1
$\begingroup$

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).

$\endgroup$
10
  • 1
    $\begingroup$ Seems your product should have $\log_c d$ as third factor [your product skips that]. $\endgroup$
    – coffeemath
    Mar 13, 2021 at 19:20
  • 1
    $\begingroup$ @coffeemath A chain with a missing link is a broken chain, thanks :-). $\endgroup$ Mar 13, 2021 at 19:24
  • 4
    $\begingroup$ 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. $\endgroup$ Mar 13, 2021 at 19:29
  • $\begingroup$ @Joe At some places, people call 12:30AM 0:30AM or 12:30 PM 0:30PM. Not sure what the reason is. $\endgroup$ Mar 13, 2021 at 19:29
  • 1
    $\begingroup$ Your displayed equation does not match the formula you say you are trying to prove. The second displayed equation leads to $\log_b(c) = \log_a(c)/\og_a(b)$. $\endgroup$ Mar 13, 2021 at 19:38

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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}

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

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

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

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

$\large \Rightarrow \large \cdots$

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

$\large \Rightarrow \large \cdots$

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

$\large \Rightarrow \large \cdots$

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

$\large \Rightarrow \large \cdots$

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

$\large \Rightarrow \large \cdots$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.