List of Logarithmic Identities

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Algebraic identities or laws

Using simpler operations

Logarithms can be used to make calculations easier. For example, two numbers can be multiplied just by using a logarithm table and adding.

$\log_b(xy) = \log_b(x) + \log_b(y) \!\,$	because	$b^c \cdot b^d = b^{c + d} \!\,$
$\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y)$	because	$b^{c-d} = \tfrac{b^c}{b^d}$
$\log_b(x^d) = d \log_b(x) \!\,$	because	$(b^c)^d = b^{cd} \!\,$
$\log_b\!\left(\!\sqrt[y]{x}\right) = \begin{matrix}\frac{\log_b(x)}{y}\end{matrix}$	because	$\sqrt[y]{x} = x^{1/y}$
$x^{\log_b(y)} = y^{\log_b(x)} \!\,$	because	$x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)} \!\,$
$c\log_b(x)+d\log_b(y) = \log_b(x^c y^d) \!\,$	because	$\log_b(x^c y^d) = \log_b(x^c) + \log_b(y^d) \!\,$

Where $b$ , $x$ , and $y$ are positive real numbers and $b \ne 1$ . Both $c$ and $d$ are real numbers.

Trivial identities

$\log_b(1) = 0 \!\,$	because	$b^0 = 1\!\,$
$\log_b(b) = 1 \!\,$	because	$b^1 = b\!\,$

Note that $\log_b(0) \!\,$ is undefined because there is no number $x \!\,$ such that $b^x = 0 \!\,$ . In fact, there is a vertical asymptote on the graph of $\log_b(x) \!\,$ at $x = 0$ .

Canceling exponentials

Logarithms and exponentials (antilogarithms) with the same base cancel each other. This is true because logarithms and exponentials are inverse operations (just like multiplication and division).

$b^{\log_b(x)} = x$	because	$\mathrm{antilog}_b(\log_b(x)) = x \!\,$
$\log_b(b^x) = x \!\,$	because	$\log_b(\mathrm{antilog}_b(x)) = x \!\,$

Changing the base

$\log_b a = {\log_d a \over \log_d b}$

This identity is needed to evaluate logarithms on calculators. For instance, most calculators have buttons for ln and for log₁₀, but not for log₂. To find log₂(3), one must calculate log₁₀(3) / log₁₀(2) (or ln(3)/ln(2), which yields the same result).

Proof

Let

c = log b a

.

Then

b c = a

.

Take

log d

on both sides:

log d b c = log d a

Simplify and solve for

c

:

c log d b = log d a

$c=\frac{\log_d a}{\log_d b}$

Since

c = log b a

, than $\log_b a=\frac{\log_d a}{\log_d b}$

This formula has several consequences:

$\log_b a = \frac {1} {\log_a b}$

$\log_{b^n} a = {{\log_b a} \over n}$

$b^{\log_a d} = d^{\log_a b}$

$- \log_b a = \log_b \left({1 \over a}\right) = \log_{1 \over b} a$

$\log_{b_1}a_1 \,\cdots\, \log_{b_n}a_n= \log_{b_{\pi(1)}}a_1\, \cdots\, \log_{b_{\pi(n)}}a_n, \,$

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

$\log_b w\cdot \log_a x\cdot \log_d c\cdot \log_d z= \log_d w\cdot \log_b x\cdot \log_a c\cdot \log_d z. \,$

Summation/subtraction

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

$\log_b (a+c) = \log_b a + \log_b (1+b^{\log_b c - \log_b a})$

$\log_b (a-c) = \log_b a + \log_b (1-b^{\log_b c - \log_b a})$

which gives the special cases:

$\log_b (a+c) = \log_b a + \log_b (1+\frac{c}{a})$

$\log_b (a-c) = \log_b a + \log_b (1-\frac{c}{a})$

Note that in practice $a$ and $c$ have to be switched on the right hand side of the equations if $c > a$ . Also note that the subtraction identity is not defined if $a = c$ since the logarithm of zero is not defined.

More generally:

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