User:Guy vandegrift/sandbox

Alternative proof
We begin with an equation that relates two exponentials involving two bases, $\{a,b\}$, that are positive and not equal to one:
 * $b^y=a^z$,

The first step is to randomly select and solve for either $a$ or $b$. If $b$ is selected:
 * $b=a^{z/y}$

Next we take the logarithm of both sides, randomly taking the logarithm's base to be either $a$ or $b$. If $a$ is selected:

Now define, $x=b^y=a^z$, and note that: Substituting these values of $y$ and $z$ into our expression, $\log_a b=z/y$, yields the desired version of the change-of-base formula:
 * $\log_a b= \dfrac{\log_a x}{\log_b x}\Longrightarrow $$\boxed{\log_b x = \dfrac{\log_a x}{\log_a b}}$

If other choices between $a$ and $b$ are made
This proof achieved the desired result only because two consecutive decisions were made regarding the selection of $a$ or $b$. Fortunately, there is little reason to remember these choices, provided the goal is to derive a useful formula. Suppose, for example, that we chose to solve for $b=a^{z/y}$, but instead take the logarithm of both sides at base $b$ instead of $a$. The result would be slightly different change-of-base formula:
 * $\log_b b=1=\log_b\left(a^{z/y}\right)= \dfrac{\log_a x}{\log_b x}\log_b a$$\Longrightarrow \boxed{\log_b x = (\log_b a)\cdot(\log_a x)}$

Since both calculations are valid, we conclude that
 * $\boxed{\left(\log_a b\right)\cdot\left(\log_b a\right)=1}$.

To understand why this identity is important, imagine that you are converting between base-2 and base-10, and have a numerical value for $\log_{10}2$, while attempting to use a change-of-base formula that requires a numerical value of $\log_2 10.$

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