MHB Calculating a Logarithmic Product Series

[karush]

compute the product.

$\left(\log_{2}\left({3}\right)\right)\cdot
\left(\log_{3}\left({4}\right)\right)\cdot
\left(\log_{4}\left({5}\right)\right)\cdots
\left(\log_{126}\left({127}\right)\right)\cdot
\left(\log_{127}\left({128}\right)\right)$

The answer to this is 7
I assume this can be done with a $\lim_{{2}\to{127}}$

or use a change of base

$\frac{\log\left({3}\right)}{\log\left({2}\right)}\cdot
\frac{\log\left({4}\right)}{\log\left({3}\right)}$ etc

but I can't seem to figure out the setup

 

[MarkFL]

You are on the right track with the change of base formula. Write the product as follows:

$$\frac{\log_2(3)\cdot\log_2(4)\cdots\log_2(127)\log_2(128)}{\log_2(3)\cdot\log_2(4)\cdots\log_2(126)\log_2(127)}$$

After cancelling, you are then left with:

$$\log_2(128)=\log_2\left(2^7\right)=7\log_2(2)=7$$

 

[karush]

Really, it's that easy, (Speechless)(Speechless)(Speechless)