# Chain Rule of Logarithm?

The students are taught the well known change base rule of logarithm:

\begin{align}\log_a b = \frac{\log_c b}{\log_c a}\end{align} Most text books proves it by invoking $$(a^x)^y=a^{xy}$$ to show: \begin{align}\log_c a\times\log_a b=\log_c b\end{align}

Question:

Why don't we call it Chain Rule of Logarithm (at least as a second name)? Is it because the way we use it is almost alway in the change of base format or something else? \begin{align} \log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z\end{align}

It is easy to remember and the students can have fun to continue chaining it (similar to the change rule of derivative).

• Seems your product should have $\log_c d$ as third factor [your product skips that]. Mar 13, 2021 at 19:20
• @coffeemath A chain with a missing link is a broken chain, thanks :-). Mar 13, 2021 at 19:24
• A few moments of googling (delete "derivative", delete "calculus" to avoid many false-positive hits) led to this question and the usage in these books, so clearly others have anticipated you. My guess for why this phase (or a similar one) hasn't been adopted is that no one needs to apply the base change over and over again in a single base change situation -- just change from the base you have to the base you want, without going through many intermediary bases. Mar 13, 2021 at 19:29
• @Joe At some places, people call 12:30AM 0:30AM or 12:30 PM 0:30PM. Not sure what the reason is. Mar 13, 2021 at 19:29
• Your displayed equation does not match the formula you say you are trying to prove. The second displayed equation leads to $\log_b(c) = \log_a(c)/\og_a(b)$. Mar 13, 2021 at 19:38

A side note:

The argument of the current factor must be equal to the base of following factor. I´ve colored the corresponding parameters.

\begin{align}\log_{\color{blue}{a}} \color{red}{b}\times\log_{\color{red}{b}} \color{orange}{c}\times\log_{\color{orange}{c}} \color{green}{d}\times\cdots\times\log_y \color{yellowgreen}{z}=\log_{\color{blue}{a}} \color{yellowgreen}z\end{align}

I hope you see the difference to your term. The chain rule is about derivatives and the concept is very different from the rule you´ve posted.

It is more related to the overall growth rate $$r$$, if you have n consecutive growth rates ($$r_i$$), with $$r_i=\frac{y_{i+1}}{y_i}-1$$.

$$1+r=\frac{y_{1}}{y_0}\cdot \frac{y_{2}}{y_1}\cdot \ldots \cdot \frac{y_{n-1}}{y_{n-2}} \cdot \frac{y_{n}}{y_{n-1}}=\frac{y_{n}}{y_{0}}$$

The rule you´ve posted is related to the concept of the geometric mean.

Well, finally I found another user or person who concurs with my idea. I discovered this relation personally.

\begin{align} \log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z\end{align}

but this is just a special case of

\begin{align} \log_{b_1}{a_1} \times \log_{b_2}{a_2} \times \log_{b_3}{a_3} \times \log_{b_4}{a_4} \times \cdots \times \log_{b_n}{a_n} = \log_{b_1}{a_{\pi_1}} \times \log_{b_2}{a_{\pi_2}} \times \log_{b_3}{a_{\pi_3}} \times \log_{b_4}{a_4} \times \cdots \times \log_{b_n}{a_{\pi_n}} \end{align}

Well, I think this "chain form" of base changing is as it another form of another identity in reversing ts proof

$$\large n^{\log_bx} = x^{\log_bn}$$

One revering its simple proof is:

$$\large c^{\log_ab} = b^{\log_ac}$$

$$\large \Rightarrow \log_bc^{log_ab}=log_bb^{log_ac}$$

$$\large \Rightarrow \log_ab\cdot \log_bc=\log_bb\cdot \log_ac$$

$$\large \Rightarrow \log_ab\cdot \log_bc=1\cdot \log_ac$$

\begin{align}\log_a b\times\log_b c=\log_a c\end{align}

One simple proof is:

$$\log_ab\cdot \log_bc=1\cdot \log_ac$$

$$\large \Rightarrow \log_ab\cdot \log_bc=\log_bb\cdot \log_ac$$

$$\large \Rightarrow \log_bc^{log_ab}=log_bb^{log_ac}$$

$$\large \Rightarrow \large c^{\log_ab} = b^{\log_ac}$$

Then how do we proceed our proof from two terms to many terms? Use \begin{align}\log_a b = \frac{\log_c b}{\log_c a}\end{align}

$$\log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z$$

$$\large \Rightarrow \large \frac{1}{\log_b a} \times \log_b c \times \frac{1}{log_dc} \times \log_d e \times \cdots\times\log_y z=\log_a z$$

$$\large \Rightarrow \large \frac{ \log_b c}{\log_b a} \times \frac{\log_d e}{log_dc} \times \frac{\log_f g}{log_fe}\times \cdots\times\log_y z=\log_a z$$

$$\large \Rightarrow \large \log_a c \times log_ce \times \log_eg \times \log_gi \times \log_ik \times \log_km \times \log_mo \times \log_oq \times \log_qs \times \log_su \times \log_uw \times \log_wy \times\log_y z=\log_a z$$

$$\large \Rightarrow \large \cdots$$

$$\large \Rightarrow \large \log_ae \times \log_ei \times \log_im \times \log_mq \times \log_qu \times \log_uy \times\log_y z=\log_a z$$

$$\large \Rightarrow \large \cdots$$

$$\large \Rightarrow \large \log_ai \times \log_iq \times \log_qy \times\log_y z=\log_a z$$

$$\large \Rightarrow \large \cdots$$

$$\large \Rightarrow \large \log_aq \times \log_qy \times\log_y z=\log_a z$$

$$\large \Rightarrow \large \cdots$$

$$\large \Rightarrow \large \log_ay \times\log_y z=\log_a z$$

$$\large \Rightarrow \large \cdots$$

$$\large \Rightarrow \large \log_a z=\log_a z$$