User:Guy vandegrift/sandbox

Alternative proof
Though more tedious than the aforementioned proof, this discussion is more thorough. It involves a series of steps that force a decision between equally promising paths. All paths lead to a useful result, and comparing these results produces an important lemma. If you remember that the logarithm is the exponent, you will guess that a good starting point is:
 * $b^y=a^z$,
 * where:
 * $y$ equals the logarithm of something that involves $b$ as the base.
 * $z$ equals the logarithm of something that involves $z$ as the base.

Before introducing logarithms, we first express one base in terms of the other. If the base is $b$:
 * $b=a^{z/y}$

Next, randomly select a base and take the logarithm of both sides. Here, we choose $a$ for as the logarithm's base:

Now define, $x=b^y=a^z=x$, and recall that

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 * $b^y = x \Rightarrow y = \log_b x,$
 * $a^z = x \Rightarrow z = \log_a x.$

Definition:General_Logarithm

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Substituting these values of $y$ and $z$ into our expression, $\log_a b=z/y$, yields a useful result:
 * $\log_a b= \dfrac{\log_a x}{\log_b x}\Rightarrow \log_b x = \dfrac{\log_a x}{\log_a b}$

Lemma
We get a slightly different result if we instead choose base $b$ for the logarithm:
 * $\log_b b=1=\log_b\left(a^{z/y}\right)= \dfrac{\log_a x}{\log_b x}\log_b a\Rightarrow \log_b x = (\log_b a)\cdot(\log_a x)$

The fact that different paths lead to different expressions leads us to an important identity:
 * $\left(\log_a b\right)\cdot\left(\log_b a\right)=1$.

map
$\map {\log_a} {b^y}$

Archives

 * Special:Permalink/551517 What is physics joke (failed to parse)
 * Special:Permalink/551563 and 551651 Change of Base of Logarithm