Logarithm

Algebra Level 3

Let $\{a_n\}$ be a sequence of positive real numbers ( $a_i\neq 1$ ).

Compute $\displaystyle\prod_{i=1}^{n-1} \log_{a_i} a_{i+1}$ .

\pi

\infty

\displaystyle\sum_{i=1}^n \log a_i

1

e

\log_{a_1}a_n

\displaystyle\prod_{i=1}^n \log a_i

0

1 solution

Rohit Udaiwal
Mar 20, 2016

Recall that $\log_{\color{#20A900}{a}}{\color{#3D99F6}{b}}=\dfrac{\log \color{#3D99F6}{b}}{\log \color{#20A900}{a}}$ ,by the base changing property of logarithms . Then,the required product $\displaystyle\prod_{i=1}^{n-1} \log_{a_i} a_{i+1}$ transforms into a telescopic product as illustrated below $\begin{aligned} \displaystyle\prod_{i=1}^{n-1} \log_{a_i} a_{i+1}=&\dfrac{\log a_{2}}{\log a_{1}}\cdot \dfrac{\log a_{3}}{\log a_{2}}\cdot \ldots \cdot \dfrac{ \log a_{n}}{\log a_{n-1}} \\ =&\dfrac{\log a_{n}}{\log a_{1}} =\boxed{\log_{a_{1}} {a_{n}}} \end{aligned}$

Always elegant solutions :-)..... $\left(+\displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}\right)$ .

Edit:- This level 5 limit calculation is left as an exercise to the readers. Appropriate hints are provided below. ;-P

Rishabh Jain - 5 years, 2 months ago

To those who didn't understand

Apply L'hopital's :P

Mehul Arora - 5 years, 2 months ago

Simple standard approach... (+1)

Rishabh Jain - 5 years, 2 months ago

@Rishabh Jain – I use $\ln(1+x)=x-\frac 12 x^2+\cdots$
:P

展豪張 - 5 years, 2 months ago

@展豪張 – Right... Using Expansion series was also a choice.. :-)

Rishabh Jain - 5 years, 2 months ago

@Rishabh Jain – Yeah... I think L'hopital's and expansion series are the two main stream approach in topic:calculus in brilliant... :D

展豪張 - 5 years, 2 months ago

Thanks,I try to write solutions learning from your methods :P

Rohit Udaiwal - 5 years, 2 months ago

(+1) Exactly the solution that I want!

展豪張 - 5 years, 2 months ago

Logarithm

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