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Basic log question, completely lost!

  1. Apr 25, 2007 #1
    Basic log question, completely lost! :(

    1. The problem statement, all variables and given/known data
    Prove that y = a(1+b)^t is equivalent to ln y = a+bt


    2. Relevant equations



    3. The attempt at a solution
    i'm really confused with how they could possibly be equal.

    y = a(1+b)^t
    t = ln y/ln a(1+b)
    ln y = t x ln (a+b)
    ln y = ln (a+b)^t

    Could someone please help.
     
  2. jcsd
  3. Apr 25, 2007 #2
    Of course they are not equivalent for all values of t. Are you sure you have the question right?

    And, BTW, steps 3 and 4 of your attempt doesn't seem right, although it doesn't matter now.
     
  4. Apr 25, 2007 #3
    ohh sorry, i made a mistake.

    This is the original question:

    prove that yt = y0(1+r)^t is equivalent to ln yt = y0 + rt

    i just made y0 = a and r = b previously

    My Attempt:
    yt = y0(1+r)^t
    t = ln yt/ln y0(1+r)

    you said i was wrong previous, where do i go from this step then?? Tanx a heap 4 helping.
     
  5. Apr 25, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First thing you should do is go back and look at the problem again! What you have stated still is not true. From yt= yo(1+r)^t the best you can say, by taking ln of both sides, is ln(yt)= ln(y0)+ t ln(1+r). In general that is NOT the same as "ln yt= y0+ rt".

    (You can say that "to first order" (ignoring higher powers of t) yt (NOT ln(yt)) is approximately equal to y0+ rt.)
     
    Last edited: Apr 26, 2007
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