自然對數

自然對數者（Natural logarithm），對數也，以歐拉數爲底，底計有：

註︰蓋當今數學之事，誠難僅以文述，而無符號，故凡數學之文，咸有漢字、拉丁字相易之事，以合文言、數學，則無論文理之人，皆可明之也。

${\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=\sum _{n=0}^{\infty }{\frac {1}{n!}}=2.7182818284......}$

表

n	ln(n)約數	整數數列線上大全
2	0.693147 180559 945309 417232 1214	A002162
3	1.098612 288668 109691 395245 2369	A002391
4	1.386294 361119 890618 834464 2429	A016627
5	1.609437 912434 100374 600759 3332	A016628
6	1.791759 469228 055000 812477 3583	A016629
7	1.945910 149055 313305 105352 7434	A016630
8	2.079441 541679 835928 251696 3643	A016631
9	2.197224 577336 219382 790490 4738	A016632
10	2.302585 092994 045684 017991 4546	A002392

式

${\displaystyle \ln(a)+\ln(b)=\ln(ab)}$反之亦然

${\displaystyle \ln(a)-ln(b)=\ln({\frac {a}{b}})}$

${\displaystyle \ln 1=0}$

${\displaystyle \ln e=1}$

${\displaystyle \ln a^{n}=n\ln a}$

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

${\displaystyle \lg n={\frac {\ln(n)}{ln10}}}$

${\displaystyle a^{n}=e^{n\ln a}}$

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

對數運算

由指數函數可知

${\displaystyle \log _{a}{N}=x|_{a^{x}=N}}$

${\displaystyle \log _{a}1=0}$

${\displaystyle \log _{a}{a}=1}$

${\displaystyle a^{\log _{a}{N}}=N}$其對數恒等式也，其至用也，可證對數運算，見其下

${\displaystyle \log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}}$

設${\displaystyle m=\log _{a}{M},n=\log _{a}{N}}$,則

${\displaystyle a^{m}\cdot a^{n}=a^{m+n}}$

由對數恒等式可知

${\displaystyle MN=a^{m+n}}$

${\displaystyle m+n=\log _{a}{(MN)}}$

即${\displaystyle \log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}}$

${\displaystyle \log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}}$

設${\displaystyle m=\log _{a}{M},n=\log _{a}{N}}$,則

${\displaystyle a^{m}\div a^{n}=a^{m-n}}$

由對數恒等式可知

${\displaystyle {\frac {M}{N}}=a^{m-n}}$

${\displaystyle m-n=\log _{a}{\frac {M}{N}}}$

即${\displaystyle \log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}}$

其亦有其推論

${\displaystyle \log _{a}{N^{m}}=m\log _{a}{N}}$

對數之易底公式

${\displaystyle \log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}}$

證：令${\displaystyle \log _{a}{b}=m}$，则

${\displaystyle a^{m}=b}$

${\displaystyle \log _{c}{b}=\log _{c}{a^{m}}=m\log _{c}{a}}$

則${\displaystyle m=\log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}={\frac {\log _{c}{a^{m}}}{\log _{c}{a}}}={\frac {m\log _{c}{a}}{\log _{c}{a}}}=m}$

故${\displaystyle \log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}}$

推論

${\displaystyle \log _{m}{n}={\frac {1}{\log _{n}{m}}}}$

${\displaystyle \log _{a^{m}}{b^{n}}={\frac {n}{m}}\log _{a}{b}}$

${\displaystyle e^{n}=10^{\frac {n}{\ln 10}}}$

對數製表公式之證明

求證${\displaystyle \ln 1=0,\ln n=\ln {\frac {n}{n-1}}+\ln {(n-1)}=2\sum _{k=1}^{\infty }[{\frac {1}{2k-1}}({\frac {1}{2n-1}})^{2k-1}]+\ln(n-1)}$

證明

${\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots }$

${\displaystyle \ln(1-x)=x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}-\cdots }$

${\displaystyle \operatorname {arth} {x}={\frac {1}{2}}\ln {\frac {1+x}{1-x}}=x+{\frac {1}{3}}x^{3}+{\frac {1}{5}}x^{5}+\cdots =\sum _{n=1}^{\infty }\left[{\frac {1}{2n-1}}x^{2n-1}\right]=\sum _{k=1}^{\infty }\left[{\frac {1}{2k-1}}x^{2k-1}\right]}$

令${\displaystyle {\frac {1+x}{1-x}}={\frac {n}{n-1}}}$，則

${\displaystyle (n-1)(1+x)=n(1-x)}$

${\displaystyle x={\frac {1}{2n-1}}}$

${\displaystyle \ln {\frac {n}{n-1}}=2\operatorname {arth} {\frac {1}{2n-1}}=2\sum _{k=1}^{\infty }\left[{\frac {1}{2k-1}}\left({\frac {1}{2n-1}}\right)^{2k-1}\right]}$

故原式得證