Natural logarithm

Definition and properties

The natural logarithm can be defined more generally as the inverse function of the exponential function ${\displaystyle e^x}$, so that ${\displaystyle e^{\ln (x)}=x}$ or ${\displaystyle \ln (e^x)=x}$. Because the exponential function ${\displaystyle e^x}$ is positive and invertible for any real input ${\displaystyle x}$, this definition of ${\displaystyle \ln (x)}$ is well defined for any positive ${\displaystyle x}$. That is, the domain of a natural logarithm is ${\displaystyle x\in (0,\mathrm{\infty })}$.

The natural logarithm can be defined as the area under the graph of a rectangular hyperbola with equation ${\displaystyle y=1/x}$ between ${\displaystyle x=1}$ and ${\displaystyle x=a}$. That is,$${\displaystyle \ln a={\int }_1^a\frac{1}{x}dx.}$$If a is in ${\displaystyle (0,1)}$, 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: $${\displaystyle \begin{array}{cc}\ln 1& =0,\\ \ln e& =1,\\ \ln (xy)& =\ln x+\ln y\text{for }x>0\text{and }y>0,\\ \ln (x/y)& =\ln x-\ln y\text{for }x>0\text{and }y>0,\\ \ln (x^y)& =y\ln x\text{for }x>0,\\ \ln (\sqrt[y]{x})& =(\ln x)/y\text{for }x>0\text{and }y\ne 0,\\ \ln x& <\ln y\text{for }0<x<y.\end{array}}$$

As ${\displaystyle x}$ approaches ${\displaystyle 0}$ from the right, the natural logarithm approaches negative infinity, ${\displaystyle -\mathrm{\infty }}$. As ${\displaystyle x}$ approaches ${\displaystyle \mathrm{\infty }}$, the natural logarithm approaches ${\displaystyle \mathrm{\infty }}$. By the concept of limit, these are: $${\displaystyle \underset{x\to 0^{+}}{lim}\ln x=-\mathrm{\infty },\underset{x\to \mathrm{\infty }}{lim}\ln x=\mathrm{\infty }}$$

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 log_e x, or, when the base e is implicitly known, as log x. The log_e x form is a specific instance of the general notation for the logarithm to base b of a number x, which is shown as log_b 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

• ${\displaystyle \underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1}$
• ${\displaystyle \underset{\alpha \to 0}{lim}\frac{x^{\alpha }-1}{\alpha }=\ln x\text{for }x>0}$
• ${\displaystyle \frac{x-1}{x}\le \ln x\le x-1\text{for}x>0}$
• ${\displaystyle \ln (1+x^{\alpha })\le \alpha x\text{for}x\ge 0\text{and }\alpha \ge 1}$

Series

Since the natural logarithm ${\displaystyle \ln (x)}$ 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 ${\displaystyle |x-1|\le 1}$ and ${\displaystyle x\ne 0,}$ then $${\displaystyle \begin{array}{cc}\ln x& ={\int }_1^x\frac{1}{t}dt={\int }_0^{x-1}\frac{1}{1+u}du\\ & ={\int }_0^{x-1}(1-u+u^2-u^3+\cdots )du\\ & =(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+\cdots \\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}(x-1{)}^k}{k}.\end{array}}$$

This is the Taylor series for ${\displaystyle \ln x}$ around 1. A change of variables yields the Mercator series: $${\displaystyle \ln (1+x)=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}}{k}{x}^k=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots ,}$$ valid for ${\displaystyle |x|\le 1}$ and ${\displaystyle x\ne -1.}$

Leonhard Euler, disregarding ${\displaystyle x\ne -1}$, nevertheless applied this series to ${\displaystyle x=-1}$ to show that the harmonic series equals the natural logarithm of $\frac{1}{1-1}$; 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 ${\displaystyle x=\frac{1}{n}}$, is: $${\displaystyle \ln \left(\frac{n+1}{n}\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}}{k{n}^k}=\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-\frac{1}{4n^4}+\cdots }$$

If ${\displaystyle \mathrm{Re}(x)\ge 1/2,}$ then $${\displaystyle \begin{array}{cc}\ln (x)& =-\ln \left(\frac{1}{x}\right)=-\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}(\frac{1}{x}-1{)}^k}{k}=\sum _{k=1}^{\mathrm{\infty }}\frac{(x-1{)}^k}{k{x}^k}\\ & =\frac{x-1}{x}+\frac{(x-1{)}^2}{2x^2}+\frac{(x-1{)}^3}{3x^3}+\frac{(x-1{)}^4}{4x^4}+\cdots \end{array}}$$

Now, taking ${\displaystyle x=\frac{n+1}{n}}$ for positive integers n, we get: $${\displaystyle \ln \left(\frac{n+1}{n}\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(n+1{)}^k}=\frac{1}{n+1}+\frac{1}{2(n+1{)}^2}+\frac{1}{3(n+1{)}^3}+\frac{1}{4(n+1{)}^4}+\cdots }$$

If ${\displaystyle \mathrm{Re}(x)\ge 0}$ and ${\displaystyle x\ne 0,}$ 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 $${\displaystyle \begin{array}{cc}\ln (1+y)-\ln (1-y)& =\sum _{i=1}^{\mathrm{\infty }}\frac{1}{i}\left((-1{)}^{i-1}{y}^i-(-1{)}^{i-1}(-y{)}^i\right)=\sum _{i=1}^{\mathrm{\infty }}\frac{y^i}{i}\left((-1{)}^{i-1}+1\right)\\ & =y\sum _{i=1}^{\mathrm{\infty }}\frac{y^{i-1}}{i}\left((-1{)}^{i-1}+1\right)\stackrel{i-1\to 2k}{=}2y\sum _{k=0}^{\mathrm{\infty }}\frac{y^{2k}}{2k+1},\end{array}}$$ we arrive at $${\displaystyle \begin{array}{cc}\ln (x)& =\frac{2(x-1)}{x+1}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2k+1}{\left(\frac{(x-1{)}^2}{(x+1{)}^2}\right)}^k\\ & =\frac{2(x-1)}{x+1}\left(\frac{1}{1}+\frac{1}{3}\frac{(x-1{)}^2}{(x+1{)}^2}+\frac{1}{5}{\left(\frac{(x-1{)}^2}{(x+1{)}^2}\right)}^2+\cdots \right).\end{array}}$$ Using the substitution ${\displaystyle x=\frac{n+1}{n}}$ again for positive integers n, we get: $${\displaystyle \begin{array}{cc}\ln \left(\frac{n+1}{n}\right)& =\frac{2}{2n+1}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1)((2n+1{)}^2{)}^k}\\ & =2\left(\frac{1}{2n+1}+\frac{1}{3(2n+1{)}^3}+\frac{1}{5(2n+1{)}^5}+\cdots \right).\end{array}}$$

This is by far the fastest-converging series described here.

The natural logarithm can also be expressed as an infinite product: $${\displaystyle \ln (x)=(x-1)\prod _{k=1}^{\mathrm{\infty }}\left(\frac{2}{1+\sqrt[2^k]{x}}\right)}$$

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

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

Integration

The natural logarithm allows simple integration of functions of the form ${\displaystyle g(x)=\frac{f^{\prime }(x)}{f(x)}}$: an antiderivative of g(x) is given by ${\displaystyle \ln (|f(x)|)}$. This is the case because of the chain rule and the following fact: $${\displaystyle \frac{d}{dx}\ln \left|x\right|=\frac{1}{x},x\ne 0}$$

In other words, when integrating over an interval of the real line that does not include ${\displaystyle x=0}$, then $${\displaystyle \int \frac{1}{x}dx=\ln |x|+C}$$ where C is an arbitrary constant of integration.

Likewise, when the integral is over an interval where ${\displaystyle f(x)\ne 0}$, $${\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C.}$$ For example, consider the integral of ${\displaystyle \tan (x)}$ over an interval that does not include points where ${\displaystyle \tan (x)}$ is infinite: $${\displaystyle \int \tan xdx=\int \frac{\sin x}{\cos x}dx=-\int \frac{\frac{d}{dx}\cos x}{\cos x}dx=-\ln \left|\cos x\right|+C=\ln \left|\sec x\right|+C.}$$

The natural logarithm can be integrated using integration by parts. Let: $${\displaystyle u=\ln x\Rightarrow du=\frac{dx}{x}}$$ $${\displaystyle dv=dx\Rightarrow v=x}$$ then: $${\displaystyle \begin{array}{cc}\int \ln xdx& =x\ln x-\int \frac{x}{x}dx\\ & =x\ln x-x+C\end{array}}$$

Efficient computation

For ${\displaystyle \ln (x)}$ 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.

Natural logarithm of 10

The natural logarithm of 10, a transcendental number approximately equal to 2.30258509, plays a role for example in the computation of natural logarithms of numbers represented in scientific notation, as a mantissa multiplied by a power of 10: $${\displaystyle \ln (a\cdot {10}^n)=\ln a+n\ln 10.}$$

This means that one can effectively calculate the logarithms of numbers with very large or very small magnitude using the logarithms of a relatively small set of decimals in the range [1, 10).

High precision

To compute the natural logarithm with many digits of precision, the Taylor series approach is not efficient since the convergence is slow. Especially if x is near 1, a good alternative is to use Halley's method or Newton's method to invert the exponential function, because the series of the exponential function converges more quickly. For finding the value of y to give ${\displaystyle \exp (y)-x=0}$ using Halley's method, or equivalently to give ${\displaystyle \exp (y/2)-x\exp (-y/2)=0}$ using Newton's method, the iteration simplifies to $${\displaystyle y_{n+1}=y_n+2\cdot \frac{x-\exp (y_n)}{x+\exp (y_n)}}$$ which has cubic convergence to ${\displaystyle \ln (x)}$.

Another alternative for extremely high precision calculation is the formula $${\displaystyle \ln x\approx \frac{\pi }{2M(1,4/s)}-m\ln 2,}$$ where M denotes the arithmetic-geometric mean of 1 and 4/s, and $${\displaystyle s=x{2}^m>2^{p/2},}$$ with m chosen so that p bits of precision is attained. (For most purposes, the value of 8 for m is sufficient.) In fact, if this method is used, Newton inversion of the natural logarithm may conversely be used to calculate the exponential function efficiently. (The constants ${\displaystyle \ln 2}$ and π can be pre-computed to the desired precision using any of several known quickly converging series.) Or, the following formula can be used: $${\displaystyle \ln x=\frac{\pi }{M\left({\theta }_2^2(1/x),{\theta }_3^2(1/x)\right)},x\in (1,\mathrm{\infty })}$$ where $${\displaystyle {\theta }_2(x)=\sum _{n\in \mathbb{Z}}{x}^{(n+1/2{)}^2},{\theta }_3(x)=\sum _{n\in \mathbb{Z}}{x}^{n^2}}$$ are the Jacobi theta functions.

Based on a proposal by William Kahan and first implemented in the Hewlett-Packard HP-41C calculator in 1979 (referred to under "LN1" in the display, only), some calculators, operating systems (for example Berkeley UNIX 4.3BSD), computer algebra systems and programming languages (for example C99) provide a special natural logarithm plus 1 function, alternatively named LNP1, or log1p to give more accurate results for logarithms close to zero by passing arguments x, also close to zero, to a function log1p(x), which returns the value ln(1+x), instead of passing a value y close to 1 to a function returning ln(y). The function log1p avoids in the floating point arithmetic a near cancelling of the absolute term 1 with the second term from the Taylor expansion of the natural logarithm. This keeps the argument, the result, and intermediate steps all close to zero where they can be most accurately represented as floating-point numbers.

In addition to base e, the IEEE 754-2008 standard defines similar logarithmic functions near 1 for binary and decimal logarithms: log₂(1 + x) and log₁₀(1 + x).

Similar inverse functions named "expm1", "expm" or "exp1m" exist as well, all with the meaning of expm1(x) = exp(x) − 1.

An identity in terms of the inverse hyperbolic tangent, $${\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}1\mathrm{p}(x)=\log (1+x)=2\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{x}{2+x}\right),}$$ gives a high precision value for small values of x on systems that do not implement log1p(x).

Computational complexity

The computational complexity of computing the natural logarithm using the arithmetic-geometric mean (for both of the above methods) is ${\displaystyle \text{O}(M(n)\ln n)}$. Here, n is the number of digits of precision at which the natural logarithm is to be evaluated, and M(n) is the computational complexity of multiplying two n-digit numbers.

Continued fractions

While no simple continued fractions are available, several generalized continued fractions exist, including: $${\displaystyle \begin{array}{cc}\ln (1+x)& =\frac{x^1}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \\ & =\frac{x}{1-0x+\frac{1^{2}x}{2-1x+\frac{2^{2}x}{3-2x+\frac{3^{2}x}{4-3x+\frac{4^{2}x}{5-4x+\ddots }}}}}\end{array}}$$ $${\displaystyle \begin{array}{cc}\ln \left(1+\frac{x}{y}\right)& =\frac{x}{y+\frac{1x}{2+\frac{1x}{3y+\frac{2x}{2+\frac{2x}{5y+\frac{3x}{2+\ddots }}}}}}\\ & =\frac{2x}{2y+x-\frac{(1x{)}^2}{3(2y+x)-\frac{(2x{)}^2}{5(2y+x)-\frac{(3x{)}^2}{7(2y+x)-\ddots }}}}\end{array}}$$

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: $${\displaystyle \begin{array}{cc}\ln 2& =3\ln \left(1+\frac{1}{4}\right)+\ln \left(1+\frac{3}{125}\right)\\ & =\frac{6}{9-\frac{1^2}{27-\frac{2^2}{45-\frac{3^2}{63-\ddots }}}}+\frac{6}{253-\frac{3^2}{759-\frac{6^2}{1265-\frac{9^2}{1771-\ddots }}}}.\end{array}}$$

Furthermore, since 10 = 1.25¹⁰ × 1.024³, even the natural logarithm of 10 can be computed similarly as: $${\displaystyle \begin{array}{cc}\ln 10& =10\ln \left(1+\frac{1}{4}\right)+3\ln \left(1+\frac{3}{125}\right)\\ & =\frac{20}{9-\frac{1^2}{27-\frac{2^2}{45-\frac{3^2}{63-\ddots }}}}+\frac{18}{253-\frac{3^2}{759-\frac{6^2}{1265-\frac{9^2}{1771-\ddots }}}}.\end{array}}$$ The reciprocal of the natural logarithm can be also written in this way: $${\displaystyle \frac{1}{\ln (x)}=\frac{2x}{x^2-1}\sqrt{\frac{1}{2}+\frac{x^2+1}{4x}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{x^2+1}{4x}}}\dots }$$

For example: $${\displaystyle \frac{1}{\ln (2)}=\frac{4}{3}\sqrt{\frac{1}{2}+\frac{5}{8}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{5}{8}}}\dots }$$

Complex logarithms

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.