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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

    Mute

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    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

    SammyS

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    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]

    etc.

    [itex]=\left(31^{(\log_{31}32)}\right)[/itex]

    [itex]=32[/itex]
     
  7. Aug 20, 2012 #6

    HallsofIvy

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    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!
     
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