Talk:Change of Base of Logarithm

A more pedagogical approach to the change of base formula

What I wrote on Wikiverity is shown below. I don't think it should replace the concise and elegant argument on Change_of_Base_of_Logarithm. Nevertheless, my calculation has some pedagogical advantages that may or may not fall within the intended scope of pr∞wiki. If you want me to put this somewhere on proofwiki, let me know.--Guy vandegrift (talk) 21:34, 18 December 2021 (UTC)

Derivations are easier to remember if they do not require identities that may see obscure to some readers. Above all, each step must seem obvious. Otherwise, one can find themselves going in circles (which often happened to me when I tried to derive the change of base formula myself.)

The following derivation relies on definitions that emerge naturally if we use the exponential, $y=b^n$, to define the exponent, $n$, as the logarithm. In other words, we base our derivation the following fundamental formulas:

$y=B^N=b^n \Rightarrow N=\log_B(y) \text{ and } n=\log_b(y).$

The advantage of this emphasis on exponents is that students can be easily convinced of two identities required in our derivation: $b^{n+m}=b^nb^m$ and $b^{kn}=\left(b^n\right)^k$. Our starting point is easy to remember because it is symmetrical and involves only exponents (which are presumably more familiar to students):

$b^n=B^N$.

Next, we randomly select one of the two bases and solve:

$b=B^{N/n}$

Now take the logarithm of both sides. It makes no difference whether the base of this logarithm is $B$ or $b$. For example if we take $log_B$, we obtain:

$\log_B(b)=\log_B(B^{N/n})=\dfrac N n \log_B (B)=\dfrac N n = \dfrac{\log_B(y)}{\log_b(y)}$

Equating the first and terms in this sequence of equalities yields our desired result,

$\log_b(y)=\dfrac{\log_B(y)}{\log_B(b)}$

We get an equivalent result the other base $(b)$ had been selected:

$\log_B(y)=\dfrac{\log_b(y)}{\log_b(B)}$

This result can also be obtained by the interchanging variables $b\rightleftarrows B$ on the previous formula. Those who pursue both lines of algebra will stumble upon the interesting fact that,

$\log_b(B)\cdot \log_B(b)=1$.