| Rationality | Transcendental |
|---|---|
| Decimal | 0.69314 71805 59945 3094... |
In mathematics, the natural logarithm of 2 is the unique real number argument such that the exponential function equals two. It appears frequently in various formulas and is also given by the alternating harmonic series. The decimal value of the natural logarithm of 2 truncated at 30 decimal places is given by:
\ln 2 \approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458.
The logarithm of 2 in other bases is obtained with the formula
\log_b 2 = \frac{\ln 2}{\ln b}.
The common logarithm in particular is ()
\log_{10} 2 \approx 0.301\,029\,995\,663\,981\,195.
The inverse of this number is the binary logarithm of 10:
\log_2 10 =\frac{1}{\log_{10} 2} \approx 3.321\,928\,095().
By the Lindemann–Weierstrass theorem, the natural logarithm of any natural number other than 0 and 1 (more generally, of any positive algebraic number other than 1) is a transcendental number. It is also contained in the ring of algebraic periods.
Series representations
Rising alternate factorial
\ln 2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots.This is the well-known "alternating harmonic series".\ln 2 = \frac{1}{2} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}.\ln 2 = \frac{5}{8} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)}.\ln 2 = \frac{2}{3} +\frac{3}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)}.\ln 2 = \frac{131}{192} +\frac{3}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}.\ln 2 = \frac{661}{960} +\frac{15}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}.\ln 2 = \frac{2}{3}\left(1+\frac{2}{4^3-4}+\frac{2}{8^3-8}+\frac{2}{12^3-12}+\dots\right) .
Binary rising constant factorial
\ln 2 = \sum_{n=1}^\infty \frac{1}{2^{n}n}.\ln 2 = 1 -\sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)}.\ln 2 = \frac{1}{2} + 2 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)} .\ln 2 = \frac{5}{6} - 6 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)} .\ln 2 = \frac{7}{12} + 24 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)} .\ln 2 = \frac{47}{60} - 120 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)} .
Other series representations
\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \ln 2.\sum_{n=1}^\infty \frac{1}{n(4n^2-1)} = 2\ln 2 -1.\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} = \ln 2 -1.\sum_{n=1}^\infty \frac{(-1)^n}{n(9n^2-1)} = 2\ln 2 -\frac{3}{2}.\sum_{n=1}^\infty \frac{1}{4n^2-2n} = \ln 2.\sum_{n=1}^\infty \frac{2(-1)^{n+1}(2n-1)+1}{8n^2-4n} = \ln 2.\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+1} = \frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}.\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2} = -\frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}.\sum_{n=0}^\infty \frac{(-1)^{n}}{(3n+1)(3n+2)} = \frac{2\ln 2}{3}.\sum_{n=1}^\infty \frac{1}{\sum_{k=1}^n k^2} = 18 - 24 \ln 2using\lim_{N\rightarrow \infty} \sum_{n=N}^{2N} \frac{1}{n} = \ln 2\sum_{n=1}^\infty \frac{1}{4n^2-3n} = \ln 2 + \frac{\pi}{6}(sums of the reciprocals of decagonal numbers)
Involving the Riemann Zeta function
\sum_{n=1}^\infty \frac{1}{n}[\zeta(2n)-1] = \ln 2.\sum_{n=2}^\infty \frac{1}{2^n}[\zeta(n)-1] = \ln 2 -\frac{1}{2}.\sum_{n=1}^\infty \frac{1}{2n+1}[\zeta(2n+1)-1] = 1-\gamma-\frac{\ln 2}{2}.\sum_{n=1}^\infty \frac{1}{2^{2n-1}(2n+1)}\zeta(2n) = 1-\ln 2.
(γ is the Euler–Mascheroni constant and ζ Riemann's zeta function.)
BBP-type representations
\ln 2 = \frac{2}{3} + \frac{1}{2} \sum_{k = 1}^\infty \left(\frac{1}{2k}+\frac{1}{4k+1}+\frac{1}{8k+4}+\frac{1}{16k+12}\right) \frac{1}{16^k} .
(See more about Bailey–Borwein–Plouffe (BBP)-type representations.)
Applying the three general series for natural logarithm to 2 directly gives:
\ln 2 = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}.\ln 2 = \sum_{n = 1}^\infty \frac{1}{2^{n}n}.\ln 2 = \frac{2}{3} \sum_{k = 0}^\infty \frac{1}{9^{k}(2k+1)}.
Applying them to \textstyle 2 = \frac{3}{2} \cdot \frac{4}{3} gives:
\ln 2 = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{2^n n} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{3^n n} .\ln 2 = \sum_{n = 1}^\infty \frac{1}{3^n n} + \sum_{n = 1}^\infty \frac{1}{4^n n} .\ln 2 = \frac{2}{5} \sum_{k = 0}^\infty \frac{1}{25^{k}(2k+1)} + \frac{2}{7} \sum_{k = 0}^\infty \frac{1}{49^{k}(2k+1)} .
Applying them to \textstyle 2 = (\sqrt{2})^2 gives:
\ln 2 = 2 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(\sqrt{2} + 1)^n n} .\ln 2 = 2 \sum_{n = 1}^\infty \frac{1}{(2 + \sqrt{2})^n n} .\ln 2 = \frac{4}{3 + 2 \sqrt{2}} \sum_{k = 0}^\infty \frac{1}{(17 + 12 \sqrt{2})^{k}(2k+1)} .
Applying them to \textstyle 2 = { \left( \frac{16}{15} \right) }^{7} \cdot { \left( \frac{81}{80} \right) }^{3} \cdot { \left( \frac{25}{24} \right) }^{5} gives:
\ln 2 = 7 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{15^n n} + 3 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{80^n n} + 5 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{24^n n} .\ln 2 = 7 \sum_{n = 1}^\infty \frac{1}{16^n n} + 3 \sum_{n = 1}^\infty \frac{1}{81^n n} + 5 \sum_{n = 1}^\infty \frac{1}{25^n n} .\ln 2 = \frac{14}{31} \sum_{k = 0}^\infty \frac{1}{961^{k}(2k+1)} + \frac{6}{161} \sum_{k = 0}^\infty \frac{1}{25921^{k}(2k+1)} + \frac{10}{49} \sum_{k = 0}^\infty \frac{1}{2401^{k}(2k+1)} .
Representation as integrals
The natural logarithm of 2 occurs frequently as the result of integration. Some explicit formulas for it include:
\int_0^1 \frac{dx}{1+x} = \int_1^2 \frac{dx}{x} = \ln 2
\int_0^\infty e^{-x}\frac{1-e^{-x}}{x} \, dx= \ln 2
\int_0^\infty 2^{-x} dx= \frac{1}{\ln 2}
\int_0^\frac{\pi}{3} \tan x \, dx=2\int_0^\frac{\pi}{4} \tan x \, dx = \ln 2
-\frac{1}{\pi i}\int_{0}^{\infty} \frac{\ln x \ln\ln x}{(x+1)^2} \, dx= \ln 2
Other representations
The Pierce expansion is
\ln 2 = 1 -\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12} -\cdots.
The Engel expansion is
\ln 2 = \frac{1}{2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3\cdot 7} + \frac{1}{2\cdot 3\cdot 7\cdot 9}+\cdots.
The cotangent expansion is
\ln 2 = \cot({\arccot(0) -\arccot(1) + \arccot(5) - \arccot(55) + \arccot(14187) -\cdots}).
The simple continued fraction expansion is
\ln 2 = \left[ 0; 1, 2, 3, 1, 6, 3, 1, 1, 2, 1, 1, 1, 1, 3, 10, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 3, 1,...\right],
which yields rational approximations, the first few of which are 0, 1, 2/3, 7/10, 9/13 and 61/88.
This generalized continued fraction:
\ln 2 = \left[ 0;1,2,3,1,5,\tfrac{2}{3},7,\tfrac{1}{2},9,\tfrac{2}{5},...,2k-1,\frac{2}{k},...\right],[1]- also expressible as
\ln 2 = \cfrac{1} {1+\cfrac{1} {2+\cfrac{1} {3+\cfrac{2} {2+\cfrac{2} {5+\cfrac{3} {2+\cfrac{3} {7+\cfrac{4} {2+\ddots}}}}}}}} = \cfrac{2} {3-\cfrac{1^2} {9-\cfrac{2^2} {15-\cfrac{3^2} {21-\ddots}}}}
The following continued fraction representation (J.L.Lagrange) gives (asymptotically) 1.53 new correct decimal places per cycle: [2][3]
\ln 2=\frac{\frac{1}{2}}{L_{0}}, L_{k}=2k+1+\frac{k+1}{2+\frac{k+1}{L_{k+1}}}
or
\ln 2=\frac{2}{G_{0}}, G_{k}=6k+3-\frac{(k+1)^2}{G_{k+1}}
Bootstrapping other logarithms
Given a value of ln 2, a scheme of computing the logarithms of other integers is to tabulate the logarithms of the prime numbers and in the next layer the logarithms of the composite numbers c based on their factorizations
c=2^i3^j5^k7^l\cdots\rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots
This employs
| Prime | Approximate natural logarithm | OEIS |
|---|---|---|
| 2 | 0.693147180559945309417232121458 | |
| 3 | 1.09861228866810969139524523692 | |
| 5 | 1.60943791243410037460075933323 | |
| 7 | 1.94591014905531330510535274344 | |
| 11 | 2.39789527279837054406194357797 | |
| 13 | 2.56494935746153673605348744157 | |
| 17 | 2.83321334405621608024953461787 | |
| 19 | 2.94443897916644046000902743189 | |
| 23 | 3.13549421592914969080675283181 | |
| 29 | 3.36729582998647402718327203236 | |
| 31 | 3.43398720448514624592916432454 | |
| 37 | 3.61091791264422444436809567103 | |
| 41 | 3.71357206670430780386676337304 | |
| 43 | 3.76120011569356242347284251335 | |
| 47 | 3.85014760171005858682095066977 | |
| 53 | 3.97029191355212183414446913903 | |
| 59 | 4.07753744390571945061605037372 | |
| 61 | 4.11087386417331124875138910343 | |
| 67 | 4.20469261939096605967007199636 | |
| 71 | 4.26267987704131542132945453251 | |
| 73 | 4.29045944114839112909210885744 | |
| 79 | 4.36944785246702149417294554148 | |
| 83 | 4.41884060779659792347547222329 | |
| 89 | 4.48863636973213983831781554067 | |
| 97 | 4.57471097850338282211672162170 |
In a third layer, the logarithms of rational numbers r = a⁄b are computed with ln(r) = ln(a) − ln(b), and logarithms of roots via ln √c = 1⁄n ln(c).
The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers 2i close to powers bj of other numbers b is comparatively easy, and series representations of ln(b) are found by coupling 2 to b with logarithmic conversions.
Example
If ps = qt + d with some small d, then ps⁄qt = 1 + d⁄qt and therefore
s\ln p -t\ln q = \ln\left(1+\frac{d}{q^t}\right) = \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m}\left(\frac{d}{q^t}\right)^m = \sum_{n=0}^\infty \frac{2}{2n+1} {\left(\frac{d}{2 q^t + d}\right)}^{2n+1} .
Selecting q = 2 represents ln p by ln 2 and a series of a parameter d⁄qt that one wishes to keep small for quick convergence. Taking 32 = 23 + 1, for example, generates
2\ln 3 = 3\ln 2 -\sum_{k\ge 1}\frac{(-1)^k}{8^{k}k} = 3\ln 2 + \sum_{n=0}^\infty \frac{2}{2n+1} {\left(\frac{1}{2 \cdot 8 + 1}\right)}^{2n+1} .
This is actually the third line in the following table of expansions of this type:
| s | p | t | q | d⁄qt |
|---|---|---|---|---|
| 1 | 3 | 1 | 2 | 1⁄2 = 0.50000000… |
| 1 | 3 | 2 | 2 | −1⁄4 = −0.25000000… |
| 2 | 3 | 3 | 2 | 1⁄8 = 0.12500000… |
| 5 | 3 | 8 | 2 | −13⁄256 = −0.05078125… |
| 12 | 3 | 19 | 2 | 7153⁄524,288 = 0.01364326… |
| 1 | 5 | 2 | 2 | 1⁄4 = 0.25000000… |
| 3 | 5 | 7 | 2 | −3⁄128 = −0.02343750… |
| 1 | 7 | 2 | 2 | 3⁄4 = 0.75000000… |
| 1 | 7 | 3 | 2 | −1⁄8 = −0.12500000… |
| 5 | 7 | 14 | 2 | 423⁄16,384 = 0.02581787… |
| 1 | 11 | 3 | 2 | 3⁄8 = 0.37500000… |
| 2 | 11 | 7 | 2 | −7⁄128 = −0.05468750… |
| 11 | 11 | 38 | 2 | 10,433,763,667⁄274,877,906,944 = 0.03795781… |
| 1 | 13 | 3 | 2 | 5⁄8 = 0.62500000… |
| 1 | 13 | 4 | 2 | −3⁄16 = −0.18750000… |
| 3 | 13 | 11 | 2 | 149⁄2048 = 0.07275391… |
| 7 | 13 | 26 | 2 | −4,360,347⁄67,108,864 = −0.06497423… |
| 10 | 13 | 37 | 2 | 419,538,377⁄137,438,953,472 = 0.00305254… |
| 1 | 17 | 4 | 2 | 1⁄16 = 0.06250000… |
| 1 | 19 | 4 | 2 | 3⁄16 = 0.18750000… |
| 4 | 19 | 17 | 2 | −751⁄131,072 = −0.00572968… |
| 1 | 23 | 4 | 2 | 7⁄16 = 0.43750000… |
| 1 | 23 | 5 | 2 | −9⁄32 = −0.28125000… |
| 2 | 23 | 9 | 2 | 17⁄512 = 0.03320312… |
| 1 | 29 | 4 | 2 | 13⁄16 = 0.81250000… |
| 1 | 29 | 5 | 2 | −3⁄32 = −0.09375000… |
| 7 | 29 | 34 | 2 | 70,007,125⁄17,179,869,184 = 0.00407495… |
| 1 | 31 | 5 | 2 | −1⁄32 = −0.03125000… |
| 1 | 37 | 5 | 2 | 5⁄32 = 0.15625000… |
| 4 | 37 | 21 | 2 | −222,991⁄2,097,152 = −0.10633039… |
| 5 | 37 | 26 | 2 | 2,235,093⁄67,108,864 = 0.03330548… |
| 1 | 41 | 5 | 2 | 9⁄32 = 0.28125000… |
| 2 | 41 | 11 | 2 | −367⁄2048 = −0.17919922… |
| 3 | 41 | 16 | 2 | 3385⁄65,536 = 0.05165100… |
| 1 | 43 | 5 | 2 | 11⁄32 = 0.34375000… |
| 2 | 43 | 11 | 2 | −199⁄2048 = −0.09716797… |
| 5 | 43 | 27 | 2 | 12,790,715⁄134,217,728 = 0.09529825… |
| 7 | 43 | 38 | 2 | −3,059,295,837⁄274,877,906,944 = −0.01112965… |
Starting from the natural logarithm of q = 10 one might use these parameters:
| s | p | t | q | d⁄qt |
|---|---|---|---|---|
| 10 | 2 | 3 | 10 | 3⁄125 = 0.02400000… |
| 21 | 3 | 10 | 10 | 460,353,203⁄10,000,000,000 = 0.04603532… |
| 3 | 5 | 2 | 10 | 1⁄4 = 0.25000000… |
| 10 | 5 | 7 | 10 | −3⁄128 = −0.02343750… |
| 6 | 7 | 5 | 10 | 17,649⁄100,000 = 0.17649000… |
| 13 | 7 | 11 | 10 | −3,110,989,593⁄100,000,000,000 = −0.03110990… |
| 1 | 11 | 1 | 10 | 1⁄10 = 0.10000000… |
| 1 | 13 | 1 | 10 | 3⁄10 = 0.30000000… |
| 8 | 13 | 9 | 10 | −184,269,279⁄1,000,000,000 = −0.18426928… |
| 9 | 13 | 10 | 10 | 604,499,373⁄10,000,000,000 = 0.06044994… |
| 1 | 17 | 1 | 10 | 7⁄10 = 0.70000000… |
| 4 | 17 | 5 | 10 | −16,479⁄100,000 = −0.16479000… |
| 9 | 17 | 11 | 10 | 18,587,876,497⁄100,000,000,000 = 0.18587876… |
| 3 | 19 | 4 | 10 | −3141⁄10,000 = −0.31410000… |
| 4 | 19 | 5 | 10 | 30,321⁄100,000 = 0.30321000… |
| 7 | 19 | 9 | 10 | −106,128,261⁄1,000,000,000 = −0.10612826… |
| 2 | 23 | 3 | 10 | −471⁄1000 = −0.47100000… |
| 3 | 23 | 4 | 10 | 2167⁄10,000 = 0.21670000… |
| 2 | 29 | 3 | 10 | −159⁄1000 = −0.15900000… |
| 2 | 31 | 3 | 10 | −39⁄1000 = −0.03900000… |
Known digits
This is a table of recent records in calculating digits of ln 2. As of December 2018, it has been calculated to more digits than any other natural logarithm[4][5] of a natural number, except that of 1.
| Date | Name | Number of digits |
|---|---|---|
| January 7, 2009 | A.Yee & R.Chan | 15,500,000,000 |
| February 4, 2009 | A.Yee & R.Chan | 31,026,000,000 |
| February 21, 2011 | Alexander Yee | 50,000,000,050 |
| May 14, 2011 | Shigeru Kondo | 100,000,000,000 |
| February 28, 2014 | Shigeru Kondo | 200,000,000,050 |
| July 12, 2015 | Ron Watkins | 250,000,000,000 |
| January 30, 2016 | Ron Watkins | 350,000,000,000 |
| April 18, 2016 | Ron Watkins | 500,000,000,000 |
| December 10, 2018 | Michael Kwok | 600,000,000,000 |
| April 26, 2019 | Jacob Riffee | 1,000,000,000,000 |
| August 19, 2020 | Seungmin Kim[6][7] | 1,200,000,000,100 |
| September 9, 2021 | William Echols[8] | 1,500,000,000,000 |
| February 12, 2024 | Jordan Ranous[9] | 3,000,000,000,000 |
| November 15, 2025 | Mamdouh Barakat[10] | 3,100,000,000,000 |
See also
- Rule of 72#Continuous compounding, in which ln 2 figures prominently
- Half-life#Formulas for half-life in exponential decay, in which ln 2 figures prominently
- Erdős–Moser equation: all solutions must come from a convergent of ln 2.
References
- Brent, Richard P. (1976). "Fast multiple-precision evaluation of elementary functions". J. ACM. 23 (2): 242–251. doi:10.1145/321941.321944. MR 0395314. S2CID 6761843
- Uhler, Horace S. (1940). "Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17". Proc. Natl. Acad. Sci. U.S.A.. 26 (3): 205–212. Bibcode:1940PNAS...26..205U. doi:10.1073/pnas.26.3.205. MR 0001523. PMC 1078033. PMID 16588339
- Sweeney, Dura W. (1963). "On the computation of Euler's constant". Mathematics of Computation. 17 (82): 170–178. doi:10.1090/S0025-5718-1963-0160308-X. MR 0160308.
- Chamberland, Marc (2003). "Binary BBP-formulae for logarithms and generalized Gaussian–Mersenne primes". Journal of Integer Sequences. 6: 03.3.7. Bibcode:2003JIntS...6...37C. MR 2046407. Archived from the original on 2011-06-06. Retrieved 2010-04-29.
- Gourévitch, Boris & Jesús Goyanes Guillera (2007). "Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas". Applied Math. E-Notes. 7: 237–246. MR 2346048.
- Wu, Qiang (2003). "On the linear independence measure of logarithms of rational numbers". Mathematics of Computation. 72 (242): 901–911. doi:10.1090/S0025-5718-02-01442-4
- ^ Borwein, J.; Crandall, R.; Free, G. (2004). "On the Ramanujan AGM Fraction, I: The Real-Parameter Case". Exper. Math.. 13 (3): 278–280. doi:10.1080/10586458.2004.10504540. S2CID 17758274
- ^ A. N. Khovanski, The applications of continued fractions and their Generalisation to problemes in approximation theory,1963, Noordhoff, Groningen, The Netherlands
- ^ Mathematical Analysis Functions, Limits, Series, Continued Fractions. Edited by L.A. Lyusternik and A. R. Yanpol’Skii Translated by D. E. Brown, Translation edited by E. Spence, 1965, p.273. Pergamon Press, Oxford, London, Edinburgh, New York
- ^ "y-cruncher". numberworld.org. Retrieved 10 December 2018.
- ^ "Natural log of 2". numberworld.org. Retrieved 10 December 2018.
- ^ "Records set by y-cruncher". Archived 2020-09-15 at the Wayback Machine. Retrieved September 15, 2020.
- ^ "Natural logarithm of 2 (Log(2)) world record by Seungmin Kim". 19 August 2020. Retrieved September 15, 2020.
- ^ "William Echols". Retrieved March 15, 2025.
- ^ "Records set by y-cruncher". Retrieved March 15, 2025.
- ^ "Records set by y-cruncher". Retrieved December 28, 2025.
External links
- Gourdon, Xavier & Sebah, Pascal. "The logarithm constant:log 2"