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Finding log something, in terms of A and B.

  1. Oct 19, 2009 #1

    Suy

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    1. The problem statement, all variables and given/known data

    If Logb2=A and Logb49=B, what is logb397, in terms of A and B.

    This is one of the bonus question in my geometric quiz and i don't remember if the number is 397. I wonder if anyone get this?
    2. Relevant equations



    3. The attempt at a solution
    here is what i did
    bA=2 and bB=49
    b=21/A b=491/B
    491/B=21/A --> log(49)/log(2)=B/A

    because this is geometric quiz, so i assume this one have a geometric
    so i use arn-1
    r : log(49)/log(2)=B/A
    a : Logb2

    Logb2(B/A)n-1=logb397
    am i right?
    ty!
     
  2. jcsd
  3. Oct 19, 2009 #2

    Mark44

    Staff: Mentor

    I'm guessing that the number is 392, not 397. 392 = 8*49 = 23*49.
    This makes no sense whatever. From your presentation of the problem, it has nothing to do with a geometric sequence, or anything else having to do with geometry. This problem is strictly concerned with the properties of logarithms.
    I'm assuming that this is the actual problem description:
    If Logb2=A and Logb49=B, what is logb392, in terms of A and B.​

    logb 392 = logb (8 * 49) = logb (23 * 49)

    Now, use the properties of logs on the last expression above to get quantities that are in terms of A and B.
     
  4. Oct 19, 2009 #3

    Suy

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    but i remember it is a odd number
     
  5. Oct 19, 2009 #4

    Mark44

    Staff: Mentor

    Well, if you can't remember exactly what the problem is, I can't help you.
     
  6. Oct 19, 2009 #5

    Suy

    User Avatar

    but what happen when it is a odd number? like 397, is it possible to solve it?
     
  7. Oct 19, 2009 #6

    Mark44

    Staff: Mentor

    If you don't know what the number is, you can't work the problem - that's what happens.
     
  8. Oct 19, 2009 #7

    Suy

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    ok, thx
     
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