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Logarithmic identities

In mathematics, there are several logarithmic identities.

Contents

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^m \cdot b^n = b^{m + n}
\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y) because \begin{matrix}\frac{b^m}{b^n}\end{matrix} = b^{m - n}
\log_b(x^y) = y \log_b(x) \!\, because (b^n)^y = b^{ny} \!\,
\log_b\!\left(\!\sqrt[y]{x}\right) = \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 \!\,

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}

Trivial identities

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

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".

Derivatives of logarithmic functions

{d \over dx} \log_a x = {1 \over x \ln a} = {\log_a e \over x }

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 Hn 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{1}{2} \end{matrix} \, x^2
x^{\left [ 3 \right ]} = x^3 \log(x) - \begin{matrix} \frac{3}{4} \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} + C

01-04-2007 01:32:10
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