x because antilog b ( log b ⁡ ( x ) )x (x)}=x{\text{ because }}{\mbox{antilog}}_{b}(\log _{b}(x))=x} log b ⁡ ( b x )x because log b ⁡ ( antilog b ( x ) )x (b^{x})=x{\text{ because }}\log _{b}({\mbox{antilog}}_{b}(x))=x} Both of the above are derived from the following two equations that define a logarithm: (note that in this explanation, the variables of x and x may not be referring to the same number) log b ⁡ ( y )
Secara matematis ini mengartikan, b b log ⁡ xx \!\log x}=x} karena b antilog ⁡ ( b log ⁡ x )x \!\operatorname {antilog} (\,^{b}\!\log x)=x} ; dan b log ⁡ ( b x )x \!\log(b^{x})=x} karena b log ⁡ ( b antilog ⁡ x )x \!\log(\,^{b}\!\operatorname {antilog} x)=x} .
Looking at the equation log b ⁡ ( y )x (y)=x} , and substituting the value for y of b xyy} , we get the following equation: log b ⁡ ( y )
b log ⁡ ( x y )b log ⁡ x − b log ⁡ y \!\log \left({\frac {x}{y}}\right)=\,^{b}\!\log x-\,^{b}\!\log y}
yb y log b ⁡ ( x ) ⇒ log b ⁡ ( x y )
d xx log b ⁡ ( x ) − x ln ⁡ ( b ) + Cx ( log b ⁡ ( x )
y log b ⁡ ( x )(b^{\log _{b}(x)})^{y}=b^{y\log _{b}(x)}\Rightarrow \log _{b}(x^{y})=y\log _{b}(x)} The law relating to quotients then follows: log b ⁡ ( x y )
b log ⁡ ( x y )b log ⁡ x − b log ⁡ y \!\log \left({\frac {x}{y}}\right)=\,^{b}\!\log x-\,^{b}\!\log y}
log b ⁡ ( x y − 1 )log b ⁡ ( x ) + log b ⁡ ( y − 1 )log b ⁡ ( x ) − log b ⁡ ( y ) {\bigg (}{\frac {x}{y}}{\bigg )}=\log _{b}(xy^{-1})=\log _{b}(x)+\log _{b}(y^{-1})=\log _{b}(x)-\log _{b}(y)} log b ⁡ ( 1 y )log b ⁡ ( y − 1 )− log b ⁡ ( y ) {\bigg (}{\frac {1}{y}}{\bigg )}=\log _{b}(y^{-1})=-\log _{b}(y)} Similarly, the root law is derived by rewriting the root as a reciprocal power: log b ⁡ ( x y )
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
We want to relate the expressions log b ⁡ ( x ) (x)} and log b ⁡ ( y ) (y)} .
Pertukaran basis pada logaritma dapat dirumuskan sebagai b log ⁡ x1 x log ⁡ b \!\log x={\frac {1}{^{x}\!\log b}}} .
b m ⋅ b nb m + n )(b^{n})=b^{m}\cdot b^{n}=b^{m+n}} To recover the logarithms, we apply log b } to both sides of the equality.
Perhatikan bahwa sifat logaritma di atas dapat kita pakai untuk membuktikan bahwa b log ⁡ x 1 n1 n b log ⁡ x \!\log x^{\frac {1}{n}}={\frac {1}{n}}\,^{b}\!\log x} .
log b ⁡ ( b m + n ) (xy)=\log _{b}(b^{m+n})}
b log ⁡ b nn \!\log b^{n}=n} .
The right side may be simplified using one of the logarithm properties from before: we know that log b ⁡ ( b m + n )
Perhatikan bahwa sifat logaritma di atas dapat kita pakai untuk membuktikan bahwa b log ⁡ x 1 n1 n b log ⁡ x \!\log x^{\frac {1}{n}}={\frac {1}{n}}\,^{b}\!\log x} .
m + n (b^{m+n})=m+n} , giving log b ⁡ ( x y )
b log ⁡ b nn \!\log b^{n}=n} .
b m b nb m − n {y}}={\frac {(b^{m})}{(b^{n})}}={\frac {b^{m}}{b^{n}}}=b^{m-n}} To recover the logarithms, we apply log b } to both sides of the equality.
Perhatikan bahwa sifat logaritma di atas dapat kita pakai untuk membuktikan bahwa b log ⁡ x 1 n1 n b log ⁡ x \!\log x^{\frac {1}{n}}={\frac {1}{n}}\,^{b}\!\log x} .
m − n (b^{m-n})=m-n} , giving log b ⁡ ( x y )
b log ⁡ b nn \!\log b^{n}=n} .
log b ⁡ ( x y )log b ⁡ ( x ) − log b ⁡ ( y ) \left({\frac {x}{y}}\right)=\log _{b}(x)-\log _{b}(y)} This completes the derivation.
b log ⁡ ( x y )b log ⁡ x − b log ⁡ y \!\log \left({\frac {x}{y}}\right)=\,^{b}\!\log x-\,^{b}\!\log y}
log b ⁡ ( x r ) log b ⁡ ( x r )
d xx log b ⁡ ( x ) − x ln ⁡ ( b ) + Cx ( log b ⁡ ( x )
As briefly mentioned in the § Trivial identities section, the definition of a logarithm is: log b ⁡ ( y )
Berikut adalah daftar identitas logaritma beserta dengan pembuktian-pembuktiannya, antara lain: Salah satu yang paling mendasar dalam identitas logaritma, ialah b log ⁡ b1 \!\log b=1} , karena b 1bb} .
=y\iff x=\log _{b}(y)={\frac {\log(y)}{\log(b)}}={\frac {\ln(y)}{\ln(b)}}.}
b log ⁡ ( x y )b log ⁡ x − b log ⁡ y \!\log \left({\frac {x}{y}}\right)=\,^{b}\!\log x-\,^{b}\!\log y}
log a ⁡ ( x ) (b^{m})=\log _{a}(x)} Now, using the logarithm of a power property, which states that log a ⁡ ( b m )
Pertukaran basis pada logaritma dapat dirumuskan sebagai b log ⁡ x1 x log ⁡ b \!\log x={\frac {1}{^{x}\!\log b}}} .
m log a ⁡ ( b ) (b^{m})=m\log _{a}(b)} , m log a ⁡ ( b )log a ⁡ ( x ) (b)=\log _{a}(x)} Isolating m , we get the following: mlog a ⁡ ( x ) log a ⁡ ( b ) (x)}{\log _{a}(b)}}} Resubstituting mlog b ⁡ ( x ) (x)} back into the equation, log b ⁡ ( x )log a ⁡ ( x ) log a ⁡ ( b ) (x)={\frac {\log _{a}(x)}{\log _{a}(b)}}} This completes the proof that log b ⁡ ( x )log a ⁡ ( x ) log a ⁡ ( b ) (x)={\frac {\log _{a}(x)}{\log _{a}(b)}}} .
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
This formula has several consequences: log b ⁡ a1 log a ⁡ b a={\frac {1}{\log _{a}b}}} log b n ⁡ alog b ⁡ a n }a={\log _{b}a \over n}} log b ⁡ alog b ⁡ e ⋅ log e ⁡ alog b ⁡ e ⋅ ln ⁡ a a=\log _{b}e\cdot \log _{e}a=\log _{b}e\cdot \ln a} b log a ⁡ dd log a ⁡ b d}=d^{\log _{a}b}} − log b ⁡ alog b ⁡ ( 1 a )
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
log 1 / b ⁡ a a=\log _{b}\left({1 \over a}\right)=\log _{1/b}a} log b 1 ⁡ a 1 ⋯ log b n ⁡ a nlog b π ( 1 ) ⁡ a 1 ⋯ log b π ( n ) ⁡ a n , }a_{1}\,\cdots \,\log _{b_{n}}a_{n}=\log _{b_{\pi (1)}}a_{1}\,\cdots \,\log _{b_{\pi (n)}}a_{n},} where π {\textstyle \pi } is any permutation of the subscripts 1, ..., n.
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
For example log b ⁡ w ⋅ log a ⁡ x ⋅ log d ⁡ c ⋅ log d ⁡ zlog d ⁡ w ⋅ log b ⁡ x ⋅ log a ⁡ c ⋅ log d ⁡ z .
d xx log b ⁡ ( x ) − x ln ⁡ ( b ) + Cx ( log b ⁡ ( x )
log b ⁡ a 0 + log b ⁡ ( 1 + ∑ i1 N b ( log b ⁡ a i − log b ⁡ a 0 ) ) \sum _{i=0}^{N}a_{i}=\log _{b}a_{0}+\log _{b}\left(1+\sum _{i=1}^{N}{\frac {a_{i}}{a_{0}}}\right)=\log _{b}a_{0}+\log _{b}\left(1+\sum _{i=1}^{N}b^{\left(\log _{b}a_{i}-\log _{b}a_{0}\right)}\right)}
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
A useful identity involving exponents: x log ⁡ ( log ⁡ ( x ) ) log ⁡ ( x )log ⁡ ( x ) {\log(x)}}=\log(x)} or more universally: x log ⁡ ( a ) log ⁡ ( x )
xlog ⁡ x {\log x}}=\log x} atau x log ⁡ a log ⁡ xa {\log x}}=a} Sama halnya dengan penambahan dan pengurangan, maupun perkalian dan pembagian, logaritma dapat membatalkan eksponen karena kedua operasi tersebut saling invers.
a {\log(x)}}=a} 1 1 log x ⁡ ( a ) + 1 log y ⁡ ( a )log x y ⁡ ( a ) {{\frac {1}{\log _{x}(a)}}+{\frac {1}{\log _{y}(a)}}}}=\log _{xy}(a)} 1 1 log x ⁡ ( a ) − 1 log y ⁡ ( a )log x y ⁡ ( a ) {{\frac {1}{\log _{x}(a)}}-{\frac {1}{\log _{y}(a)}}}}=\log _{\frac {x}{y}}(a)} Based on, x 1 + x ≤ ln ⁡ ( 1 + x ) ≤ x ( 6 + x ) 6 + 4 x ≤ x for all − 1 < x {1+x}}\leq \ln(1+x)\leq {\frac {x(6+x)}{6+4x}}\leq x{\mbox{ for all }}{-1}<x} 2 x 2 + x ≤ 3 − 27 3 + 2 x ≤ x 1 + x + x 2 / 12 ≤ ln ⁡ ( 1 + x ) ≤ x 1 + x ≤ x 2 2 + x 1 + x for 0 ≤ x , reverse for − 1 < x ≤ 0 {\frac {2x}{2+x}}&\leq 3-{\sqrt {\frac {27}{3+2x}}}\leq {\frac {x}{\sqrt {1+x+x^{2}/12}}}\\[4pt]&\leq \ln(1+x)\leq {\frac {x}{\sqrt {1+x}}}\leq {\frac {x}{2}}{\frac {2+x}{1+x}}\\[4pt]&{\text{ for }}0\leq x{\text{, reverse for }}{-1}<x\leq 0\end{aligned}}} All are accurate around x0 , but not for large numbers.
b log ⁡ x + b log ⁡ ( 1 − y x ) \!\log(x-y)=\,^{b}\!\log x+\,^{b}\!\log \left(1-{\frac {y}{x}}\right)} Lebih umumnya lagi, b log ⁡ ( ∑ i0 n x i )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n a i a 0 )b log ⁡ x 0 + b log ⁡ ( 1 + ∑ i1 n b ( b log ⁡ x i − b log ⁡ x 0 ) ) \!\log \left(\sum _{i=0}^{n}x_{i}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}{\frac {a_{i}}{a_{0}}}\right)=\,^{b}\!\log x_{0}+\,^{b}\!\log \left(1+\sum _{i=1}^{n}b^{\left(^{b}\!\log x_{i}-^{b}\!\log x_{0}\right)}\right)} .
∫ ln ⁡ x d xx ln ⁡ x − x + Cx ( ln ⁡ x − 1 ) + C ∫ log a ⁡ x d xx log a ⁡ x − x ln ⁡ a + Cx ( ln ⁡ x − 1 )
Turunan dalam basis lain, antara lain d d x ln ⁡ x1 x }{\mathrm {d} x}}\ln x={\frac {1}{x}}} Integral logaritma dalam kalkulus dapat dirumuskan sebagai ∫ log b ⁡ ( x )
ln ⁡ a + C x\,dx=x\log _{a}x-{\frac {x}{\ln a}}+C={\frac {x(\ln x-1)}{\ln a}}+C} To remember higher integrals, it is convenient to define: x [ n ]x n ( log ⁡ ( x )
− 1 ln ⁡ ( b ) ) + C (x)\,dx=x\log _{b}(x)-{\frac {x}{\ln(b)}}+C=x\left(\log _{b}(x)-{\frac {1}{\ln(b)}}\right)+C} Integral dalam basis lain, antara lain ∫ ln ⁡ ( x )
− xx\log(x)-x} x [ 2 ]x 2 log ⁡ ( x )
≈ xBentuk pecahan berlanjut
− 11 6 x 3x^{3}\log(x)-{\begin{matrix}{\frac {11}{6}}\end{matrix}}x^{3}} Then: d d x x [ n ]n x [ n − 1 ] {dx}}\,x^{\left[n\right]}=nx^{\left[n-1\right]}} ∫ x [ n ] d
− ( 3 x ) 2 7 ( 2 y + x ) − ⋱ {y}}\right)={\cfrac {x}{y+{\cfrac {1x}{2+{\cfrac {1x}{3y+{\cfrac {2x}{2+{\cfrac {2x}{5y+{\cfrac {3x}{2+\ddots }}}}}}}}}}}}={\cfrac {2x}{2y+x-{\cfrac {(1x)^{2}}{3(2y+x)-{\cfrac {(2x)^{2}}{5(2y+x)-{\cfrac {(3x)^{2}}{7(2y+x)-\ddots }}}}}}}}} Daftar identitas eksponensiasi
1 + 2 π i k Principal value forms: Log ⁡ ( z 1 ) + Log ⁡ ( z 2 )
ln ⁡ ( 1 + x )∑ n ≥ 1 ( − 1 ) n − 1 x n n {\frac {(-1)^{n-1}x^{n}}{n}}} log ⁡ x ≈ x x − 1 x -1}{x}}} log ⁡ ( 1 + x )