit, these are:
lim x → 0 + ln x = − ∞ , lim x → ∞ ln x = ∞ {\displaystyle \lim _{x\to 0^{+}}\ln x=-\infty ,\quad \lim _{x\to \infty }\ln x=\infty }
Notation The mathematical notation for the natural logarithm of x, or the logarithm to the base e of a number x, can be written as ln x, as loge x, or, when the base e is implicitly known, as log x. The loge x form is a specific instance of the general notation for the logarithm to base b of a number x, which is shown as logb x. (For example, the base-2 logarithm of 8 can be written as log2 8 = 3.) Some authors use log x without an explicit base to refer to the natural logarithm. An example can be commonly found in prime number theorem. In addition to mathematics, this usage is commonplace in some programming languages. However, in some other contexts such as chemistry, log x can be used to denote the common (base 10) logarithm. It may also refer to the binary (base 2) logarithm in the context of computer science, particularly in the context of time complexity.
In calculus
Several identities
lim x → 0 ln ( 1 + x ) x = 1 {\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=1}
lim α → 0 x α − 1 α = ln x for x > 0 {\displaystyle \lim _{\alpha \to 0}{\frac {x^{\alpha }-1}{\alpha }}=\ln x\quad {\text{for }}\;x>0}
x − 1 x ≤ ln x ≤ x − 1 for x > 0 {\displaystyle {\frac {x-1}{x}}\leq \ln x\leq x-1\quad {\text{for}}\quad x>0}
ln ( 1 + x α ) ≤ α x for x ≥ 0 and α ≥ 1 {\displaystyle \ln {(1+x^{\alpha })}\leq \alpha x\quad {\text{for}}\quad x\geq 0\;{\text{and }}\;\alpha \geq 1}
Series
Since the natural logarithm ln ( x ) {\displaystyle \ln(x)} is undefined at 0, the function itself does not have a Maclaurin series, unlike many other elementary functions. Instead, one looks for Taylor expansions around other points. For example, if | x − 1 | ≤ 1 {\displaystyle \vert x-1\vert \leq 1} and x ≠ 0 , {\displaystyle x\neq 0,} then
ln x = ∫ 1 x 1 t d t = ∫ 0 x − 1 1 1 + u d u = ∫ 0 x − 1 ( 1 − u + u 2 − u 3 + ⋯ ) d u = ( x − 1 ) − ( x − 1 ) 2 2 + ( x − 1 ) 3 3 − ( x − 1 ) 4 4 + ⋯ = ∑ k = 1 ∞ ( − 1 ) k − 1 ( x − 1 ) k k . {\displaystyle {\begin{aligned}\ln x&=\int _{1}^{x}{\frac {1}{t}}\,dt=\int _{0}^{x-1}{\frac {1}{1+u}}\,du\\&=\int _{0}^{x-1}(1-u+u^{2}-u^{3}+\cdots )\,du\\&=(x-1)-{\frac {(x-1)^{2}}{2}}+{\frac {(x-1)^{3}}{3}}-{\frac {(x-1)^{4}}{4}}+\cdots \\&=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(x-1)^{k}}{k}}.\end{aligned}}}
This is the Taylor series for ln x {\displaystyle \ln x} around 1. A change of variables yields the Mercator series:
ln ( 1 + x ) = ∑ k = 1 ∞ ( − 1 ) k − 1 k x k = x − x 2 2 + x 3 3 − ⋯ , {\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}x^{k}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}
valid for | x | ≤ 1 {\displaystyle |x|\leq 1} and x ≠ − 1. {\displaystyle x\neq -1.}
Leonhard Euler, disregarding x ≠ − 1 {\displaystyle x\neq -1} , nevertheless applied this series to x = − 1 {\displaystyle x=-1} to show that the harmonic series equals the natural logarithm of 1 1 − 1 {\textstyle {\frac {1}{1-1}}} ; that is, the logarithm of infinity. Nowadays, more formally, one can prove that the harmonic series truncated at N is close to the logarithm of N, when N is large, with the difference converging to the Euler–Mascheroni constant. The figure is a graph of ln(1 + x) and some of its Taylor polynomials around 0. These approximations converge to the function only in the region −1 < x ≤ 1; outside this region, the higher-degree Taylor polynomials devolve to worse approximations for the function. A useful special case for positive integers n, taking x = 1 n {\displaystyle x={\tfrac {1}{n}}} , is:
ln ( n + 1 n ) = ∑ k = 1 ∞ ( − 1 ) k − 1 k n k = 1 n − 1 2 n 2 + 1 3 n 3 − 1 4 n 4 + ⋯ {\displaystyle \ln \left({\frac {n+1}{n}}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{kn^{k}}}={\frac {1}{n}}-{\frac {1}{2n^{2}}}+{\frac {1}{3n^{3}}}-{\frac {1}{4n^{4}}}+\cdots }
If Re ( x ) ≥ 1 / 2 , {\displaystyle \operatorname {Re} (x)\geq 1/2,} then
ln ( x ) = − ln ( 1 x ) = − ∑ k = 1 ∞ ( − 1 ) k − 1 ( 1 x − 1 ) k k = ∑ k = 1 ∞ ( x − 1 ) k k x k = x − 1 x + ( x − 1 ) 2 2 x 2 + ( x − 1 ) 3 3 x 3 + ( x − 1 ) 4 4 x 4 + ⋯ {\displaystyle {\begin{aligned}\ln(x)&=-\ln \left({\frac {1}{x}}\right)=-\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}({\frac {1}{x}}-1)^{k}}{k}}=\sum _{k=1}^{\infty }{\frac {(x-1)^{k}}{kx^{k}}}\\&={\frac {x-1}{x}}+{\frac {(x-1)^{2}}{2x^{2}}}+{\frac {(x-1)^{3}}{3x^{3}}}+{\frac {(x-1)^{4}}{4x^{4}}}+\cdots \end{aligned}}}
Now, taking x = n + 1 n {\displaystyle x={\tfrac {n+1}{n}}} for positive integers n, we get:
ln ( n + 1 n ) = ∑ k = 1 ∞ 1 k ( n + 1 ) k = 1 n + 1 + 1 2 ( n + 1 ) 2 + 1 3 ( n + 1 ) 3 + 1 4 ( n + 1 ) 4 + ⋯ {\displaystyle \ln \left({\frac {n+1}{n}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k(n+1)^{k}}}={\frac {1}{n+1}}+{\frac {1}{2(n+1)^{2}}}+{\frac {1}{3(n+1)^{3}}}+{\frac {1}{4(n+1)^{4}}}+\cdots }
If Re ( x ) ≥ 0 {\displaystyle \operatorname {Re} (x)\geq 0} and x ≠ 0 , {\displaystyle x\neq 0,} then
ln ( x ) = ln ( 2 x 2 ) = ln ( 1 + x − 1 x + 1 1 − x − 1 x + 1 ) = ln ( 1 + x − 1 x + 1 ) − ln ( 1 − x − 1 x + 1 ) . {\displaystyle \ln(x)=\ln \left({\frac {2x}{2}}\right)=\ln \left({\frac {1+{\frac {x-1}{x+1}}}{1-{\frac {x-1}{x+1}}}}\right)=\ln \left(1+{\frac {x-1}{x+1}}\right)-\ln \left(1-{\frac {x-1}{x+1}}\right).}
Since
ln ( 1 + y ) − ln ( 1 − y ) = ∑ i = 1 ∞ 1 i ( ( − 1 ) i − 1 y i − ( − 1 ) i − 1 ( − y ) i ) = ∑ i = 1 ∞ y i i ( ( − 1 ) i − 1 + 1 ) = y ∑ i = 1 ∞ y i − 1 i ( ( − 1 ) i − 1 + 1 ) = i − 1 → 2 k 2 y ∑ k = 0 ∞ y 2 k 2 k + 1 , {\displaystyle {\begin{aligned}\ln(1+y)-\ln(1-y)&=\sum _{i=1}^{\infty }{\frac {1}{i}}\left((-1)^{i-1}y^{i}-(-1)^{i-1}(-y)^{i}\right)=\sum _{i=1}^{\infty }{\frac {y^{i}}{i}}\left((-1)^{i-1}+1\right)\\&=y\sum _{i=1}^{\infty }{\frac {y^{i-1}}{i}}\left((-1)^{i-1}+1\right){\overset {i-1\to 2k}{=}}\;2y\sum _{k=0}^{\infty }{\frac {y^{2k}}{2k+1}},\end{aligned}}} we arrive at
ln ( x ) = 2 ( x − 1 ) x + 1 ∑