List of logarithmic identities

By definition, we know that:

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

where ${\displaystyle \color {blue}b\color {black}\neq 0}$ or ${\displaystyle \color {blue}b\color {black}\neq 1}$.

Setting ${\displaystyle \color {red}x\color {black}=0}$, we can see that: ${\displaystyle \color {blue}b\color {black}\color {red}^{x}\color {black}=\color {green}y\color {black}\iff \color {blue}b\color {black}\color {red}^{(0)}\color {black}=\color {green}y\color {black}\iff \color {blue}1\color {black}=\color {green}y\color {black}\iff \color {green}y\color {black}=\color {blue}1\color {black}}$. So, substituting these values into the formula, we see that: ${\displaystyle \color {black}\log \color {blue}_{b}\color {black}(\color {green}y\color {black})=\color {red}x\color {black}\iff \color {black}\log \color {blue}_{b}\color {black}(\color {blue}1\color {black})=\color {red}0\color {black}}$ , which gets us the first property.

Setting ${\displaystyle \color {red}x\color {black}=1}$, we can see that: ${\displaystyle \color {blue}b\color {black}\color {red}^{x}\color {black}=\color {green}y\color {black}\iff \color {blue}b\color {black}\color {red}^{(1)}\color {black}=\color {green}y\color {black}\iff \color {blue}b\color {black}=\color {green}y\color {black}\iff \color {green}y\color {black}=\color {blue}b\color {black}}$. So, substituting these values into the formula, we see that: ${\displaystyle \color {black}\log \color {blue}_{b}\color {black}(\color {green}y\color {black})=\color {red}x\color {black}\iff \color {black}\log \color {blue}_{b}\color {black}(\color {blue}b\color {black})=\color {red}1\color {black}}$ , 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.

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.

To state the logarithm of a product law formally:

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

Derivation:

Let ${\displaystyle b\in \mathbb {R} _{+}}$, where ${\displaystyle b\neq 1}$, and let ${\displaystyle x,y\in \mathbb {R} _{+}}$. 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

${\displaystyle {\begin{aligned}m&=\log _{b}(x)\iff b^{m}=x,\\n&=\log _{b}(y)\iff b^{n}=y.\end{aligned}}}$

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.

To state the logarithm of a quotient law formally:

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

Derivation:

Let ${\displaystyle b\in \mathbb {R} _{+}}$, where ${\displaystyle b\neq 1}$, and let ${\displaystyle x,y\in \mathbb {R} _{+}}$.

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:

${\displaystyle {\begin{aligned}m&=\log _{b}(x)\iff b^{m}=x,\\n&=\log _{b}(y)\iff b^{n}=y.\end{aligned}}}$

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.

To state the logarithm of a power law formally,

${\displaystyle \forall b\in \mathbb {R} _{+},b\neq 1,\forall x\in \mathbb {R} _{+},\forall r\in \mathbb {R} ,\log _{b}(x^{r})=r\log _{b}(x)}$

Derivation:

Let ${\displaystyle b\in \mathbb {R} _{+}}$, where ${\displaystyle b\neq 1}$, let ${\displaystyle x\in \mathbb {R} _{+}}$, and let ${\displaystyle r\in \mathbb {R} }$. 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

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

This completes the derivation.

Let ${\displaystyle a,b\in \mathbb {R} _{+}}$, where ${\displaystyle a,b\neq 1}$ Let ${\displaystyle x\in \mathbb {R} _{+}}$. 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. 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.

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

Applying ${\displaystyle \log _{a}}$ to both sides of the equality,

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

Now, using the logarithm of a power property, which states that ${\displaystyle \log _{a}(b^{m})=m\log _{a}(b)}$,

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

Isolating ${\displaystyle m}$, we get the following:

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

Resubstituting ${\displaystyle m=\log _{b}(x)}$ back into the equation,

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

This completes the proof that ${\displaystyle \log _{b}(x)={\frac {\log _{a}(x)}{\log _{a}(b)}}}$.

This formula has several consequences:

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

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

$${\displaystyle b^{\log _{a}d}=d^{\log _{a}b}}$$

$${\displaystyle -\log _{b}a=\log _{b}\left({1 \over a}\right)=\log _{1/b}a}$$

$${\displaystyle \log _{b_{1}}a_{1}\,\cdots \,\log _{b_{n}}a_{n}=\log _{b_{\pi (1)}}a_{1}\,\cdots \,\log _{b_{\pi (n)}}a_{n},}$$

where ${\textstyle \pi }$ is any permutation of the subscripts 1, ..., n. For example

$${\displaystyle \log _{b}w\cdot \log _{a}x\cdot \log _{d}c\cdot \log _{d}z=\log _{d}w\cdot \log _{b}x\cdot \log _{a}c\cdot \log _{d}z.}$$

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

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 ${\displaystyle c\gg a}$ 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:

$${\displaystyle \log _{b}\sum _{i=0}^{N}a_{i}=\log _{b}a_{0}+\log _{b}\left(1+\sum _{i=1}^{N}{\frac {a_{i}}{a_{0}}}\right)=\log _{b}a_{0}+\log _{b}\left(1+\sum _{i=1}^{N}b^{\left(\log _{b}a_{i}-\log _{b}a_{0}\right)}\right)}$$

A useful identity involving exponents:

$${\displaystyle x^{\frac {\log(\log(x))}{\log(x)}}=\log(x)}$$

or more universally:

$${\displaystyle x^{\frac {\log(a)}{\log(x)}}=a}$$

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

${\displaystyle \lim _{x\to 0^{+}}\log _{a}(x)=\infty \quad {\mbox{if }}0<a<1}$

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

${\displaystyle \lim _{x\to \infty }\log _{a}(x)=-\infty \quad {\mbox{if }}0<a<1}$

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

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

The last limit is often summarized as "logarithms grow more slowly than any power or root of x".

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

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

${\displaystyle {d \over dx}\log _{a}x={1 \over x\ln a},x>0,a>0,{\text{ and }}a\neq 1}$

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

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

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

To remember higher integrals, it is convenient to define

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

where ${\displaystyle H_{n}}$ is the n^th harmonic number:

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

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

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

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

Then

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

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

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].
• ${\displaystyle \operatorname {Log} (z)=\ln(|z|)+i\operatorname {Arg} (z)}$
• ${\displaystyle e^{\operatorname {Log} (z)}=z}$

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:

${\displaystyle \log(z)=\ln(|z|)+i\arg(z)}$

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

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

Principal value forms:

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

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

Multiple value forms, for any k an integer:

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

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

Principal value forms:

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

${\displaystyle \operatorname {Log} (z_{1})+\operatorname {Log} (z_{2})=\operatorname {Log} (z_{1}z_{2})\quad (-\pi <\operatorname {Arg} (z_{1})+\operatorname {Arg} (z_{2})\leq \pi$ ;{\text{ e.g., }}\operatorname {Re} z_{1}\geq 0{\text{ and }}\operatorname {Re} z_{2}>0)} [7]

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

${\displaystyle \operatorname {Log} (z_{1})-\operatorname {Log} (z_{2})=\operatorname {Log} (z_{1}/z_{2})\quad (-\pi <\operatorname {Arg} (z_{1})-\operatorname {Arg} (z_{2})\leq \pi$ ;{\text{ e.g., }}\operatorname {Re} z_{1}\geq 0{\text{ and }}\operatorname {Re} z_{2}>0)} [7]

Multiple value forms:

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

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

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

Principal value form:

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

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

Multiple value forms:

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

Where k₁, k₂ are any integers:

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

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