自然對數者（Natural logarithm），對數也，以歐拉數爲底，底計有：
e = lim n → ∞ ( 1 + 1 n ) n = ∑ n = 0 ∞ 1 n ! = 2.7182818284...... {\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=\sum _{n=0}^{\infty }{\frac {1}{n!}}=2.7182818284......}
ln ( a ) + ln ( b ) = ln ( a b ) {\displaystyle \ln(a)+\ln(b)=\ln(ab)} 反之亦然
ln ( a ) − l n ( b ) = ln ( a b ) {\displaystyle \ln(a)-ln(b)=\ln({\frac {a}{b}})}
ln 1 = 0 {\displaystyle \ln 1=0}
ln e = 1 {\displaystyle \ln e=1}
ln a n = n ln a {\displaystyle \ln a^{n}=n\ln a}
log a b = ln b ln a {\displaystyle \log _{a}b={\frac {\ln b}{\ln a}}}
lg n = ln ( n ) l n 10 {\displaystyle \lg n={\frac {\ln(n)}{ln10}}}
a n = e n ln a {\displaystyle a^{n}=e^{n\ln a}}
ln n = ln n n − 1 + ln ( n − 1 ) = 2 ∑ k = 1 ∞ [ 1 2 k − 1 ( 1 2 n − 1 ) 2 k − 1 ] + ln ( n − 1 ) {\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)}
由指數函數可知
log a N = x | a x = N {\displaystyle \log _{a}{N}=x|_{a^{x}=N}}
log a 1 = 0 {\displaystyle \log _{a}1=0}
log a a = 1 {\displaystyle \log _{a}{a}=1}
a log a N = N {\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
log a M + log a N = log a ( M N ) {\displaystyle \log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}}
設 m = log a M , n = log a N {\displaystyle m=\log _{a}{M},n=\log _{a}{N}} ,則
a m ⋅ a n = a m + n {\displaystyle a^{m}\cdot a^{n}=a^{m+n}}
由對數恒等式可知
M N = a m + n {\displaystyle MN=a^{m+n}}
m + n = log a ( M N ) {\displaystyle m+n=\log _{a}{(MN)}}
即 log a M + log a N = log a ( M N ) {\displaystyle \log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}}
log a M − log a N = log a M N {\displaystyle \log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}}
a m ÷ a n = a m − n {\displaystyle a^{m}\div a^{n}=a^{m-n}}
M N = a m − n {\displaystyle {\frac {M}{N}}=a^{m-n}}
m − n = log a M N {\displaystyle m-n=\log _{a}{\frac {M}{N}}}
即 log a M − log a N = log a M N {\displaystyle \log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}}
其亦有其推論
log a N m = m log a N {\displaystyle \log _{a}{N^{m}}=m\log _{a}{N}}
對數之易底公式
log a b = log c b log c a {\displaystyle \log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}}
證：令 log a b = m {\displaystyle \log _{a}{b}=m} ，则
a m = b {\displaystyle a^{m}=b}
log c b = log c a m = m log c a {\displaystyle \log _{c}{b}=\log _{c}{a^{m}}=m\log _{c}{a}}
則 m = log a b = log c b log c a = log c a m log c a = m log c a log c a = m {\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 log a b = log c b log c a {\displaystyle \log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}}
推論
log m n = 1 log n m {\displaystyle \log _{m}{n}={\frac {1}{\log _{n}{m}}}}
log a m b n = n m log a b {\displaystyle \log _{a^{m}}{b^{n}}={\frac {n}{m}}\log _{a}{b}}
e n = 10 n ln 10 {\displaystyle e^{n}=10^{\frac {n}{\ln 10}}}
對數製表公式之證明
求證 ln 1 = 0 , ln n = ln n n − 1 + ln ( n − 1 ) = 2 ∑ k = 1 ∞ [ 1 2 k − 1 ( 1 2 n − 1 ) 2 k − 1 ] + ln ( n − 1 ) {\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
ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ⋯ {\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots }
ln ( 1 + x ) = x − x 2 2 − x 3 3 − x 4 4 − ⋯ {\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}-\cdots }
arth x = 1 2 ln 1 + x 1 − x = x + 1 3 x 3 + 1 5 x 5 + ⋯ = ∑ n = 1 ∞ [ 1 2 n − 1 x 2 n − 1 ] = ∑ k = 1 ∞ [ 1 2 k − 1 x 2 k − 1 ] {\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 1 + x 1 − x = n n − 1 {\displaystyle {\frac {1+x}{1-x}}={\frac {n}{n-1}}} ，則
( n − 1 ) ( 1 + x ) = n ( 1 − x ) {\displaystyle (n-1)(1+x)=n(1-x)}
x = 1 2 n − 1 {\displaystyle x={\frac {1}{2n-1}}}
ln n n − 1 = 2 arth 1 2 n − 1 = 2 ∑ k = 1 ∞ [ 1 2 k − 1 ( 1 2 n − 1 ) 2 k − 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
反之亦然
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce7e4f2b3ae37ca3533e24ac32ecbac3e5429f63)
其對數恒等式也，其至用也，可證對數運算，見其下
,則
,則
，则
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/433bb3c4a985dfc77f3603748737f5350ec332ba)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fb99ebd7615851883739df2ac95506a71306a66)
，則
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9ab2bff2fc46ff264915a15104bb67875ba1fb3)
