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Logarithm expression is messy

  1. May 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Given that

    [tex]\log_{4n} 40\sqrt {3} = \log_{3n} 45[/tex]

    find

    [tex]n^3[/tex]


    2. Relevant equations

    Logarithm properties

    3. The attempt at a solution

    I can get an expression for n but looks messy, and suspect there is probably a more compact answer. This is what I did:

    [tex] \dfrac{\log 40\sqrt{3}}{\log 4n} = \dfrac{\log 45}{\log 3n}

    \leftrightarrow \dfrac{\log 40\sqrt{3}}{\log 45} = \dfrac {\log 4n}{\log 3n}

    = \dfrac{\log 4 + \log n}{\log 3 + \log n}[/tex]

    and then I solved for log n; but again, my answer is 'messy'.

    Thanks for any help.
     
    Last edited: May 16, 2007
  2. jcsd
  3. May 16, 2007 #2

    Dick

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    I think the answer is just a little 'messy'. I don't think you can simplify it significantly.
     
  4. May 16, 2007 #3
    I have been told the answer is actually very compact.
     
  5. May 16, 2007 #4

    Dick

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    Now that you mention it, if I numerically evaluate the answer, it is suspiciously close to an integer. Can you guess which one I'm thinking of? I can't figure out how to get there directly though.
     
  6. May 16, 2007 #5

    Dick

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    Ok. Got it. Take the expression you get for log(n) and factor all of the numbers into primes. Now if multiply everthing out and simplify, then, lo and behold, the result magically becomes log(N)/3. Where N was that number I was thinking of. Pretty painful - but I still can't think of a more direct way. Plenty of practice with the rules of logarithms there.
     
  7. May 16, 2007 #6
    Ok, I got the answer after some very round about math. Right now I'm trying to find a way to do it with less work. I got the answer... n^3 is a two digit integer. If I figure out a way to do this more easily, I will throw you in the right direction.
     
  8. May 16, 2007 #7

    Dick

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    Maybe we are taking this 'don't tell the answer thing' too seriously. I suspect that teleport knows the answer too. But this is getting to be sort of fun. So I'll say that the first letter of the last digit of the two digit number is 'f'. Your turn. Finding this with a lot less work would be a cool thing. But maybe somebody just cooked it to work this way.
     
  9. May 17, 2007 #8

    HallsofIvy

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    If [itex]log_{4n}(40\sqrt{3})= log_{3n}(45)[/itex], then, letting a be that mutual value, [itex](4n)^a= 40\sqrt{3}[/itex] and [itex](3n)^a= 45[/itex]. Dividing one equation by the other [itex](4/3)^a= 8\sqrt{3}/9= 2^3/3^{3/2}[/itex]. That is, 2^{2a}= 2^3 and 3^a= 3^{3/2}: a= 3/2.

    Now we have [itex](4n)^{3/2}= 40\sqrt{3}[/itex] so [itex](4n)^3= 4^3 n^3= 64 n^3= 1600(3)= 4800[itex]. n^3= 4800/64= 75.
     
  10. May 20, 2007 #9
    Nice! However, I don't think you did well in providing the solution.

    Sorry for not replying in a few days. I had a lot of workload in my summer courses. But now its done... for now!
     
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