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In mathematics, there are several logarithmic identities.

Algebraic identities


Using simpler operations

People use logarithms to make calculations easier. For instance, 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= \begin{matrix}\frac{\log_b(x)}{y}\end{matrix} because \sqrt[y{x} = x^{1/y}

Cancelling exponentials

Logarithms and exponentials (antilogarithms) with the same base cancel each other.
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 \!\,

Keep in mind that these are true only for real x; the logarithm is infinite-valued on the complex plane, and thus the complex exponential is not invertable.

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).

This formula has several consequences:

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

\log_{a^n} b = \frac{1}{n} \log_a b

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

Logarithmic "snake" identity

Any sequence

\!\ \log_a(b) \cdot \log_b(c) \cdot \log_c(d)\cdots \log_y(z)

can be simplified to...

\!\ \log_a(z)

This leads to the identity

\log_a b = {\log_c b \over \log_c a}.\,

Trivial identities

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

\log_b(0) is undefined because there is no number x such that b^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} \log_a x = {1 \over x \ln a} = {\log_a e \over x }

Integral definition

\log_e 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)
where H_n is the nth harmonic number. So, for example, the first few are:
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

Logarithms | Mathematical identities

Identidades logarítmicas | identités logarithmiques | Identità sui logaritmi | логаритамске једначине

 

This article is licensed under the GNU Free Documentation License. It uses material from the "Logarithmic identities".

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