## How do I evaluate this log function?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.

(log23)(log34)(log45) ... (log3132)
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 Recognitions: Homework Help Use the fact that $$\log_b a = \frac{\log_c a}{\log_c b}$$ for any positive, real numbers a, b and c (with c > 1).
 That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?

## How do I evaluate this log function?

 Quote by feihong47 That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
You use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the "LaTeX Reference" by clicking the Σ in the toolbar.

If you don't have to write anything complicated, you can just use the quick symbols and x2 and x2 buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.

EDIT: Here's the LaTeX guide.

Mentor
 Quote by feihong47 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that. (log23)(log34)(log45) ... (log3132)
You could also approach this as follows:

Let $\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32)$

Then, $\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$

By laws of exponents and the definition of a logarithm,

$2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

etc.

$=\left(31^{(\log_{31}32)}\right)$

$=32$
 Recognitions: Gold Member Science Advisor Staff Emeritus So, just in case others misunderstand, $2^y= 32= 2^5$ and therefore, y= 5 as the original poster said.
 Very nice approach. Thanks!