List_of_logarithmic

List of logarithmic identities

In mathematics, there are several logarithmic identities.

1 Algebraic identities
2 Calculus identities
3 Approximating large numbers
4 References

Algebraic identities

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	$b^x \cdot b^y = b^{x + y} \!\,$
$\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y)$	because	$\begin{matrix}\frac{b^x}{b^y}\end{matrix} = b^{x - y}$
$\log_b(x^y) = y \log_b(x) \!\,$	because	$(b^x)^y = b^{xy} \!\,$
$\log_b\!\left(\!\sqrt[y]{x}\right) = \begin{matrix}\frac{\log_b(x)}{y}\end{matrix}$	because	$\sqrt[y]{x} = x^{1/y}$
$x^{\log(y)} = y^{\log(x)} \!\,$	because	$x^{\log(y)} = e^{\log(x)^{\log(y)}} = e^{\log(y)^{\log(x)}} = y^{\log(x)} \!\,$
$\log_b(-x) = \log_b(x) + i\pi\log_b(e)\!\,$	because	Euler's identity： $e^{i\pi} + 1 = 0 \,\!$

Where $b$ , $x$ and $x y$ are positive real numbers. If $x y$ is positive, but $x$ is not, then $\log_b(x^y) = y \log_b(-x) \!\,$ .

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

$\log_a b = {\log_c b \over \log_c a}$

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), you have to calculate log₁₀(3) / log₁₀(2) (or ln(3)/ln(2), which is the same thing).

Proof

Let $y = log a b$ .

Then $a y = b$ .

Take $log c$ on both sides: $log c a y = log c b$

Simplify and solve for $y$ : $y log c a = log c b$

$y=\frac{\log_c b}{\log_c a}$

Since $y = log a b$ , then $\log_a b=\frac{\log_c b}{\log_c a}$

This formula has several consequences:

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

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

$a^{\log_b c} = c^{\log_b a}$

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

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

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

Changing the base (continued)

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

- log a b = log a (1 / b) = log (1 / a) b

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.

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 $\log_b(x) \!\,$ when $x = 0$ .

Calculus identities

Limits

$\lim_{x \to 0^+} \log_a x = -\infty \quad \mbox{if } a > 1$

$\lim_{x \to 0^+} \log_a x = \infty \quad \mbox{if } a < 1$

$\lim_{x \to \infty} \log_a x = \infty \quad \mbox{if } a > 1$

$\lim_{x \to \infty} \log_a x = -\infty \quad \mbox{if } a < 1$

$\lim_{x \to 0^+} x^b \log_a x = 0$

$\lim_{x \to \infty} {1 \over x^b} \log_a x = 0$

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

note: to say the limit of a function "equals infinity" is not strictly correct notation, as "infinity" is not a value. What is meant by the limits equations above is simply that the functions increase/decrease without bound.

Derivatives of logarithmic functions

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

${d \over dx} \log_b x = {1 \over x \ln b},\qquad b > 0, b \ne 1$

Integral definition

$\ln x = \int_1^x \frac {1}{t} dt$

Integrals of logarithmic functions

$\int \log_a x \, dx = x(\log_a x - \log_a e) + C$

To remember higher integrals, it's convenient to define:

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

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

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

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

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

Then,

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

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

Approximating large numbers

The identities of logarithms can be used to approximate large numbers. Note that log_b(a) + log_b(c) = log_b(a*c), 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.

References

This article does not cite any references or sources.
Please help improve this article by adding citations to reliable sources. Unverifiable material may be challenged and removed. (October 2007)

Contents