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Geometric series involving logarithms

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A geometric series has first term and common ratio both equal to ##a##, where ##a>1##
    Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

    Hence, evaluate
    [itex]log_{3}(\frac{3}{2} a^{2}+ a^{4}+...+ a^{58})[/itex]

    Giving your answer in the form of ##A-log_{3} B##, where ##A## and ##B## are positive integers to be determined.


    2. Relevant equations
    Geometric series, logarithms


    3. The attempt at a solution
    For the first part, I have to write them as
    ##S_{12} =28 S_{6}##. Then I have to apply the formula for geometric sum...
    Then I arrived at ##a^{13} -28a^{7} +27a=0##
    Factoring give me ##a(a^{12}-28a^{6}+27)=0##

    How should I solve for a in this case?
     
  2. jcsd
  3. Jan 28, 2014 #2

    jedishrfu

    Staff: Mentor

    If you replace a^6 by b then you'll notice you have a quadratic b^2 - 28b + 27 = 0 and you should be able to solve for b and by extension for a.
     
  4. Jan 28, 2014 #3
    Ok, now I have a=sqrt(3)... But how about the second part? :/
     
  5. Jan 29, 2014 #4
    You have the value of a, just plug it in the second part. It's a geometric progress inside the log. :)
     
  6. Jan 29, 2014 #5
    But it doesn't look like one if I plugged in...
    I got 9/2 + 9 + 27+...
     
  7. Jan 29, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The sum is that of a geometric progression + something.
    (1/2 a2+a2+a4+....+a58))

    ehild
     
  8. Jan 29, 2014 #7
    I get ##30-\log_3 {2}##. Is this correct?
     
  9. Jan 29, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes.


    ehild
     
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