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Logarithmic Functions

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve 3(6x-1) = 1 + 4/6x giving your answer in the form x = a - log6b where a,b are integers (Z)


    2. Relevant equations



    3. The attempt at a solution
    3(6x)(6-1)(6x) = 6x + 4
    Let 6x be y,
    (1/2)y2 - y - 4 = 0
    y = 1 ± √(1+8) = 1 ± 3
    3x = 4
    xlog63 = log64

    What should I do next? :confused:
     
  2. jcsd
  3. May 12, 2012 #2
    Read the bolded statements carefully. :wink:
     
  4. May 12, 2012 #3
    OH NO! What is IB HL Maths doing to me :eek:
    Thanks! :approve:
     
  5. May 12, 2012 #4
    Its normal, after lots of studying!!
    Take short breaks, now and then :biggrin:
     
  6. May 12, 2012 #5

    NascentOxygen

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    Staff: Mentor

    Isn't it 6x = 4?
     
  7. May 13, 2012 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, now take the logarithm, with respect to any base, of both sides.

    By the way, going back to your first formula, [itex]x log_6(3)= log_6(4)[/itex], how would you solve any equation of the form Ax= B for x?
     
  8. May 13, 2012 #7
    What do you mean by "form Ax= B for x?" ? :eek:
     
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