List of logarithmic identities – Wikipedia

In mathematics, many logarithmic identities exist. The following is a compilation of the notable of these, many of which are used for computational purposes.

Trivial identities[edit]

${displaystyle log _{b}(1)=0}$	because	${displaystyle b^{0}=1}$
${displaystyle log _{b}(b)=1}$	because	${displaystyle b^{1}=b}$

Explanations[edit]

By definition, we know that:

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}$

,

where

${displaystyle color {blue}bcolor {black}neq 0}$

or

${displaystyle color {blue}bcolor {black}neq 1}$

.

Setting

${displaystyle color {red}xcolor {black}=0}$

,
we can see that:

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{(0)}color {black}=color {green}ycolor {black}iff color {blue}1color {black}=color {green}ycolor {black}iff color {green}ycolor {black}=color {blue}1color {black}}$

. So, substituting these values into the formula, we see that:

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {blue}1color {black})=color {red}0color {black}}$

, which gets us the first property.

Setting

${displaystyle color {red}xcolor {black}=1}$

,
we can see that:

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{(1)}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}=color {green}ycolor {black}iff color {green}ycolor {black}=color {blue}bcolor {black}}$

. So, substituting these values into the formula, we see that:

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {blue}bcolor {black})=color {red}1color {black}}$

, which gets us the second property.

Many mathematical identities are called trivial , only because they are relatively simple (typically from the perspective of an experienced mathematician).
This is not to say that calling an identity or formula trivial means that it’s not important.

Cancelling exponentials[edit]

Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operations—much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations.

${displaystyle b^{log _{b}(x)}=x{text{ because }}{mbox{antilog}}_{b}(log _{b}(x))=x}$

${displaystyle log _{b}(b^{x})=x{text{ because }}log _{b}({mbox{antilog}}_{b}(x))=x}$

^[1]

Both of the above are derived from the following two equations that define a logarithm:
(note that in this explanation, the variables of

${displaystyle color {red}xcolor {black}}$

and

${displaystyle x}$

may not be referring to the same number)

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}$

Looking at the equation

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}$

, and substituting the value for

${displaystyle color {red}xcolor {black}}$

of

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}}$

, we get the following equation:

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{log _{b}(y)}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})}color {black}=color {green}ycolor {black}}$

, which gets us the first equation.
Another more rough way to think about it is that

${displaystyle color {blue}bcolor {black}color {red}^{text{something}}color {black}=color {green}ycolor {black}}$

,
and that that “

${displaystyle color {red}{text{something}}}$

” is

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})}$

.

Looking at the equation

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}}$

, and substituting the value for

${displaystyle color {green}ycolor {black}}$

of

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}$

, we get the following equation:

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {green}b^{x}color {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}({color {blue}bcolor {black}color {red}^{x}color {black}}color {black})=color {red}xcolor {black}}$

, which gets us the second equation.
Another more rough way to think about it is that

${displaystyle color {black}log color {blue}_{b}color {black}(color {green}{text{something}}color {black})=color {red}xcolor {black}}$

,
and that that something “

${displaystyle color {green}{text{something}}}$

” is

${displaystyle color {blue}bcolor {black}color {red}^{x}color {black}}$

.

Using simpler operations[edit]

Logarithms can be used to make calculations easier. For example, two numbers can be multiplied just by using a logarithm table and adding. These are often known as logarithmic properties, which are documented in the table below.^[2] The first three operations below assume that x = b^c and/or y = b^d, so that log_b(x) = c and log_b(y) = d. Derivations also use the log definitions x = b^log_b(x) and x = log_b(b^x).

Where

${displaystyle b}$

,

${displaystyle x}$

, and

${displaystyle y}$

are positive real numbers and

${displaystyle bneq 1}$

, and

${displaystyle c}$

and

${displaystyle d}$

are real numbers.

The laws result from canceling exponentials and the appropriate law of indices. Starting with the first law:

${displaystyle xy=b^{log _{b}(x)}b^{log _{b}(y)}=b^{log _{b}(x)+log _{b}(y)}Rightarrow log _{b}(xy)=log _{b}(b^{log _{b}(x)+log _{b}(y)})=log _{b}(x)+log _{b}(y)}$

The law for powers exploits another of the laws of indices:

${displaystyle x^{y}=(b^{log _{b}(x)})^{y}=b^{ylog _{b}(x)}Rightarrow log _{b}(x^{y})=ylog _{b}(x)}$

The law relating to quotients then follows:

${displaystyle log _{b}{bigg (}{frac {x}{y}}{bigg )}=log _{b}(xy^{-1})=log _{b}(x)+log _{b}(y^{-1})=log _{b}(x)-log _{b}(y)}$

${displaystyle log _{b}{bigg (}{frac {1}{y}}{bigg )}=log _{b}(y^{-1})=-log _{b}(y)}$

Similarly, the root law is derived by rewriting the root as a reciprocal power:

${displaystyle log _{b}({sqrt[{y}]{x}})=log _{b}(x^{frac {1}{y}})={frac {1}{y}}log _{b}(x)}$

Derivations of product, quotient, and power rules[edit]

These are the three main logarithm laws/rules/principles,^[3] from which the other properties listed above can be proven. Each of these logarithm properties correspond to their respective exponent law, and their derivations/proofs will hinge on those facts. There are multiple ways to derive/prove each logarithm law – this is just one possible method.

Logarithm of a product[edit]

To state the logarithm of a product law formally:

${displaystyle forall bin mathbb {R} _{+},bneq 1,forall x,y,in mathbb {R} _{+},log _{b}(xy)=log _{b}(x)+log _{b}(y)}$

Derivation:

Let

${displaystyle bin mathbb {R} _{+}}$

, where

${displaystyle bneq 1}$

,
and let

${displaystyle x,yin mathbb {R} _{+}}$

. We want to relate the expressions

${displaystyle log _{b}(x)}$

and

${displaystyle log _{b}(y)}$

. This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to

${displaystyle log _{b}(x)}$

and

${displaystyle log _{b}(y)}$

quite often, we will give them some variable names to make working with them easier: Let

${displaystyle m=log _{b}(x)}$

, and let

${displaystyle n=log _{b}(y)}$

.

Rewriting these as exponentials, we see that

${displaystyle m=log _{b}(x)iff b^{m}=x}$

and

${displaystyle n=log _{b}(y)iff b^{n}=y}$

. From here, we can relate

${displaystyle b^{m}}$

(i.e.

${displaystyle x}$

) and

${displaystyle b^{n}}$

(i.e.

${displaystyle y}$

) using exponent laws as

${displaystyle xy=(b^{m})(b^{n})=b^{m}cdot b^{n}=b^{m+n}}$

To recover the logarithms, we apply

${displaystyle log _{b}}$

to both sides of the equality.

${displaystyle log _{b}(xy)=log _{b}(b^{m+n})}$

The right side may be simplified using one of the logarithm properties from before: we know that

${displaystyle log _{b}(b^{m+n})=m+n}$

, giving

${displaystyle log _{b}(xy)=m+n}$

We now resubstitute the values for

${displaystyle m}$

and

${displaystyle n}$

into our equation, so our final expression is only in terms of

${displaystyle x}$

,

${displaystyle y}$

, and

${displaystyle b}$

.

${displaystyle log _{b}(xy)=log _{b}(x)+log _{b}(y)}$

This completes the derivation.

Logarithm of a quotient[edit]

To state the logarithm of a quotient law formally:

${displaystyle forall bin mathbb {R} _{+},bneq 1,forall x,y,in mathbb {R} _{+},log _{b}left({frac {x}{y}}right)=log _{b}(x)-log _{b}(y)}$

Derivation:

Let

${displaystyle bin mathbb {R} _{+}}$

, where

${displaystyle bneq 1}$

,
and let

${displaystyle x,yin mathbb {R} _{+}}$

.

We want to relate the expressions

${displaystyle log _{b}(x)}$

and

${displaystyle log _{b}(y)}$

. This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to

${displaystyle log _{b}(x)}$

and

${displaystyle log _{b}(y)}$

quite often, we will give them some variable names to make working with them easier: Let

${displaystyle m=log _{b}(x)}$

, and let

${displaystyle n=log _{b}(y)}$

.

Rewriting these as exponentials, we see that:

${displaystyle m=log _{b}(x)iff b^{m}=x}$

and

${displaystyle n=log _{b}(y)iff b^{n}=y}$

. From here, we can relate

${displaystyle b^{m}}$

(i.e.

${displaystyle x}$

) and

${displaystyle b^{n}}$

(i.e.

${displaystyle y}$

) using exponent laws as

${displaystyle {frac {x}{y}}={frac {(b^{m})}{(b^{n})}}={frac {b^{m}}{b^{n}}}=b^{m-n}}$

To recover the logarithms, we apply

${displaystyle log _{b}}$

to both sides of the equality.

${displaystyle log _{b}left({frac {x}{y}}right)=log _{b}left(b^{m-n}right)}$

The right side may be simplified using one of the logarithm properties from before: we know that

${displaystyle log _{b}(b^{m-n})=m-n}$

, giving

${displaystyle log _{b}left({frac {x}{y}}right)=m-n}$

We now resubstitute the values for

${displaystyle m}$

and

${displaystyle n}$

into our equation, so our final expression is only in terms of

${displaystyle x}$

,

${displaystyle y}$

, and

${displaystyle b}$

.

${displaystyle log _{b}left({frac {x}{y}}right)=log _{b}(x)-log _{b}(y)}$

This completes the derivation.

Logarithm of a power[edit]

To state the logarithm of a power law formally,

${displaystyle forall bin mathbb {R} _{+},bneq 1,forall xin mathbb {R} _{+},forall rin mathbb {R} ,log _{b}(x^{r})=rlog _{b}(x)}$

Derivation:

Let

${displaystyle bin mathbb {R} _{+}}$

, where

${displaystyle bneq 1}$

, let

${displaystyle xin mathbb {R} _{+}}$

, and let

${displaystyle rin mathbb {R} }$

. For this derivation, we want to simplify the expression

${displaystyle log _{b}(x^{r})}$

. To do this, we begin with the simpler expression

${displaystyle log _{b}(x)}$

. Since we will be using

${displaystyle log _{b}(x)}$

often, we will define it as a new variable: Let

${displaystyle m=log _{b}(x)}$

.

To more easily manipulate the expression, we rewrite it as an exponential. By definition,

${displaystyle m=log _{b}(x)iff b^{m}=x}$

, so we have

${displaystyle b^{m}=x}$

Similar to the derivations above, we take advantage of another exponent law. In order to have

${displaystyle x^{r}}$

in our final expression, we raise both sides of the equality to the power of

${displaystyle r}$

:

${displaystyle {begin{aligned}(b^{m})^{r}&=(x)^{r}\b^{mr}&=x^{r}end{aligned}}}$

where we used the exponent law

${displaystyle (b^{m})^{r}=b^{mr}}$

.

To recover the logarithms, we apply

${displaystyle log _{b}}$

to both sides of the equality.

${displaystyle log _{b}(b^{mr})=log _{b}(x^{r})}$

The left side of the equality can be simplified using a logarithm law, which states that

${displaystyle log _{b}(b^{mr})=mr}$

.

${displaystyle mr=log _{b}(x^{r})}$

Substituting in the original value for

${displaystyle m}$

, rearranging, and simplifying gives

${displaystyle {begin{aligned}left(log _{b}(x)right)r&=log _{b}(x^{r})\rlog _{b}(x)&=log _{b}(x^{r})\log _{b}(x^{r})&=rlog _{b}(x)end{aligned}}}$

This completes the derivation.

Changing the base[edit]

To state the change of base logarithm formula formally:

This identity is useful to evaluate logarithms on calculators. For instance, most calculators have buttons for ln and for log₁₀, but not all calculators have buttons for the logarithm of an arbitrary base.

Proof/derivation[edit]

Let

${displaystyle a,bin mathbb {R} _{+}}$

, where

${displaystyle a,bneq 1}$

Let

${displaystyle xin mathbb {R} _{+}}$

. Here,

${displaystyle a}$

and

${displaystyle b}$

are the two bases we will be using for the logarithms. They cannot be 1, because the logarithm function is not well defined for the base of 1.^{[citation needed]} The number

${displaystyle x}$

will be what the logarithm is evaluating, so it must be a positive number. Since we will be dealing with the term

${displaystyle log _{b}(x)}$

quite frequently, we define it as a new variable: Let

${displaystyle m=log _{b}(x)}$

.

To more easily manipulate the expression, it can be rewritten as an exponential.

Applying

${displaystyle log _{a}}$

to both sides of the equality,

Now, using the logarithm of a power property, which states that

${displaystyle log _{a}(b^{m})=mlog _{a}(b)}$

,

Isolating

${displaystyle m}$

, we get the following:

Resubstituting

${displaystyle m=log _{b}(x)}$

back into the equation,

This completes the proof that

${displaystyle log _{b}(x)={frac {log _{a}(x)}{log _{a}(b)}}}$

.

This formula has several consequences:

where

${textstyle pi }$

is any permutation of the subscripts 1, …, n. For example

Summation/subtraction[edit]

The following summation/subtraction rule is especially useful in probability theory when one is dealing with a sum of log-probabilities:

${displaystyle log _{b}(a+c)=log _{b}a+log _{b}left(1+{frac {c}{a}}right)}$	because	${displaystyle left(a+cright)=atimes left(1+{frac {c}{a}}right)}$
${displaystyle log _{b}(a-c)=log _{b}a+log _{b}left(1-{frac {c}{a}}right)}$	because	${displaystyle left(a-cright)=atimes left(1-{frac {c}{a}}right)}$

Note that the subtraction identity is not defined if

${displaystyle a=c}$

, since the logarithm of zero is not defined. Also note that, when programming,

${displaystyle a}$

and

${displaystyle c}$

may have to be switched on the right hand side of the equations if

${displaystyle cgg a}$

to avoid losing the “1 +” due to rounding errors. Many programming languages have a specific log1p(x) function that calculates

${displaystyle log _{e}(1+x)}$

without underflow (when

${displaystyle x}$

is small).

More generally:

Exponents[edit]

A useful identity involving exponents:

or more universally:

Other/resulting identities[edit]

Inequalities[edit]

Based on,^[4][5] and ^[6]

${displaystyle {frac {x}{1+x}}leq ln(1+x)leq {frac {x(6+x)}{6+4x}}leq x{mbox{ for all }}{-1}$

${displaystyle {begin{aligned}{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 }}0leq x{text{, reverse for }}{-1}$

All are accurate around

${displaystyle x=0}$

, but not for large numbers.

Calculus identities[edit]

Limits[edit]

${displaystyle lim _{xto 0^{+}}log _{a}(x)=-infty quad {mbox{if }}a>1}$

${displaystyle lim _{xto 0^{+}}log _{a}(x)=infty quad {mbox{if }}0$

${displaystyle lim _{xto infty }log _{a}(x)=infty quad {mbox{if }}a>1}$

${displaystyle lim _{xto infty }log _{a}(x)=-infty quad {mbox{if }}0$

${displaystyle lim _{xto 0^{+}}x^{b}log _{a}(x)=0quad {mbox{if }}b>0}$

${displaystyle lim _{xto infty }{frac {log _{a}(x)}{x^{b}}}=0quad {mbox{if }}b>0}$

Derivatives of logarithmic functions[edit]

${displaystyle {d over dx}ln x={1 over x},x>0}$

${displaystyle {d over dx}ln |x|={1 over x},xneq 0}$

${displaystyle {d over dx}log _{a}x={1 over xln a},x>0,a>0,{text{ and }}aneq 1}$

Integral definition[edit]

${displaystyle ln x=int _{1}^{x}{frac {1}{t}} dt}$

Integrals of logarithmic functions[edit]

${displaystyle int ln x,dx=xln x-x+C=x(ln x-1)+C}$

${displaystyle int log _{a}x,dx=xlog _{a}x-{frac {x}{ln a}}+C={frac {x(ln x-1)}{ln a}}+C}$

To remember higher integrals, it is convenient to define

${displaystyle x^{left[nright]}=x^{n}(log(x)-H_{n})}$

where

${displaystyle H_{n}}$

is the n^thharmonic number:

${displaystyle x^{left[0right]}=log x}$

${displaystyle x^{left[1right]}=xlog(x)-x}$

${displaystyle x^{left[2right]}=x^{2}log(x)-{begin{matrix}{frac {3}{2}}end{matrix}}x^{2}}$

${displaystyle x^{left[3right]}=x^{3}log(x)-{begin{matrix}{frac {11}{6}}end{matrix}}x^{3}}$

Then

${displaystyle {frac {d}{dx}},x^{left[nright]}=nx^{left[n-1right]}}$

${displaystyle int x^{left[nright]},dx={frac {x^{left[n+1right]}}{n+1}}+C}$

Approximating large numbers[edit]

The identities of logarithms can be used to approximate large numbers. Note that log_b(a) + log_b(c) = log_b(ac), where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, 2^32,582,657 −1. To get the base-10 logarithm, we would multiply 32,582,657 by log₁₀(2), getting 9,808,357.09543 = 9,808,357 + 0.09543. We can then get 10^9,808,357 × 10^0.09543 ≈ 1.25 × 10^9,808,357.

Similarly, factorials can be approximated by summing the logarithms of the terms.

Complex logarithm identities[edit]

The complex logarithm is the complex number analogue of the logarithm function. No single valued function on the complex plane can satisfy the normal rules for logarithms. However, a multivalued function can be defined which satisfies most of the identities. It is usual to consider this as a function defined on a Riemann surface. A single valued version, called the principal value of the logarithm, can be defined which is discontinuous on the negative x axis, and is equal to the multivalued version on a single branch cut.

Definitions[edit]

In what follows, a capital first letter is used for the principal value of functions, and the lower case version is used for the multivalued function. The single valued version of definitions and identities is always given first, followed by a separate section for the multiple valued versions.

• ln(r) is the standard natural logarithm of the real number r.
• Arg(z) is the principal value of the arg function; its value is restricted to (−π, π]. It can be computed using Arg(x + iy) = atan2(y, x).
• Log(z) is the principal value of the complex logarithm function and has imaginary part in the range (−π, π].
• 
  ${displaystyle operatorname {Log} (z)=ln(|z|)+ioperatorname {Arg} (z)}$

  

• 
  ${displaystyle e^{operatorname {Log} (z)}=z}$

  

The multiple valued version of log(z) is a set, but it is easier to write it without braces and using it in formulas follows obvious rules.

• log(z) is the set of complex numbers v which satisfy e^v = z
• arg(z) is the set of possible values of the arg function applied to z.

When k is any integer:

${displaystyle log(z)=ln(|z|)+iarg(z)}$

${displaystyle log(z)=operatorname {Log} (z)+2pi ik}$

${displaystyle e^{log(z)}=z}$

Constants[edit]

Principal value forms:

${displaystyle operatorname {Log} (1)=0}$

${displaystyle operatorname {Log} (e)=1}$

Multiple value forms, for any k an integer:

${displaystyle log(1)=0+2pi ik}$

${displaystyle log(e)=1+2pi ik}$

Summation[edit]

Principal value forms:

${displaystyle operatorname {Log} (z_{1})+operatorname {Log} (z_{2})=operatorname {Log} (z_{1}z_{2}){pmod {2pi i}}}$

${displaystyle operatorname {Log} (z_{1})+operatorname {Log} (z_{2})=operatorname {Log} (z_{1}z_{2})quad (-pi 0)}$

[7]

${displaystyle operatorname {Log} (z_{1})-operatorname {Log} (z_{2})=operatorname {Log} (z_{1}/z_{2}){pmod {2pi i}}}$

${displaystyle operatorname {Log} (z_{1})-operatorname {Log} (z_{2})=operatorname {Log} (z_{1}/z_{2})quad (-pi 0)}$

[7]

Multiple value forms:

${displaystyle log(z_{1})+log(z_{2})=log(z_{1}z_{2})}$

${displaystyle log(z_{1})-log(z_{2})=log(z_{1}/z_{2})}$

Powers[edit]

A complex power of a complex number can have many possible values.

Principal value form:

${displaystyle {z_{1}}^{z_{2}}=e^{z_{2}operatorname {Log} (z_{1})}}$

${displaystyle operatorname {Log} {left({z_{1}}^{z_{2}}right)}=z_{2}operatorname {Log} (z_{1}){pmod {2pi i}}}$

Multiple value forms:

${displaystyle {z_{1}}^{z_{2}}=e^{z_{2}log(z_{1})}}$

Where k₁, k₂ are any integers:

${displaystyle log {left({z_{1}}^{z_{2}}right)}=z_{2}log(z_{1})+2pi ik_{2}}$

${displaystyle log {left({z_{1}}^{z_{2}}right)}=z_{2}operatorname {Log} (z_{1})+z_{2}2pi ik_{1}+2pi ik_{2}}$

See also[edit]

References[edit]

External links[edit]