Natural logarithm

The concept of the natural logarithm was worked out by Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa before 1649.[5] 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,[6] although the mathematics teacher John Speidell had already compiled a table of what in fact were effectively natural logarithms in 1619.[7] 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.[7]^: 152

The notations ln x and log_e x both refer unambiguously to the natural logarithm of x, and log x without an explicit base may also refer to the natural logarithm. This usage is common in mathematics, along with some scientific contexts as well as in many programming languages.[nb 1] In some other contexts such as chemistry, however, 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.

The natural logarithm can be defined in several equivalent ways.

Integral definition

ln a as the area of the shaded region under the curve f(x) = 1/x from 1 to a. If a is less than 1, the area taken to be negative.

The area under the hyperbola satisfies the logarithm rule. Here A(s,t) denotes the area under the hyperbola between s and t.

The natural logarithm of a positive, real number a may be defined as the area under the graph of the hyperbola with equation y = 1/x between x = 1 and x = a. This is the integral[3]

${\displaystyle \ln a=\int _{1}^{a}{\frac {1}{x}}\,dx.}$

If a is less than 1, then this area is considered to be negative.

This function is a logarithm because it satisfies the fundamental multiplicative property of a logarithm:[4]

${\displaystyle \ln(ab)=\ln a+\ln b.}$

This can be demonstrated by splitting the integral that defines ln ab into two parts, and then making the variable substitution x = at (so dx = a dt) in the second part, as follows:

${\displaystyle {\begin{aligned}\ln ab=\int _{1}^{ab}{\frac {1}{x}}\,dx&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{a}^{ab}{\frac {1}{x}}\,dx\\[5pt]&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{at}}a\,dt\\[5pt]&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{t}}\,dt\\[5pt]&=\ln a+\ln b.\end{aligned}}}$

In elementary terms, this is simply scaling by 1/a in the horizontal direction and by a in the vertical direction. Area does not change under this transformation, but the region between a and ab is reconfigured. Because the function a/(ax) is equal to the function 1/x, the resulting area is precisely ln b.

The number e can then be defined to be the unique real number a such that ln a = 1.

The natural logarithm also has an improper integral representation [8]

${\displaystyle \ln \left(x\right)=\int _{0}^{\infty }{\frac {e^{-t}-e^{-tx}}{t}}dt}$

• ${\displaystyle \ln 1=0}$
• ${\displaystyle \ln e=1}$
• ${\displaystyle \ln(xy)=\ln x+\ln y\quad {\text{for }}\;x>0\;{\text{and }}\;y>0}$
• ${\displaystyle \ln(x/y)=\ln x-\ln y}$
• ${\displaystyle \ln(x^{y})=y\ln x\quad {\text{for }}\;x>0}$
• ${\displaystyle \ln x<\ln y\quad {\text{for }}\;0<x<y}$
• ${\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=1}$
• ${\displaystyle \lim _{\alpha \to 0}{\frac {x^{\alpha }-1}{\alpha }}=\ln x\quad {\text{for }}\;x>0}$
• ${\displaystyle {\frac {x-1}{x}}\leq \ln x\leq x-1\quad {\text{for}}\quad x>0}$
• ${\displaystyle \ln {(1+x^{\alpha })}\leq \alpha x\quad {\text{for}}\quad x\geq 0\;{\text{and }}\;\alpha \geq 1}$

The derivative of the natural logarithm as a real-valued function on the positive reals is given by[3]

${\displaystyle {\frac {d}{dx}}\ln x={\frac {1}{x}}.}$

How to establish this derivative of the natural logarithm depends on how it is defined firsthand. If the natural logarithm is defined as the integral

${\displaystyle \ln x=\int _{1}^{x}{\frac {1}{t}}\,dt,}$

then the derivative immediately follows from the first part of the fundamental theorem of calculus.

On the other hand, if the natural logarithm is defined as the inverse of the (natural) exponential function, then the derivative (for x > 0) can be found by using the properties of the logarithm and a definition of the exponential function. From the definition of the number ${\displaystyle e=\lim _{u\to 0}(1+u)^{1/u},}$ the exponential function can be defined as ${\displaystyle e^{x}=\lim _{u\to 0}(1+u)^{x/u}=\lim _{h\to 0}(1+hx)^{1/h}}$, where ${\displaystyle u=hx,h=u/x.}$ The derivative can then be found from first principles.

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\ln x&=\lim _{h\to 0}{\frac {\ln(x+h)-\ln x}{h}}\\&=\lim _{h\to 0}\left[{\frac {1}{h}}\ln \left({\frac {x+h}{x}}\right)\right]\\&=\lim _{h\to 0}\left[\ln \left(\left(1+{\frac {h}{x}}\right)^{\frac {1}{h}}\right)\right]\quad &&{\text{all above for logarithmic properties}}\\&=\ln \left[\lim _{h\to 0}\left(1+{\frac {h}{x}}\right)^{\frac {1}{h}}\right]\quad &&{\text{for continuity of the logarithm}}\\&=\ln e^{1/x}\quad &&{\text{for the definition of }}e^{x}=\lim _{h\to 0}(1+hx)^{1/h}\\&={\frac {1}{x}}\quad &&{\text{for the definition of the ln as inverse function.}}\end{aligned}}}$

Also, we have:

${\displaystyle {\frac {d}{dx}}\ln ax={\frac {d}{dx}}(\ln a+\ln x)={\frac {d}{dx}}\ln a+{\frac {d}{dx}}\ln x={\frac {1}{x}}.}$

so, unlike its inverse function ${\displaystyle e^{ax}}$, a constant in the function doesn't alter the differential.

The Taylor polynomials for ln(1 + x) only provide accurate approximations in the range −1 < x ≤ 1. Beyond some x > 1, the Taylor polynomials of higher degree are increasingly worse approximations.

Since the natural logarithm is undefined at 0, ${\displaystyle \ln(x)}$ itself does not have a Maclaurin series, unlike many other elementary functions. Instead, one looks for Taylor expansions around other points. For example, if ${\displaystyle \vert x-1\vert \leq 1{\text{ and }}x\neq 0,}$ then[9]

${\displaystyle {\begin{aligned}\ln x&=\int _{1}^{x}{\frac {1}{t}}\,dt=\int _{0}^{x-1}{\frac {1}{1+u}}\,du\\&=\int _{0}^{x-1}(1-u+u^{2}-u^{3}+\cdots )\,du\\&=(x-1)-{\frac {(x-1)^{2}}{2}}+{\frac {(x-1)^{3}}{3}}-{\frac {(x-1)^{4}}{4}}+\cdots \\&=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(x-1)^{k}}{k}}.\end{aligned}}}$

This is the Taylor series for ln x around 1. A change of variables yields the Mercator series:

${\displaystyle \ln(1+x)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}x^{k}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}$

valid for |x| ≤ 1 and x ≠ −1.

Leonhard Euler,[10] disregarding ${\displaystyle x\neq -1}$, nevertheless applied this series to x = −1, in order to show that the harmonic series equals the (natural) logarithm of 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.

At right is a picture 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 of this region the higher-degree Taylor polynomials evolve to worse approximations for the function.

A useful special case for positive integers n, taking ${\displaystyle x={\tfrac {1}{n}}}$, is:

${\displaystyle \ln \left({\frac {n+1}{n}}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{kn^{k}}}={\frac {1}{n}}-{\frac {1}{2n^{2}}}+{\frac {1}{3n^{3}}}-{\frac {1}{4n^{4}}}+\cdots }$

If ${\displaystyle \operatorname {Re} (x)\geq 1/2,}$ then

${\displaystyle {\begin{aligned}\ln(x)&=-\ln \left({\frac {1}{x}}\right)=-\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}({\frac {1}{x}}-1)^{k}}{k}}=\sum _{k=1}^{\infty }{\frac {(x-1)^{k}}{kx^{k}}}\\&={\frac {x-1}{x}}+{\frac {(x-1)^{2}}{2x^{2}}}+{\frac {(x-1)^{3}}{3x^{3}}}+{\frac {(x-1)^{4}}{4x^{4}}}+\cdots \end{aligned}}}$

Now, taking ${\displaystyle x={\tfrac {n+1}{n}}}$ for positive integers n, we get:

${\displaystyle \ln \left({\frac {n+1}{n}}\right)=\sum _{k=1}^{\infty }{\frac {1}{k(n+1)^{k}}}={\frac {1}{n+1}}+{\frac {1}{2(n+1)^{2}}}+{\frac {1}{3(n+1)^{3}}}+{\frac {1}{4(n+1)^{4}}}+\cdots }$

If ${\displaystyle \operatorname {Re} (x)\geq 0{\text{ and }}x\neq 0,}$ then

${\displaystyle \ln(x)=\ln \left({\frac {2x}{2}}\right)=\ln \left({\frac {1+{\frac {x-1}{x+1}}}{1-{\frac {x-1}{x+1}}}}\right)=\ln \left(1+{\frac {x-1}{x+1}}\right)-\ln \left(1-{\frac {x-1}{x+1}}\right).}$

Since

${\displaystyle {\begin{aligned}\ln(1+y)-\ln(1-y)&=\sum _{i=1}^{\infty }{\frac {1}{i}}\left((-1)^{i-1}y^{i}-(-1)^{i-1}(-y)^{i}\right)=\sum _{i=1}^{\infty }{\frac {y^{i}}{i}}\left((-1)^{i-1}+1\right)\\&=y\sum _{i=1}^{\infty }{\frac {y^{i-1}}{i}}\left((-1)^{i-1}+1\right){\overset {i-1\to 2k}{=}}\;2y\sum _{k=0}^{\infty }{\frac {y^{2k}}{2k+1}},\end{aligned}}}$

we arrive at

${\displaystyle {\begin{aligned}\ln(x)&={\frac {2(x-1)}{x+1}}\sum _{k=0}^{\infty }{\frac {1}{2k+1}}{\left({\frac {(x-1)^{2}}{(x+1)^{2}}}\right)}^{k}\\&={\frac {2(x-1)}{x+1}}\left({\frac {1}{1}}+{\frac {1}{3}}{\frac {(x-1)^{2}}{(x+1)^{2}}}+{\frac {1}{5}}{\left({\frac {(x-1)^{2}}{(x+1)^{2}}}\right)}^{2}+\cdots \right).\end{aligned}}}$

Using the substitution ${\displaystyle x={\tfrac {n+1}{n}}}$ again for positive integers n, we get:

${\displaystyle {\begin{aligned}\ln \left({\frac {n+1}{n}}\right)&={\frac {2}{2n+1}}\sum _{k=0}^{\infty }{\frac {1}{(2k+1)((2n+1)^{2})^{k}}}\\&=2\left({\frac {1}{2n+1}}+{\frac {1}{3(2n+1)^{3}}}+{\frac {1}{5(2n+1)^{5}}}+\cdots \right).\end{aligned}}}$

This is, by far, the fastest converging of the series described here.

The natural logarithm can also be expressed as an infinite product:[11]

${\displaystyle \ln(x)=(x-1)\prod _{k=1}^{\infty }\left({\frac {2}{1+{\sqrt[{2^{k}}]{x}}}}\right)}$

Two examples might be:

${\displaystyle \ln(2)=\left({\frac {2}{1+{\sqrt {2}}}}\right)\left({\frac {2}{1+{\sqrt[{4}]{2}}}}\right)\left({\frac {2}{1+{\sqrt[{8}]{2}}}}\right)\left({\frac {2}{1+{\sqrt[{16}]{2}}}}\right)...}$

${\displaystyle \pi =(2i+2)\left({\frac {2}{1+{\sqrt {i}}}}\right)\left({\frac {2}{1+{\sqrt[{4}]{i}}}}\right)\left({\frac {2}{1+{\sqrt[{8}]{i}}}}\right)\left({\frac {2}{1+{\sqrt[{16}]{i}}}}\right)...}$

From this identity, we can easily get that:

${\displaystyle {\frac {1}{\ln(x)}}={\frac {x}{x-1}}-\sum _{k=1}^{\infty }{\frac {2^{-k}x^{2^{-k}}}{1+x^{2^{-k}}}}}$

For example:

${\displaystyle {\frac {1}{\ln(2)}}=2-{\frac {\sqrt {2}}{2+2{\sqrt {2}}}}-{\frac {\sqrt[{4}]{2}}{4+4{\sqrt[{4}]{2}}}}-{\frac {\sqrt[{8}]{2}}{8+8{\sqrt[{8}]{2}}}}\cdots }$

The natural logarithm allows simple integration of functions of the form g(x) = f '(x)/f(x): an antiderivative of g(x) is given by ln(|f(x)|). This is the case because of the chain rule and the following fact:

${\displaystyle {\frac {d}{dx}}\ln \left|x\right|={\frac {1}{x}}.}$

In other words, if ${\displaystyle x}$ is a real number with ${\displaystyle x\not =0}$, then

${\displaystyle \int {\frac {1}{x}}\,dx=\ln |x|+C}$[12]

and

${\displaystyle \int {{\frac {f'(x)}{f(x)}}\,dx}=\ln |f(x)|+C.}$

Here is an example in the case of g(x) = tan(x):

${\displaystyle {\begin{aligned}&\int \tan x\,dx=\int {\frac {\sin x}{\cos x}}\,dx\\[6pt]&\int \tan x\,dx=\int {\frac {-{\frac {d}{dx}}\cos x}{\cos x}}\,dx.\end{aligned}}}$

Letting f(x) = cos(x):

${\displaystyle \int \tan x\,dx=-\ln \left|\cos x\right|+C}$

${\displaystyle \int \tan x\,dx=\ln \left|\sec x\right|+C}$

where C is an arbitrary constant of integration.

The natural logarithm can be integrated using integration by parts:

${\displaystyle \int \ln x\,dx=x\ln x-x+C.}$

Let:

${\displaystyle u=\ln x\Rightarrow du={\frac {dx}{x}}}$

${\displaystyle dv=dx\Rightarrow v=x}$

then:

${\displaystyle {\begin{aligned}\int \ln x\,dx&=x\ln x-\int {\frac {x}{x}}\,dx\\&=x\ln x-\int 1\,dx\\&=x\ln x-x+C\end{aligned}}}$

For ln(x) where x > 1, the closer the value of x is to 1, the faster the rate of convergence of its Taylor series centered at 1. The identities associated with the logarithm can be leveraged to exploit this:

${\displaystyle {\begin{aligned}\ln 123.456&=\ln(1.23456\cdot 10^{2})\\&=\ln 1.23456+\ln(10^{2})\\&=\ln 1.23456+2\ln 10\\&\approx \ln 1.23456+2\cdot 2.3025851.\end{aligned}}}$

Such techniques were used before calculators, by referring to numerical tables and performing manipulations such as those above.

While no simple continued fractions are available, several generalized continued fractions are, including:

${\displaystyle {\begin{aligned}\ln(1+x)&={\frac {x^{1}}{1}}-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-\cdots \\[5pt]&={\cfrac {x}{1-0x+{\cfrac {1^{2}x}{2-1x+{\cfrac {2^{2}x}{3-2x+{\cfrac {3^{2}x}{4-3x+{\cfrac {4^{2}x}{5-4x+\ddots }}}}}}}}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}\ln \left(1+{\frac {x}{y}}\right)&={\cfrac {x}{y+{\cfrac {1x}{2+{\cfrac {1x}{3y+{\cfrac {2x}{2+{\cfrac {2x}{5y+{\cfrac {3x}{2+\ddots }}}}}}}}}}}}\\[5pt]&={\cfrac {2x}{2y+x-{\cfrac {(1x)^{2}}{3(2y+x)-{\cfrac {(2x)^{2}}{5(2y+x)-{\cfrac {(3x)^{2}}{7(2y+x)-\ddots }}}}}}}}\end{aligned}}}$

These continued fractions—particularly the last—converge rapidly for values close to 1. However, the natural logarithms of much larger numbers can easily be computed, by repeatedly adding those of smaller numbers, with similarly rapid convergence.

For example, since 2 = 1.25³ × 1.024, the natural logarithm of 2 can be computed as:

${\displaystyle {\begin{aligned}\ln 2&=3\ln \left(1+{\frac {1}{4}}\right)+\ln \left(1+{\frac {3}{125}}\right)\\[8pt]&={\cfrac {6}{9-{\cfrac {1^{2}}{27-{\cfrac {2^{2}}{45-{\cfrac {3^{2}}{63-\ddots }}}}}}}}+{\cfrac {6}{253-{\cfrac {3^{2}}{759-{\cfrac {6^{2}}{1265-{\cfrac {9^{2}}{1771-\ddots }}}}}}}}.\end{aligned}}}$

Furthermore, since 10 = 1.25¹⁰ × 1.024³, even the natural logarithm of 10 can be computed similarly as:

${\displaystyle {\begin{aligned}\ln 10&=10\ln \left(1+{\frac {1}{4}}\right)+3\ln \left(1+{\frac {3}{125}}\right)\\[10pt]&={\cfrac {20}{9-{\cfrac {1^{2}}{27-{\cfrac {2^{2}}{45-{\cfrac {3^{2}}{63-\ddots }}}}}}}}+{\cfrac {18}{253-{\cfrac {3^{2}}{759-{\cfrac {6^{2}}{1265-{\cfrac {9^{2}}{1771-\ddots }}}}}}}}.\end{aligned}}}$

The reciprocal of the natural logarithm can be also written in this way:

${\displaystyle {\frac {1}{\ln(x)}}={\frac {2x}{x^{2}-1}}{\sqrt {{\frac {1}{2}}+{\frac {x^{2}+1}{4x}}}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {x^{2}+1}{4x}}}}}}\ldots }$

For example:

${\displaystyle {\frac {1}{\ln(2)}}={\frac {4}{3}}{\sqrt {{\frac {1}{2}}+{\frac {5}{8}}}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {5}{8}}}}}}\ldots }$

The exponential function can be extended to a function which gives a complex number as e^z for any arbitrary complex number z; simply use the infinite series with x=z complex. This exponential function can be inverted to form a complex logarithm that exhibits most of the properties of the ordinary logarithm. There are two difficulties involved: no x has e^x = 0; and it turns out that e^2iπ = 1 = e⁰. Since the multiplicative property still works for the complex exponential function, e^z = e^z+2kiπ, for all complex z and integers k.

So the logarithm cannot be defined for the whole complex plane, and even then it is multi-valued—any complex logarithm can be changed into an "equivalent" logarithm by adding any integer multiple of 2iπ at will. The complex logarithm can only be single-valued on the cut plane. For example, ln i = iπ/2 or 5iπ/2 or -3iπ/2, etc.; and although i⁴ = 1, 4 ln i can be defined as 2iπ, or 10iπ or −6iπ, and so on.

• Plots of the natural logarithm function on the complex plane (principal branch)
• 
  z = Re(ln(x + yi))
• 
  z = |(Im(ln(x + yi)))|
• 
  z = |(ln(x + yi))|
• 
  Superposition of the previous three graphs

• Approximating natural exponents (log base e)
• Iterated logarithm
• Napierian logarithm
• List of logarithmic identities
• Logarithm of a matrix
• Logarithmic differentiation
• Logarithmic integral function
• Nicholas Mercator – first to use the term natural logarithm
• Polylogarithm
• Von Mangoldt function