Editing List of logarithmic identities (section)

=== 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>