Editing List of logarithmic identities (section)

== Changing the base ==

To state the change of base logarithm formula formally:
<math display="block">\forall a, b \in \mathbb{R}_+, a, b \neq 1,  \forall x \in \mathbb{R}_+, \log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>

This identity is useful to evaluate logarithms on calculators. For instance, most calculators have buttons for [[Natural logarithm|ln]] and for [[Common logarithm|log<sub>10</sub>]], but not all calculators have buttons for the logarithm of an arbitrary base.

=== Proof/derivation ===

Let <math>a, b \in \mathbb{R}_+</math>, where <math>a, b \neq 1</math> Let <math>x \in \mathbb{R}_+</math>. Here, <math>a</math> and <math>b</math> 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.{{Citation needed|date=July 2022}} The number <math>x</math> will be what the logarithm is evaluating, so it must be a positive number. Since we will be dealing with the term <math>\log_b(x)</math> quite frequently, we define it as a new variable: Let <math>m = \log_b(x)</math>.

To more easily manipulate the expression, it can be rewritten as an exponential.
<math display="block">b^m = x </math>

Applying <math>\log_a</math> to both sides of the equality,
<math display="block">\log_a(b^m) = \log_a(x) </math>

Now, using the logarithm of a power property, which states that <math>\log_a(b^m) = m\log_a(b)</math>,
<math display="block">m\log_a(b) = \log_a(x)</math>

Isolating <math>m</math>, we get the following:
<math display="block">m = \frac{\log_a(x)}{\log_a(b)}</math>

Resubstituting <math>m = \log_b(x)</math> back into the equation,
<math display="block">\log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>

This completes the proof that <math>\log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>.

This formula has several consequences:

<math display="block"> \log_b a = \frac 1 {\log_a b} </math>

<math display="block"> \log_{b^n} a =  {\log_b a \over n} </math>

<math display="block"> \log_{b} a =  \log_b e \cdot \log_e a = \log_b e \cdot \ln a </math>

<math display="block"> b^{\log_a d} = d^{\log_a b} </math>

<math display="block"> -\log_b a = \log_b \left({1 \over a}\right) = \log_{1/b} a</math>

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<math display="block"> \log_{b_1}a_1 \,\cdots\, \log_{b_n}a_n
= \log_{b_{\pi(1)}}a_1\, \cdots\, \log_{b_{\pi(n)}}a_n, </math>

where <math display="inline">\pi</math> is any [[permutation]] of the subscripts {{math|1, ..., ''n''}}.  For example
<math display="block"> \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. </math>

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

{|
|<math>\log_b (a+c) = \log_b a + \log_b \left(1 + \frac{c}{a}\right)</math>
|because
|<math>\left(a + c \right) = a \times \left(1 + \frac{c}{a} \right)</math>
|-
|<math>\log_b (a-c) = \log_b a + \log_b \left(1 - \frac{c}{a}\right)</math>
|because
|<math>\left(a - c \right) = a \times \left(1 - \frac{c}{a} \right)</math>
|}

Note that the subtraction identity is not defined if <math>a=c</math>, since the logarithm of zero is not defined. Also note that, when programming, <math>a</math> and <math>c</math> may have to be switched on the right hand side of the equations if <math>c \gg a</math> to avoid losing the "1 +" due to rounding errors.  Many programming languages have a specific <code>log1p(x)</code> function that calculates <math>\log_e (1+x)</math> without underflow (when <math>x</math> is small).

More generally:
<math display="block">\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)</math>

=== Exponents ===
A useful identity involving exponents:
<math display="block"> x^{\frac{\log(\log(x))}{\log(x)}} = \log(x)</math>
or more universally:
<math display="block"> x^{\frac{\log(a)}{\log(x)}} = a</math>

=== Other/resulting identities ===
<math display="block"> \frac{1}{\frac{1}{\log_x(a)} + \frac{1}{\log_y(a)}} = \log_{xy}(a)</math>
<math display="block"> \frac{1}{\frac{1}{\log_x(a)}-\frac{1}{\log_y(a)}} = \log_{\frac{x}{y}}(a)</math>