1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How do I evaluate this log function?

  1. Aug 19, 2012 #1
    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)
  2. jcsd
  3. Aug 19, 2012 #2


    User Avatar
    Homework Helper

    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).
  4. Aug 19, 2012 #3
    That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
  5. Aug 20, 2012 #4
    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.
  6. Aug 20, 2012 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You could also approach this as follows:

    Let [itex]\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32) [/itex]

    Then, [itex]\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)} [/itex]

    By laws of exponents and the definition of a logarithm,

    [itex]2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]
    [itex]=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

    [itex]=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

    [itex]=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

    [itex]=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]



  7. Aug 20, 2012 #6


    User Avatar
    Science Advisor

    So, just in case others misunderstand, [itex]2^y= 32= 2^5[/itex] and therefore, y= 5 as the original poster said.
  8. Aug 22, 2012 #7
    Very nice approach. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook