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).
 A: 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.
A: 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}
where ${\pi}$ is any permutation of the subscripts 1, ..., n.
A: 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}

$\log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z$
$\large \Rightarrow \large  \frac{1}{\log_b a} \times \log_b c \times \frac{1}{log_dc} \times \log_d e \times \cdots\times\log_y z=\log_a z$
$\large \Rightarrow \large  \frac{ \log_b c}{\log_b a} \times \frac{\log_d e}{log_dc} \times \frac{\log_f g}{log_fe}\times \cdots\times\log_y z=\log_a z$
$\large \Rightarrow \large \log_a c \times log_ce \times \log_eg \times \log_gi \times \log_ik \times \log_km \times \log_mo \times \log_oq \times \log_qs \times \log_su \times \log_uw \times \log_wy \times\log_y z=\log_a z$
$\large \Rightarrow \large \cdots$
$\large \Rightarrow \large \log_ae \times \log_ei \times \log_im \times \log_mq \times \log_qu \times \log_uy \times\log_y z=\log_a z$
$\large \Rightarrow \large \cdots$
$\large \Rightarrow \large \log_ai \times \log_iq \times \log_qy \times\log_y z=\log_a z$
$\large \Rightarrow \large \cdots$
$\large \Rightarrow \large \log_aq \times \log_qy \times\log_y z=\log_a z$
$\large \Rightarrow \large \cdots$
$\large \Rightarrow \large \log_ay \times\log_y z=\log_a z$
$\large \Rightarrow \large \cdots$
$\large \Rightarrow \large \log_a z=\log_a z$

