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Couple of log questions

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data
    1.) 3logx2y + 2logxy
    2.) 4logabc - 2loga2b - 3logbc




    2. Relevant equations



    3. The attempt at a solution
    I know that 3logx2 is the same as 6logx but I don't know what to do since theres a y there
     
  2. jcsd
  3. Jan 10, 2009 #2
    What is the question? If it's to simplify as much as possible, you can use log(xy) = log(x) + log(y).
     
  4. Jan 10, 2009 #3
    Yep thats the question. I'm just confused about what to do with all these coefficients.
    Is 3logx2y the same as logx6y3??

    Can I write logx6y3 as logx6[/sup + log]y3 or do I have to get rid of them powers first?
     
  5. Jan 10, 2009 #4
    Yes assuming x^2y are both an argument of your log
     
  6. Jan 10, 2009 #5

    tiny-tim

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    Hi MadmanMurray! :smile:

    I'm guessing that they want it in the form 6logx + 3logy. :wink:

    (after all, how would you look up logx6 in log-tables if x = 2.345? … you'd have to find 2.3456 first, and the only way of doing that is to :rolleyes: … yes!! :biggrin:)
     
  7. Jan 10, 2009 #6
    Thanks a lot.

    I have 2 more log questions in front of me that are confusing as hell too:
    1.) log (x2 + 2) = 2.6

    and

    2.) 2x + 1 = 32x - 1

    For the first one there I was wondering if I can express it like this 2logx + 2? Can I do that?
     
    Last edited: Jan 10, 2009
  8. Jan 10, 2009 #7

    tiny-tim

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    Nooo … that woud be (logx2) +2 :frown:

    Hint: if loga = b, then a = eb :wink:
     
  9. Jan 10, 2009 #8

    jgens

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    Careful tiny-tim: log(a) doesn't necessarily refer to the natural logarithm. log(a) commonly refers to the logarithm base 10 as well.
     
  10. Jan 10, 2009 #9

    Delphi51

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    if log X = Y, then X = B^Y (where B is the base of the log)
    If it doesn't make sense in logarithm land, transform it to power land!
    And vice versa.
     
  11. Jan 11, 2009 #10
    Well, where I study lg is decimal logarithm, ln is natural logarithm and log refers to a logarithm with any other base which is shown in subscript right after the log symbol. For example log[tex]_{2}[/tex]8 = 3 ( I don't know why, but using LaTex here shows a subscript as a superscript on my machine. The 2 is supposed to be as a subscript. I hope you get the idea), lg100 = 2 and ln(e[tex]^{2}[/tex]) = 2.
     
  12. Jan 11, 2009 #11

    tiny-tim

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    Hi kbaumen! :smile:

    You have to use "inline" LaTeX (typing "itex" instead of "tex") if you're inserting into a line of text (see just above) …

    but it's much better, on this forum, to use the X2 or X2 tags (just above the reply box), especially since any LaTeX takes up a lot of space on the server. :wink:
     
  13. Jan 11, 2009 #12
    Oh. Thanks a lot for the explanation.
     
  14. Jan 11, 2009 #13

    HallsofIvy

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    Then x2+ 2= a2.6 where "a" is the base of the logarithm (probably 10 or e).

    Since there exponentials are to different bases, which cannot be converted to one another, there is no easy way to solve this equation.

     
  15. Jan 11, 2009 #14
    I decided to plug in some random numbers and the first one I plugged in (1) happened to work. Since theres no simple way to solve it maybe thats what I was meant to do.
     
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