Lihat sumber Logaritma alami dari 2

Nilai desimal dari [[logaritma natural]] dari [[2 (angka)|2]] (urutan {{OEIS|id=A002162}} kira-kira

:<math>\ln 2 \approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458</math>.

Logaritma dari 2 dalam basis lainnya diperoleh dengan [[Identitas logaritma|rumus]]

:<math>\log_b 2 = \frac{\ln 2}{\ln b}</math>

[[Logaritma umum]] secara khusus adalah {{OEIS2C|id=A007524}}

:<math>\log_{10} 2 \approx 0.301\,029\,995\,663\,981\,195</math>.

Invers dari bilangannya ini adalah [[logaritma biner]] dari 10:

:<math>\log_2 10 =\frac{1}{\log_{10} 2} \approx 3.321\,928\,095</math> {{OEIS2C|id=A020862}}

Dengan menggunakan [[teorema Lindemann–Weierstrass]], logaritma natural dari setiap [[bilangan asli]] selain 0 dan 1 (lebih umumnya, dari setiap positif [[bilangan aljabar]] selain 1) adalah sebuah [[bilangan transenden]].

== Wakilan deret ==

=== Faktorial bolak-balik menaik ===

* <math>\ln 2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots</math>. Ini dikenal "[[Deret harmonik (matematika)#Deret harmonik bolak-balik|deret harmonik bolak-balik]]".
* <math>\ln 2 = \frac{1}{2} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}</math>.
* <math>\ln 2 = \frac{5}{8} +\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)}</math>.
* <math>\ln 2 = \frac{2}{3} +\frac{3}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)}</math>.
* <math>\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)}</math>.
* <math>\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)}</math>.

=== Faktorial konstanta menaik biner ===

* <math>\ln 2 = \sum_{n=1}^\infty \frac{1}{2^{n}n}</math>.
* <math>\ln 2 = 1 -\sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)}</math>.
* <math>\ln 2 = \frac{1}{2} + 2 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)}</math>.
* <math>\ln 2 = \frac{5}{6} - 6 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)}</math>
* <math>\ln 2 = \frac{7}{12} + 24 \sum_{n=1}^\infty \frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}</math>.
* <math>\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)}</math>.

=== Wakilan deret lainnya ===

* <math>\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \ln 2</math>.
* <math>\sum_{n=1}^\infty \frac{1}{n(4n^2-1)} = 2\ln 2 -1</math>.
* <math>\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} = \ln 2 -1</math>.
* <math>\sum_{n=1}^\infty \frac{(-1)^n}{n(9n^2-1)} = 2\ln 2 -\frac{3}{2}</math>.
* <math>\sum_{n=1}^\infty \frac{1}{4n^2-2n} = \ln 2</math>.
* <math>\sum_{n=1}^\infty \frac{2(-1)^{n+1}(2n-1)+1}{8n^2-4n} = \ln 2</math>.
* <math>\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+1} = \frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}</math>.
* <math>\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2} = -\frac{\ln 2}{3}+\frac{\pi}{3\sqrt{3}}</math>.
* <math>\sum_{n=0}^\infty \frac{(-1)^{n}}{(3n+1)(3n+2)} = \frac{2\ln 2}{3}</math>.
* <math>\sum_{n=1}^\infty \frac{1}{\sum_{k=1}^n k^2} = 18 - 24 \ln 2</math> menggunakan <math>\lim_{N\rightarrow \infty} \sum_{n=N}^{2N} \frac{1}{n} = \ln 2</math>.
* <math>\sum_{n=1}^\infty \frac{1}{4k^2-3k} = \ln 2 + \frac{\pi}{6} </math> (jumlah timbal-balik dari [[bilangan dekagonal]]).

=== Melibatkan fungsi zeta Riemann ===

* <math>\sum_{n=2}^\infty \frac{1}{2^n}[\zeta(n)-1] = \ln 2 -\frac{1}{2}</math>.
* <math>\sum_{n=2}^\infty \frac{1}{2n+1}[\zeta(n)-1] = 1-\gamma-\frac{\ln 2}{2}</math>.
* <math>\sum_{n=1}^\infty \frac{1}{2^{2n-1}(2n+1)}\zeta(2n) = 1-\ln 2</math>.

(<math>\gamma</math> adalah [[konstanta Euler−Mascheroni]] dan <math>\zeta</math> adalah [[fungsi zeta Riemann]].)

=== Wakilan tipe-BBP ===
:<math>\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}</math>

(Lihat lebih banyak mengenai [[wakilan tipe Bailey−Borwein−Plouffe]] (BBP).) 

Menerapkan ketiga deret umum untuk logaritma natural ke 2 secara langsung memberikan:

* <math>\ln 2 = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}</math>.
* <math>\ln 2 = \sum_{n = 1}^\infty \frac{1}{2^{n}n}</math>.
* <math>\ln 2 = \frac{2}{3} \sum_{k = 0}^\infty \frac{1}{9^{k}(2k+1)}</math>.

Menerapkannya untuk <math>\textstyle{2 = \frac{3}{2} \cdot \frac{4}{3}}</math> memberikan:

* <math>\ln 2 = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{2^n n} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{3^n n}</math>.
* <math>\ln 2 = \sum_{n = 1}^\infty \frac{1}{3^n n} + \sum_{n = 1}^\infty \frac{1}{4^n n}</math>.
* <math>\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)}</math>.

Menerapkannya untuk <math>2 = \left(\sqrt{2}\right)^2</math> memberikan:

* <math>\ln 2 = 2 \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(\sqrt{2} + 1)^n n}</math>.
* <math>\ln 2 = 2 \sum_{n = 1}^\infty \frac{1}{(2 + \sqrt{2})^n n}</math>.
* <math>\ln 2 = \frac{4}{3 + 2 \sqrt{2}} \sum_{k = 0}^\infty \frac{1}{(17 + 12 \sqrt{2})^{k}(2k+1)}</math>.

Menerapkannya untuk <math>\textstyle 2 = { \left( \frac{16}{15} \right) }^{7} \cdot { \left( \frac{81}{80} \right) }^{3} \cdot { \left( \frac{25}{24} \right) }^{5}</math> memberikan:

* <math>\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}</math>.
* <math>\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}</math>.
* <math>\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)}</math>.

== Wakilan sebagai integral ==
Logaritma natural dari 2 sering terjadi sebagai hasil integrasi. Beberapa rumus eksplisit untuknya termasuk

* <math>\int_0^1 \frac{dx}{1+x} = \int_1^2 \frac{dx}{x} = \ln 2</math>.

* <math>\int_0^\infty e^{-x}\frac{1-e^{-x}}{x} \, dx= \ln 2</math>.

* <math>\int_0^\frac{\pi}{3} \tan x \, dx=2\int_0^\frac{\pi}{4} \tan x \, dx = \ln 2</math>.

== Wakilan lainnya ==
[[Pengembangan Pierce]]<nowiki/>nya adalah {{OEIS2C|A091846}}

:<math>\ln 2 = 1 -\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12} -\cdots</math>

[[Pengembangan Engel]]<nowiki/>nya adalah {{OEIS2C|id=A059180}}

:<math>\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</math>

Pengembangan kotangennya adalah {{OEIS2C|id=A081785}}

:<math>\ln 2 = \cot({\arccot(0) -\arccot(1) + \arccot(5) - \arccot(55) + \arccot(14187) -\cdots})</math>

Pengembangan [[pecahan berlanjut]]<nowiki/>nya adalah {{OEIS2C|id=A016730}}

:<math>\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]</math>

yang menghasilkan aproksimasi rasional, beberapa yang pertama adalah <math>0</math>, <math>1</math>, <math>\textstyle{\frac{2}{3}}</math>, <math>\textstyle{\frac{7}{10}}</math>, <math>\textstyle{\frac{9}{13}}</math>, dan <math>\textstyle{\frac{61}{88}}</math>.

[[Pecahan berlanjut yang digeneralisasi]] ini:

:<math>\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]</math>,<ref>{{cite journal|last1=Borwein|first1=J.|last2=Crandall|first2=R.|last3=Free|first3=G.|year=2004|title=On the Ramanujan AGM Fraction , I: The Real-Parameter Case|url=http://www.kurims.kyoto-u.ac.jp/EMIS/journals/EM/expmath/volumes/13/13.3/BorweinCrandallFee.pdf|journal=Exper. Math.|volume=13|pages=278–280|doi=10.1080/10586458.2004.10504540|number=3}}</ref> dapat diekspresikan sebagai

:<math>\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}}}}</math>

== Bootstrap logaritma lainnya ==
Diberikan sebuah nilai dari <math>\ln 2</math>, sebuah skema menghitung logaritma dari [[bilangan bulat]] lainnya adalah untuk mentabulasi logaritma dari [[bilangan prima]] dan di lapisan berikutnya, logaritma dari bilangan [[komposit]] <math>c</math> berdasarkan [[Tabel faktor bilangan prima|faktorisasi]]<nowiki/>nya

:<math>c=2^i3^j5^k7^l\cdots\rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots</math>

Ini memakai
{| class="wikitable sortable"
!Bilangan prima
!Memperkirakan logaritma natural
![[OEIS]]
|-
|[[2 (angka)|2]]
|<math>0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458</math>
|{{OEIS link|id=A002162}}
|-
|[[3 (angka)|3]]
|<math>1.098\,612\,288\,668\,109\,691\,395\,245\,236\,92</math>
|{{OEIS link|id=A002391}}
|-
|[[5 (angka)|5]]
|<math>1.609\,437\,912\,434\,100\,374\,600\,759\,333\,23</math>
|{{OEIS link|id=A016628}}
|-
|[[7 (angka)|7]]
|<math>1.945\,910\,149\,055\,313\,305\,105\,352\,743\,44</math>
|{{OEIS link|id=A016630}}
|-
|[[11 (angka)|11]]
|<math>2.397\,895\,272\,798\,370\,544\,061\,943\,577\,97</math>
|{{OEIS link|id=A016634}}
|-
|[[13 (angka)|13]]
|<math>2.564\,949\,357\,461\,536\,736\,053\,487\,441\,57</math>
|{{OEIS link|id=A016636}}
|-
|[[17 (angka)|17]]
|<math>2.833\,213\,344\,056\,216\,080\,249\,534\,617\,87</math>
|{{OEIS link|id=A016640}}
|-
|[[19 (angka)|19]]
|<math>2.944\,438\,979\,166\,440\,460\,009\,027\,431\,89</math>
|{{OEIS link|id=A016642}}
|-
|[[23 (angka)|23]]
|<math>3.135\,494\,215\,929\,149\,690\,806\,752\,831\,81</math>
|{{OEIS link|id=A016646}}
|-
|[[29 (angka)|29]]
|<math>3.367\,295\,829\,986\,474\,027\,183\,272\,032\,36</math>
|{{OEIS link|id=A016652}}
|-
|[[31 (angka)|31]]
|<math>3.433\,987\,204\,485\,146\,245\,929\,164\,324\,54</math>
|{{OEIS link|id=A016654}}
|-
|[[37 (angka)|37]]
|<math>3.610\,917\,912\,644\,224\,444\,368\,095\,671\,03</math>
|{{OEIS link|id=A016660}}
|-
|[[41 (angka)|41]]
|<math>3.713\,572\,066\,704\,307\,803\,866\,763\,373\,04</math>
|{{OEIS link|id=A016664}}
|-
|[[43 (angka)|43]]
|<math>3.761\,200\,115\,693\,562\,423\,472\,842\,513\,35</math>
|{{OEIS link|id=A016666}}
|-
|[[47 (angka)|47]]
|<math>3.850\,147\,601\,710\,058\,586\,820\,950\,669\,77</math>
|{{OEIS link|id=A016670}}
|-
|[[53 (angka)|53]]
|<math>3.970\,291\,913\,552\,121\,834\,144\,469\,139\,03</math>
|{{OEIS link|id=A016676}}
|-
|[[59 (angka)|59]]
|<math>4.077\,537\,443\,905\,719\,450\,616\,050\,373\,72</math>
|{{OEIS link|id=A016682}}
|-
|[[61 (angka)|61]]
|<math>4.110\,873\,864\,173\,311\,248\,751\,389\,103\,43</math>
|{{OEIS link|id=A016684}}
|-
|[[67 (angka)|67]]
|<math>4.204\,692\,619\,390\,966\,059\,670\,071\,996\,36</math>
|{{OEIS link|id=A016690}}
|-
|[[71 (angka)|71]]
|<math>4.262\,679\,877\,041\,315\,421\,329\,454\,532\,51</math>
|{{OEIS link|id=A016694}}
|-
|[[73 (angka)|73]]
|<math>4.290\,459\,441\,148\,391\,129\,092\,108\,857\,44</math>
|{{OEIS link|id=A016696}}
|-
|[[79 (angka)|79]]
|<math>4.369\,447\,852\,467\,021\,494\,172\,945\,541\,48</math>
|{{OEIS link|id=A016702}}
|-
|[[83 (angka)|83]]
|<math>4.418\,840\,607\,796\,597\,923\,475\,472\,223\,29</math>
|{{OEIS link|A016706}}
|-
|[[89 (angka)|89]]
|<math>4.488\,636\,369\,732\,139\,838\,317\,815\,540\,67</math>
|{{OEIS link|id=A016712}}
|-
|[[97 (angka)|97]]
|<math>4.574\,710\,978\,503\,382\,822\,116\,721\,621\,70</math>
|{{OEIS link|id=A016720}}
|}
DI lapisan ketiga, logaritma bilangan rasional <math>r = \frac{a}{b}</math> dihitung dengan menggunakan <math>\ln(r) = \ln(a) - \ln(b)</math>, dan logaritma akar melalui <math>\ln \sqrt[n]{c} = \frac{1}{n} \ln c</math>.

Logaritma dari 2 berguna dalam arti bahwa pangkat dari 2 tersebar agak padat, mencari <math>2^i</math> yang mendekati dengan pangkat <math>b^j</math> dari bilangan <math>b  </math> lainnya relatif mudah, dan representasi deret <math>\ln b</math> dengan menggabungkan <math>2</math> ke <math>b</math> dengan [[Identitas logaritmik|perubahan logaritmik]].

=== Contoh ===
Jika <math>p^s = q^t + d</math> dengan beberapa <math>d</math>, maka <math>\frac{p^s}{q^t} = 1 + \frac{d}{q^t}</math> dan karena itu

:<math>s\ln(p)-t\ln(q) = \ln\left(1+\frac{d}{q^t}\right) = \sum_{m=1}^\infty (-1)^{m+1}\frac{(\frac{d}{q^t})^m}{m} = \sum_{n=0}^\infty \frac{2}{2n+1} {\left(\frac{d}{2 q^t + d}\right)}^{2n+1}</math>

Memilih <math>q = 2</math> mewakili <math>\ln(p)</math> oleh <math>\ln 2 </math> dan sebuah deret dari sebuah parameter <math>\frac{d}{q^t}</math> yang ingin tetap kecil untuk konvergen cepat. Mengambil <math>3^2 = 2^3 + 1</math>, sebagai contoh, menghasilkan

:<math>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}</math>:

Ini sebenarnya baris ketiga dalam tabel ekspansi tipe ini:
{| class="wikitable sortable"
!<math>s</math>
!<math>p</math>
!<math>t</math>
!<math>q</math>
!<math>\frac{d}{q^t}</math>
|-
|1
|3
|1
|2
|<math>\frac 1 2 = 0.500\,000\,00\dots</math>
|-
|1
|3
|2
|2
|<math>-\frac 1 4 = -0.250\,000\,00\dots</math>
|-
|2
|3
|3
|2
|<math>\frac 1 8 = 0.125\,000\,00\dots</math>
|-
|5
|3
|8
|2
|<math>-\frac{13}{256} = -0.050\,781\,25 
\dots</math>
|-
|12
|3
|19
|2
|<math>\frac{7153}{524288} = 0.013\,643\,26\dots</math>
|-
|1
|5
|2
|2
|<math>\frac 1 4 = 0.250\,000\,00\dots</math>
|-
|3
|5
|7
|2
|<math>-\frac{3}{128} = -0.023\,437\,50\dots</math>
|-
|1
|7
|2
|2
|<math>\frac 3 4 = 0.750\,000\,00\dots</math>
|-
|1
|7
|3
|2
|<math>-\frac 1 8 = -0.125\,000\,00\dots</math>
|-
|5
|7
|14
|2
|<math>\frac{423}{16384} = 0.025\,817\,87\dots </math>
|-
|1
|1
|3
|2
|<math>\frac{3}{8} = 0.375\,000\,00\dots</math>
|-
|2
|11
|7
|2
|<math>-\frac{7}{128} = -0.054\,687\,50\dots</math>
|-
|11
|11
|38
|2
|<math>\frac{10433763667}{274877906944} = 0.037\,957\,81\dots</math>
|-
|1
|13
|3
|2
|<math>\frac{5}{8} = 0.625\,000\,00\dots</math>
|-
|1
|13
|4
|2
|<math>-\frac{3}{16} = -0.187\,500\,00\dots</math>
|-
|3
|13
|11
|2
|<math>\frac{149}{2048} = 0.072\,753\,91\dots</math>
|-
|7
|13
|26
|2
|<math>-\frac{4360347}{67108864} = -0.064\,974\,23\dots</math>
|-
|10
|13
|37
|2
|<math>\frac{419538377}{137438953472} = 0.003\,052\,54\dots</math>
|-
|1
|17
|4
|2
|<math>\frac{1}{16} = 0.062\,500\,00\dots</math>
|-
|1
|19
|4
|2
|<math>\frac{3}{16} = 0.187\,500\,00\dots</math>
|-
|4
|19
|17
|2
|<math>-\frac{751}{131072} = -0.005\,729\,68\dots</math>
|-
|1
|23
|4
|2
|<math>\frac{7}{16} = 0.437\,500\,00\dots</math>
|-
|1
|23
|5
|2
|<math>-\frac{9}{32}= -0.281\,250\,00\dots</math>
|-
|2
|23
|9
|2
|<math>\frac{17}{512} = 0.033\,203\,12\dots</math>
|-
|1
|29
|4
|2
|<math>\frac{13}{16} = 0.812\,500\,00\dots</math>
|-
|1
|29
|5
|2
|<math>-\frac{3}{32} = -0.093\,750\,00\dots</math>
|-
|7
|29
|34
|2
|<math>\frac{70007125}{17179869184} = 0.004\,074\,95\dots</math>
|-
|1
|31
|5
|2
|<math>-\frac{1}{32} = -0.031\,250\,00\dots</math>
|-
|1
|37
|5
|2
|<math>\frac{5}{32} = 0.156\,250\,00\dots</math>
|-
|4
|37
|21
|2
|<math>-\frac{222\,991}{2\,097\,152} = -0.106\,330\,39\dots</math>
|-
|5
|37
|26
|2
|<math>\frac{2\,235\,093}{67\,108\,864} = 0.033\,305\,48\dots</math>
|-
|1
|41
|5
|2
|<math>\frac{9}{32} = 0.281\,250\,00\dots</math>
|-
|2
|41
|11
|2
|<math>-\frac{367}{2048} = -0.179\,199\,22\dots</math>
|-
|3
|41
|16
|2
|<math>\frac{3385}{65\,536} = 0.051\,651\,00\dots</math>
|-
|1
|43
|5
|2
|<math>\frac{11}{32} = 0.343\,750\,00\dots</math>
|-
|2
|43
|11
|2
|<math>-\frac{199}{2048} = -0.097\,167\,97\dots</math>
|-
|5
|43
|27
|2
|<math>\frac{12\,790\,715}{134\,217\,728} = 0.095\,298\,25\dots</math>
|-
|7
|43
|38
|2
|<math>\frac{3\,059\,295\,837}{274\,877\,906\,944} = 0.011\,129\,65\dots</math>
|}
Dimulai dari logaritma natural dari <math>q = 10</math>, salah satunya dapat menggunakan parameter-parameter ini:
{| class="wikitable sortable"
!<math>s  </math>
!<math>p  </math>
!<math>t  </math>
!<math>q  </math>
!<math>\frac d {q^t}</math>
|-
|10
|2
|3
|10
|<math>\frac{3}{125} = 0.024\,000\,00\dots</math>
|-
|21
|3
|10
|10
|<math>\frac{460\,353\,203}{10\,000\,000\,000} = 0.046\,035\,32\dots</math>
|-
|3
|5
|2
|10
|<math>\frac{1}{4} = 0.250\,000\,\dots</math>
|-
|10
|5
|7
|10
|<math>-\frac{3}{128} = -0.023\,437\,50\dots </math>
|-
|6
|7
|5
|10
|<math>\frac{17\,649}{100\,000} = 0.176\,490\,00\dots</math>
|-
|13
|7
|11
|10
|<math>-\frac{3\,110\,989\,593}{100\,000\,000\,000} = -0.031\,109\,90\dots</math>
|-
|1
|11
|1
|10
|<math>\frac{1}{10} = 0.100\,000\,00\dots</math>
|-
|1
|13
|1
|10
|<math>\frac{3}{10} = 0.300\,000\,00\dots</math>
|-
|8
|13
|9
|10
|<math>-\frac{184\,269\,279}{1\,000\,000\,000} = -0.184\,269\,28\dots</math>
|-
|9
|13
|10
|10
|<math>\frac{604\,499\,373}{10\,000\,000\,000} = 0.060\,449\,94\dots</math>
|-
|1
|17
|1
|10
|<math>\frac{7}{10} = 0.700\,000\,00\dots</math>
|-
|4
|17
|5
|10
|<math>-\frac{16\,479}{100\,000} = -0.164\,790\,00\dots</math>
|-
|9
|17
|11
|10
|<math>\frac{18\,587\,876\,497}{100\,000\,000\,000} = 0.185\,878\,76\dots</math>
|-
|3
|19
|4
|10
|<math>-\frac{3141}{10\,000} = -0.314\,100\,00\dots</math>
|-
|4
|19
|5
|10
|<math>\frac{30\,321}{100\,000} = 0.303\,210\,00\dots</math>
|-
|7
|19
|9
|10
|<math>-\frac{106\,128\,261}{1\,000\,000\,000} = -0.106\,128\,26\dots</math>
|-
|2
|23
|3
|10
|<math>-\frac{471}{1000} = -0.471\,000\,00\dots</math>
|-
|3
|23
|4
|10
|<math>\frac{2167}{10\,000} = 0.216\,700\,00\dots</math>
|-
|2
|29
|3
|10
|<math>-\frac{159}{1000} = -0.159\,000\,00\dots</math>
|-
|2
|31
|3
|10
|<math>\frac{39}{1000} = -0.039\,000\,00\dots</math>
|}

== Digit yang diketahui ==
Ini adalah sebuah tabel catatan terbaru dalam menghitung digit <math>\ln 2  </math>. Mulai Desember 2018, ini telah dihitung lebih banyak digit dari setiap logaritma natural <ref>{{cite web|author=|date=|title=y-cruncher|url=http://www.numberworld.org/y-cruncher/|work=numberworld.org|accessdate=10 December 2018}}</ref><ref>{{cite web|author=|date=|title=Natural log of 2|url=http://www.numberworld.org/digits/Log(2)/|work=numberworld.org|accessdate=10 December 2018}}</ref> dari sebuah bilangan asli, kecuali 1.
{| class="wikitable"
|+
!Tanggal
!Nama
!Jumlah digit
|-
|7 Januari 2009
|A Yee & R Chan
|15,500,000,000
|-
|4 Februari 2009
|A Yee & R Chan
|31,026,000,000
|-
|21 Februari 2011
|Alexander Yee
|50,000,000,050
|-
|14 Maret, 2011
|Shigeru Kondo
|100,000,000,000
|-
|28 Februari 2014
|Shigeru Kondo
|200,000,000,050
|-
|12 Juli 2015
|Ron Watkins
|250,000,000,000
|-
|30 Januari 2016
|Ron Watkins
|350,000,000,000
|-
|18 April 2016
|Ron Watkins
|500,000,000,000
|-
|10 Desember 2018
|Michael Kwok
|600,000,000,000
|-
|26 April 2019
|Jacob Riffee
|1,000,000,000,000
|-
|19 Agustus 2020
|Seungmin Kim<ref>{{Cite web|title=Records set by y-cruncher|url=http://www.numberworld.org/y-cruncher/|archive-url=https://web.archive.org/web/20200915070251/http://www.numberworld.org/y-cruncher/|archive-date=2020-09-15|access-date=September 15, 2020}}</ref><ref>{{Cite web|title=Natural logarithm of 2 (Log(2)) world record by Seungmin Kim|url=https://ehfd.github.io/world-record/natural-logarithm-of-2-log2/|access-date=September 15, 2020}}</ref>
|1,2000,000,000,100
|}

== Lihat pula ==

* [[Aturan 72#Penggabungan kontinu]], di mana <math>\ln 2  </math> sangat menonjol
* [[Waktu-paruh#Turunan|Waktu-paruh#Rumus untuk waktu-paruh dalam peluruhan eksponensial]], di mana <math>\ln 2  </math> sangat menonjol
* [[Persamanan Erdős–Moser]], semua penyelesaian harus datang dari sebuah konvergen dari <math>\ln 2  </math>.

== Referensi ==

* {{cite journal|last1=Brent|first1=Richard P.|year=1976|title=Fast multiple-precision evaluation of elementary functions|journal=J. ACM|volume=23|issue=2|pages=242–251|doi=10.1145/321941.321944|mr=0395314}}
* {{cite journal|last1=Uhler|first1=Horace S.|year=1940|title=Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17|journal=Proc. Natl. Acad. Sci. U.S.A.|volume=26|issue=3|pages=205–212|doi=10.1073/pnas.26.3.205|mr=0001523|pmc=1078033|pmid=16588339}}
* {{cite journal|last1=Sweeney|first1=Dura W.|year=1963|title=On the computation of Euler's constant|journal=Mathematics of Computation|volume=17|issue=82|pages=170–178|doi=10.1090/S0025-5718-1963-0160308-X|mr=0160308|doi-access=free}}
* {{cite journal|last1=Chamberland|first1=Marc|year=2003|title=Binary BBP-formulae for logarithms and generalized Gaussian–Mersenne primes|url=http://www.emis.ams.org/journals/JIS/VOL6/Chamberland/chamberland60.pdf|journal=Journal of Integer Sequences|volume=6|page=03.3.7|mr=2046407|archive-url=https://web.archive.org/web/20110606014342/http://www.emis.ams.org/journals/JIS/VOL6/Chamberland/chamberland60.pdf|archive-date=2011-06-06|access-date=2010-04-29|url-status=dead}}
* {{cite journal|last1=Gourévitch|first1=Boris|last2=Guillera Goyanes|first2=Jesús|year=2007|title=Construction of binomial sums for {{pi}} and polylogarithmic constants inspired by BBP formulas|url=http://www.math.nthu.edu.tw/~amen/2007/061028-2.pdf|journal=Applied Math. E-Notes|volume=7|pages=237–246|mr=2346048}}
* {{cite journal|last1=Wu|first1=Qiang|year=2003|title=On the linear independence measure of logarithms of rational numbers|journal=Mathematics of Computation|volume=72|issue=242|pages=901–911|doi=10.1090/S0025-5718-02-01442-4|doi-access=free}}
{{reflist}}

== Pranala luar ==

* {{MathWorld|urlname=NaturalLogarithmof2|title=Natural logarithm of 2}}
* {{planetmath reference|id=10699|title=table of natural logarithms}}
* {{cite web|last1=Gourdon|first1=Xavier|last2=Sebah|first2=Pascal|title=The logarithm constant:log 2|url=http://numbers.computation.free.fr/Constants/Log2/log2.html}}

[[Kategori:Bilangan transendental riil]]
[[Kategori:Konstanta matematika]]
[[Kategori:Logaritma]]