Logaritma alami dari 2

Nilai desimal dari logaritma natural dari 2 (urutan (barisan A002162 pada OEIS) kira-kira

${\displaystyle \ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458}$.

Logaritma dari 2 dalam basis lainnya diperoleh dengan rumus

${\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}}$

Logaritma umum secara khusus adalah  A007524

${\displaystyle \log _{10}2\approx 0.301\,029\,995\,663\,981\,195}$.

Invers dari bilangannya ini adalah logaritma biner dari 10:

${\displaystyle \log _{2}10={\frac {1}{\log _{10}2}}\approx 3.321\,928\,095}$  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.

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots }$. Ini dikenal "deret harmonik bolak-balik".
• ${\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}}$.
• ${\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}}$.
• ${\displaystyle \ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}}$.
• ${\displaystyle \ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}}$.
• ${\displaystyle \ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}}$.

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}$.
• ${\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}}$.
• ${\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}}$.
• ${\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}}$
• ${\displaystyle \ln 2={\frac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}}$.
• ${\displaystyle \ln 2={\frac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}}$.

• ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2}$ menggunakan ${\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4k^{2}-3k}}=\ln 2+{\frac {\pi }{6}}}$ (jumlah timbal-balik dari bilangan dekagonal).

• ${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}}$.
• ${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2}$.

(${\displaystyle \gamma }$ adalah konstanta Euler−Mascheroni dan ${\displaystyle \zeta }$ adalah fungsi zeta Riemann.)

${\displaystyle \ln 2={\frac {2}{3}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}}$

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

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

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}}$.
• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}}$.

Menerapkannya untuk ${\displaystyle \textstyle {2={\frac {3}{2}}\cdot {\frac {4}{3}}}}$ memberikan:

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}}$.
• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {2}{5}}\sum _{k=0}^{\infty }{\frac {1}{25^{k}(2k+1)}}+{\frac {2}{7}}\sum _{k=0}^{\infty }{\frac {1}{49^{k}(2k+1)}}}$.

Menerapkannya untuk ${\displaystyle 2=\left({\sqrt {2}}\right)^{2}}$ memberikan:

• ${\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}}$.
• ${\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {4}{3+2{\sqrt {2}}}}\sum _{k=0}^{\infty }{\frac {1}{(17+12{\sqrt {2}})^{k}(2k+1)}}}$.

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

• ${\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{15^{n}n}}+3\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{80^{n}n}}+5\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{24^{n}n}}}$.
• ${\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {1}{16^{n}n}}+3\sum _{n=1}^{\infty }{\frac {1}{81^{n}n}}+5\sum _{n=1}^{\infty }{\frac {1}{25^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}+{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}+{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}}}$.

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

• ${\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\int _{1}^{2}{\frac {dx}{x}}=\ln 2}$.

• ${\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}\,dx=\ln 2}$.

• ${\displaystyle \int _{0}^{\frac {\pi }{3}}\tan x\,dx=2\int _{0}^{\frac {\pi }{4}}\tan x\,dx=\ln 2}$.

Pengembangan Piercenya adalah  A091846

${\displaystyle \ln 2=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots }$

Pengembangan Engelnya adalah  A059180

${\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\cdots }$

Pengembangan kotangennya adalah  A081785

${\displaystyle \ln 2=\cot({\operatorname {arccot}(0)-\operatorname {arccot}(1)+\operatorname {arccot}(5)-\operatorname {arccot}(55)+\operatorname {arccot}(14187)-\cdots })}$

Pengembangan pecahan berlanjutnya adalah  A016730

${\displaystyle \ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]}$

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

Pecahan berlanjut yang digeneralisasi ini:

${\displaystyle \ln 2=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]}$,^[1] dapat diekspresikan sebagai

${\displaystyle \ln 2={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+{\cfrac {3}{7+{\cfrac {4}{2+\ddots }}}}}}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}$

Diberikan sebuah nilai dari ${\displaystyle \ln 2}$, sebuah skema menghitung logaritma dari bilangan bulat lainnya adalah untuk mentabulasi logaritma dari bilangan prima dan di lapisan berikutnya, logaritma dari bilangan komposit ${\displaystyle c}$ berdasarkan faktorisasinya

${\displaystyle c=2^{i}3^{j}5^{k}7^{l}\cdots \rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots }$

Ini memakai

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

DI lapisan ketiga, logaritma bilangan rasional ${\displaystyle r={\frac {a}{b}}}$ dihitung dengan menggunakan ${\displaystyle \ln(r)=\ln(a)-\ln(b)}$, dan logaritma akar melalui ${\displaystyle \ln {\sqrt[{n}]{c}}={\frac {1}{n}}\ln c}$.

Logaritma dari 2 berguna dalam arti bahwa pangkat dari 2 tersebar agak padat, mencari ${\displaystyle 2^{i}}$ yang mendekati dengan pangkat ${\displaystyle b^{j}}$ dari bilangan ${\displaystyle b}$ lainnya relatif mudah, dan representasi deret ${\displaystyle \ln b}$ dengan menggabungkan ${\displaystyle 2}$ ke ${\displaystyle b}$ dengan perubahan logaritmik.

Jika ${\displaystyle p^{s}=q^{t}+d}$ dengan beberapa ${\displaystyle d}$, maka ${\displaystyle {\frac {p^{s}}{q^{t}}}=1+{\frac {d}{q^{t}}}}$ dan karena itu

${\displaystyle s\ln(p)-t\ln(q)=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }(-1)^{m+1}{\frac {({\frac {d}{q^{t}}})^{m}}{m}}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}}$

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

${\displaystyle 2\ln(3)=3\ln 2-\sum _{k\geq 1}{\frac {(-1)^{k}}{8^{k}k}}=3\ln 2+\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {1}{2\cdot 8+1}}\right)}^{2n+1}}$:

Ini sebenarnya baris ketiga dalam tabel ekspansi tipe ini:

${\displaystyle s}$	${\displaystyle p}$	${\displaystyle t}$	${\displaystyle q}$	${\displaystyle {\frac {d}{q^{t}}}}$
1	3	1	2	${\displaystyle {\frac {1}{2}}=0.500\,000\,00\dots }$
1	3	2	2	${\displaystyle -{\frac {1}{4}}=-0.250\,000\,00\dots }$
2	3	3	2	${\displaystyle {\frac {1}{8}}=0.125\,000\,00\dots }$
5	3	8	2	${\displaystyle -{\frac {13}{256}}=-0.050\,781\,25\dots }$
12	3	19	2	${\displaystyle {\frac {7153}{524288}}=0.013\,643\,26\dots }$
1	5	2	2	${\displaystyle {\frac {1}{4}}=0.250\,000\,00\dots }$
3	5	7	2	${\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }$
1	7	2	2	${\displaystyle {\frac {3}{4}}=0.750\,000\,00\dots }$
1	7	3	2	${\displaystyle -{\frac {1}{8}}=-0.125\,000\,00\dots }$
5	7	14	2	${\displaystyle {\frac {423}{16384}}=0.025\,817\,87\dots }$
1	1	3	2	${\displaystyle {\frac {3}{8}}=0.375\,000\,00\dots }$
2	11	7	2	${\displaystyle -{\frac {7}{128}}=-0.054\,687\,50\dots }$
11	11	38	2	${\displaystyle {\frac {10433763667}{274877906944}}=0.037\,957\,81\dots }$
1	13	3	2	${\displaystyle {\frac {5}{8}}=0.625\,000\,00\dots }$
1	13	4	2	${\displaystyle -{\frac {3}{16}}=-0.187\,500\,00\dots }$
3	13	11	2	${\displaystyle {\frac {149}{2048}}=0.072\,753\,91\dots }$
7	13	26	2	${\displaystyle -{\frac {4360347}{67108864}}=-0.064\,974\,23\dots }$
10	13	37	2	${\displaystyle {\frac {419538377}{137438953472}}=0.003\,052\,54\dots }$
1	17	4	2	${\displaystyle {\frac {1}{16}}=0.062\,500\,00\dots }$
1	19	4	2	${\displaystyle {\frac {3}{16}}=0.187\,500\,00\dots }$
4	19	17	2	${\displaystyle -{\frac {751}{131072}}=-0.005\,729\,68\dots }$
1	23	4	2	${\displaystyle {\frac {7}{16}}=0.437\,500\,00\dots }$
1	23	5	2	${\displaystyle -{\frac {9}{32}}=-0.281\,250\,00\dots }$
2	23	9	2	${\displaystyle {\frac {17}{512}}=0.033\,203\,12\dots }$
1	29	4	2	${\displaystyle {\frac {13}{16}}=0.812\,500\,00\dots }$
1	29	5	2	${\displaystyle -{\frac {3}{32}}=-0.093\,750\,00\dots }$
7	29	34	2	${\displaystyle {\frac {70007125}{17179869184}}=0.004\,074\,95\dots }$
1	31	5	2	${\displaystyle -{\frac {1}{32}}=-0.031\,250\,00\dots }$
1	37	5	2	${\displaystyle {\frac {5}{32}}=0.156\,250\,00\dots }$
4	37	21	2	${\displaystyle -{\frac {222\,991}{2\,097\,152}}=-0.106\,330\,39\dots }$
5	37	26	2	${\displaystyle {\frac {2\,235\,093}{67\,108\,864}}=0.033\,305\,48\dots }$
1	41	5	2	${\displaystyle {\frac {9}{32}}=0.281\,250\,00\dots }$
2	41	11	2	${\displaystyle -{\frac {367}{2048}}=-0.179\,199\,22\dots }$
3	41	16	2	${\displaystyle {\frac {3385}{65\,536}}=0.051\,651\,00\dots }$
1	43	5	2	${\displaystyle {\frac {11}{32}}=0.343\,750\,00\dots }$
2	43	11	2	${\displaystyle -{\frac {199}{2048}}=-0.097\,167\,97\dots }$
5	43	27	2	${\displaystyle {\frac {12\,790\,715}{134\,217\,728}}=0.095\,298\,25\dots }$
7	43	38	2	${\displaystyle {\frac {3\,059\,295\,837}{274\,877\,906\,944}}=0.011\,129\,65\dots }$

Dimulai dari logaritma natural dari ${\displaystyle q=10}$, salah satunya dapat menggunakan parameter-parameter ini:

${\displaystyle s}$	${\displaystyle p}$	${\displaystyle t}$	${\displaystyle q}$	${\displaystyle {\frac {d}{q^{t}}}}$
10	2	3	10	${\displaystyle {\frac {3}{125}}=0.024\,000\,00\dots }$
21	3	10	10	${\displaystyle {\frac {460\,353\,203}{10\,000\,000\,000}}=0.046\,035\,32\dots }$
3	5	2	10	${\displaystyle {\frac {1}{4}}=0.250\,000\,\dots }$
10	5	7	10	${\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }$
6	7	5	10	${\displaystyle {\frac {17\,649}{100\,000}}=0.176\,490\,00\dots }$
13	7	11	10	${\displaystyle -{\frac {3\,110\,989\,593}{100\,000\,000\,000}}=-0.031\,109\,90\dots }$
1	11	1	10	${\displaystyle {\frac {1}{10}}=0.100\,000\,00\dots }$
1	13	1	10	${\displaystyle {\frac {3}{10}}=0.300\,000\,00\dots }$
8	13	9	10	${\displaystyle -{\frac {184\,269\,279}{1\,000\,000\,000}}=-0.184\,269\,28\dots }$
9	13	10	10	${\displaystyle {\frac {604\,499\,373}{10\,000\,000\,000}}=0.060\,449\,94\dots }$
1	17	1	10	${\displaystyle {\frac {7}{10}}=0.700\,000\,00\dots }$
4	17	5	10	${\displaystyle -{\frac {16\,479}{100\,000}}=-0.164\,790\,00\dots }$
9	17	11	10	${\displaystyle {\frac {18\,587\,876\,497}{100\,000\,000\,000}}=0.185\,878\,76\dots }$
3	19	4	10	${\displaystyle -{\frac {3141}{10\,000}}=-0.314\,100\,00\dots }$
4	19	5	10	${\displaystyle {\frac {30\,321}{100\,000}}=0.303\,210\,00\dots }$
7	19	9	10	${\displaystyle -{\frac {106\,128\,261}{1\,000\,000\,000}}=-0.106\,128\,26\dots }$
2	23	3	10	${\displaystyle -{\frac {471}{1000}}=-0.471\,000\,00\dots }$
3	23	4	10	${\displaystyle {\frac {2167}{10\,000}}=0.216\,700\,00\dots }$
2	29	3	10	${\displaystyle -{\frac {159}{1000}}=-0.159\,000\,00\dots }$
2	31	3	10	${\displaystyle {\frac {39}{1000}}=-0.039\,000\,00\dots }$

Ini adalah sebuah tabel catatan terbaru dalam menghitung digit ${\displaystyle \ln 2}$. Mulai Desember 2018, ini telah dihitung lebih banyak digit dari setiap logaritma natural ^[2][3] dari sebuah bilangan asli, kecuali 1.

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[4][5]	1,2000,000,000,100

• Aturan 72#Penggabungan kontinu, di mana ${\displaystyle \ln 2}$ sangat menonjol
• Waktu-paruh#Rumus untuk waktu-paruh dalam peluruhan eksponensial, di mana ${\displaystyle \ln 2}$ sangat menonjol
• Persamanan Erdős–Moser, semua penyelesaian harus datang dari sebuah konvergen dari ${\displaystyle \ln 2}$.

• Brent, Richard P. (1976). "Fast multiple-precision evaluation of elementary functions". J. ACM. 23 (2): 242–251. doi:10.1145/321941.321944. MR 0395314. 
• 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. 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" (PDF). Journal of Integer Sequences. 6: 03.3.7. MR 2046407. Diarsipkan dari versi asli (PDF) tanggal 2011-06-06. Diakses tanggal 2010-04-29.  Parameter |url-status= yang tidak diketahui akan diabaikan (bantuan)
• Gourévitch, Boris; Guillera Goyanes, Jesús (2007). "Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas" (PDF). 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 . 

• (Inggris) Weisstein, Eric W. "Natural logarithm of 2". MathWorld. 
• table of natural logarithms, PlanetMath.org.
• Gourdon, Xavier; Sebah, Pascal. "The logarithm constant:log 2".