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# Logarithms

(Redirected from Decimal logarithm)

The base ${\displaystyle b}$ logarithm is the inverse of the base ${\displaystyle b}$ exponential, i.e.

${\displaystyle \log _{b}b^{y}:=y,\quad b>0,\,b\neq 1.\,}$

For example, ${\displaystyle \log _{7}2401=4}$, since ${\displaystyle 7^{4}=2401}$. If the base is not specified, in mathematics it is assumed to be Euler's number ${\displaystyle \scriptstyle e\,=\,2.71828\ldots \,}$ since it is the base of the natural logarithm, although among scientists and engineers the tacit base might be 10 (decimal logarithm, common logarithm); they then use ${\displaystyle \ln x}$ when the base is ${\displaystyle e}$. The notation Leonhard Euler himself used was ${\displaystyle l\,x}$,[1] which thankfully has been changed to something a little clearer. In computer science, base 2 (binary logarithm) is often considered.

## Formulae

Since ${\displaystyle x=b^{y}}$ implies ${\displaystyle \log x=y\log b}$, we have

${\displaystyle y=\log _{b}x={\frac {\log x}{\log b}},\quad b>0,\,b\neq 1,\,}$

where ${\displaystyle \log }$ denotes the natural logarithm.

## Maclaurin series expansions

Since for geometric series we have

${\displaystyle {\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n},\quad |x|<1,}$

thus

${\displaystyle \log \left({\frac {1}{1-x}}\right)=-\log(1-x)=\int _{0}^{x}{\frac {du}{1-u}}=\int _{0}^{x}\sum _{n=0}^{\infty }u^{n}=\left.\sum _{n=0}^{\infty }{\frac {u^{n+1}}{n+1}}\right|_{0}^{x}=\sum _{n=0}^{\infty }{\frac {x^{n+1}}{n+1}}=\sum _{n=1}^{\infty }{\frac {x^{n}}{n}},\quad |x|<1,}$

is the generating function of the harmonic sequence (unit fractions)

${\displaystyle \{{\tfrac {1}{1}},{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},{\tfrac {1}{5}},{\tfrac {1}{6}},{\tfrac {1}{7}},{\tfrac {1}{8}},{\tfrac {1}{9}},{\tfrac {1}{10}},{\tfrac {1}{11}},{\tfrac {1}{12}},\ldots \},}$

and (replacing ${\displaystyle x}$ by ${\displaystyle -x}$)

${\displaystyle \log(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}\,{\frac {x^{n}}{n}},\quad |x|<1,}$

is the generating function of the alternating harmonic sequence

${\displaystyle \{{\tfrac {1}{1}},-{\tfrac {1}{2}},{\tfrac {1}{3}},-{\tfrac {1}{4}},{\tfrac {1}{5}},-{\tfrac {1}{6}},{\tfrac {1}{7}},-{\tfrac {1}{8}},{\tfrac {1}{9}},-{\tfrac {1}{10}},{\tfrac {1}{11}},-{\tfrac {1}{12}},\ldots \},}$

which sums to log(2), obtained by setting ${\displaystyle x}$ to 1, the convergence being assured by the alternating series test.

Also (since the fractions with even denominators cancel out)

${\displaystyle \log {\sqrt {\frac {1+x}{1-x}}}={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right)=\sum _{n=1}^{\infty }{\frac {x^{2n-1}}{2n-1}},\quad |x|<1,}$

is the generating function of the unit factions with odd denominators

${\displaystyle \{{\tfrac {1}{1}},{\tfrac {1}{3}},{\tfrac {1}{5}},{\tfrac {1}{7}},{\tfrac {1}{9}},{\tfrac {1}{11}},{\tfrac {1}{13}},{\tfrac {1}{15}},{\tfrac {1}{17}},{\tfrac {1}{19}},{\tfrac {1}{21}},{\tfrac {1}{23}},\ldots \},}$

and (since the fractions with odd denominators cancel out)

${\displaystyle \log {\sqrt {\frac {1}{1-x^{2}}}}={\frac {1}{2}}\log \left({\frac {1}{(1+x)(1-x)}}\right)=\sum _{n=1}^{\infty }{\frac {x^{2n}}{2n}},\quad |x|<1,}$

is the generating function of the unit factions with even denominators

${\displaystyle \{{\tfrac {1}{2}},{\tfrac {1}{4}},{\tfrac {1}{6}},{\tfrac {1}{8}},{\tfrac {1}{10}},{\tfrac {1}{12}},{\tfrac {1}{14}},{\tfrac {1}{16}},{\tfrac {1}{18}},{\tfrac {1}{20}},{\tfrac {1}{22}},{\tfrac {1}{24}},\ldots \}.}$

#### Hierarchical list of operations pertaining to numbers [2] [3]

##### 1st iteration
• Addition:  S(S(⋯ "a times" ⋯ (S(n))))
, the sum n  +  a
, where  n
is the augend and  a
is the addend. (When addition is commutative both are simply called terms.)
• Subtraction:  P(P(⋯ "s times" ⋯ (P(n))))
, the difference n  −  s
, where  n
is the minuend and  s
is the subtrahend.
##### 2nd iteration
• Multiplication:  n + (n + (⋯ "k times" ⋯ (n + (n))))
, the product m  ⋅   k
, where  m
is the multiplicand and  k
is the multiplier.[4] (When multiplication is commutative both are simply called factors.)
• Division: the ratio n  /  d
, where  n
is the dividend and  d
is the divisor.
##### 3rd iteration
• Exponentiation (  d
as "degree",  b
as "base",  n
as "variable").
• Powers:  n  ⋅   (n  ⋅   (⋯ "d times" ⋯ (n  ⋅   (n))))
, written  n d
.
• Exponentials:  b  ⋅   (b  ⋅   (⋯ "n times" ⋯ (b  ⋅   (b))))
, written  b n
.
• Exponentiation inverses (  d
as "degree",  b
as "base",  n
as "variable").
##### 5th iteration
• Pentation (  d
as "degree",  b
as "base",  n
as "variable").
• Penta-powers:  n ^^ (n ^^ (⋯ "d times" ⋯ (n ^^ (n ^^ (n)))))
, written  n ^^^ d or n ↑↑↑ d
.
• Penta-exponentials:  b ^^ (b ^^ (⋯ "n times" ⋯ (b ^^ (b ^^ (b)))))
, written  b ^^^ n or b ↑↑↑ n
.
• Pentation inverses
##### 6th iteration
• Hexation (  d
as "degree",  b
as "base",  n
as "variable").
• Hexa-powers:  n ^^^ (n ^^^ (⋯ "d times" ⋯ (n ^^^ (n))))
, written  n ^^^^ d or n ↑↑↑↑ d
.
• Hexa-exponentials:  b ^^^ (b ^^^ (⋯ "n times" ⋯ (b ^^^ (b))))
, written  b ^^^^ n or b ↑↑↑↑ n
.
• Hexation inverses
##### 7th iteration
• Heptation (  d
as "degree",  b
as "base",  n
as "variable").
• Hepta-powers:  n ^^^^ (n ^^^^ (⋯ "d times" ⋯ (n ^^^^ (n))))
, written  n ^^^^^ d or n ↑↑↑↑↑ d
.
• Hepta-exponentials:  b ^^^^ (b ^^^^ (⋯ "n times" ⋯ (b ^^^^ (b))))
, written  b ^^^^^ n or b ↑↑↑↑↑ n
.
• Heptation inverses
##### 8th iteration
• Octation (  d
as "degree",  b
as "base",  n
as "variable").
• Octa-powers:  n ^^^^^ (n ^^^^^ (⋯ "d times" ⋯ (n ^^^^^ (n))))
, written  n ^^^^^^ d or n ↑↑↑↑↑↑ d
.
• Octa-exponentials:  b ^^^^^ (b ^^^^^ (⋯ "n times" ⋯ (b ^^^^^ (b))))
, written  b ^^^^^^ n or b ↑↑↑↑↑↑ n
.
• Octation inverses

## Notes

1. Ed Sandifer, "How Euler did it: Finding logarithms by hand," July 2005.
2. There is a lack of consensus on which comes first. Having the multiplier come second makes it consistent with the definitions for exponentiation and higher operations. This is also the convention used with transfinite ordinals:
 ω  ×  2 := ω  +  ω
.