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Rearranging equation

  1. Dec 14, 2003 #1
    How do I calculate t from the following equation?

    z=a+b(e^k*t -1)

    Guess this should be right to start with:

    t=ln(z/b)

    but what to do with the a and -1? and how do I calculate ln(z/b)? as lnz/lnb? I've tried thinking (didn't work *smirk*) and trial and error with simple numers (didn't work either)

    thanks a lot!
     
  2. jcsd
  3. Dec 14, 2003 #2

    Hurkyl

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    Staff Emeritus
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    Gold Member

    First, am I correct in presuming that you mean:

    [tex]
    z = a + b (e^{kt} - 1)
    [/tex]

    ? If so, then you need to group the exponent with parentheses; the correct way to write it is z=a+b(e^(k*t) -1).

    (Order of operations says you do exponentiation before multiplication)


    Anyways, this is a chance to use stuff you've learned previously. :smile: It looks like you want to use the fact:

    [tex]
    x = b^y \rightarrow \log_b x = y
    [/tex]

    right? However, note that the exponentiation has to be by itself on one side of the equation. Do you know how to do that?


    Compute z/b then take the log, just like the expression states.
     
  4. Dec 14, 2003 #3
    Yes, that equation of yours is what I meant. Should have read the faw on how to write equations in a forum.

    I also understand the equation you wrote further down, but I guess I'm really to stupid to rearrange teh whole equation to t=....
     
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