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Homework Help: Log question

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    1. Use the fact that loga(10)= 2.4197 and loga(13)=2.6955 to solve the following

    a. loga(130) b. loga(1.3) c. loga(1000)

    2. Relevant equations



    3. The attempt at a solution
    I'm not really sure how to do this i missed yesterdays class i dont really want the anwser just some help on how to do it becuase i dont understand it out of the book at all. any help would be greatly appreciated
     
  2. jcsd
  3. May 26, 2010 #2

    Dick

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    Do you know some rules of logarithms? Like 130=13*10. How is the log of a product related to the log of the terms?
     
  4. May 26, 2010 #3
    well i think you getting at something like this
    so if i have logam+logan

    i get loga(m)(n)
     
  5. May 26, 2010 #4

    Dick

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    Sure. log_a(m)+log_a(n)=log_a(m*n). So what's the answer to the first one?
     
  6. May 26, 2010 #5
    well i'm not sure if im correct but i multiplied the two numbers 2.4197*2.6955 = 6.522
     
  7. May 26, 2010 #6

    Dick

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    Why did you multiply them? If log_a(m)+log_a(n)=log_a(mn) and you put m=10 and n=13, I don't think you want to multiply 2.4197*2.6955.
     
  8. May 26, 2010 #7
    well then i just get log_a(10)(13) which is 130 but i dont understand what i am supposed to do with that? im sorry im just confused
     
  9. May 26, 2010 #8

    Dick

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    What do you think log_a(10)(13) means? You are a little too confused. Look, log_a(10)+log_a(13)=log_a(10*13)=log_a(130). You know log_a(10)=2.4197 and log_a(13)=2.6955. I wouldn't multiply them.
     
  10. May 26, 2010 #9
    oh my goodness lol i add them =)
     
  11. May 26, 2010 #10

    Dick

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    Thank you. Yes. Now express 1.3 and 1000 in terms of 10 and 13 and use the rules of logs again.
     
  12. May 26, 2010 #11
    i figured it out thank you for the help =) i add the first one then divide (subtract the second one, then add add add on the third one. thanks =)
     
  13. May 27, 2010 #12

    HallsofIvy

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    Yes, log(1000)= log(10*10*10)= log(10)+ log(10)+ log(10). Of course, that's the same as 3*log(10). That's another "rule of logarithms": log(xy)= y log(x).
     
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