Dismiss Notice
Join Physics Forums Today!

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:


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First, am I correct in presuming that you mean:

    z = a + b (e^{kt} - 1)

    ? 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:

    x = b^y \rightarrow \log_b x = y

    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=....

Have something to add?

Similar Discussions: Rearranging equation
  1. Rearranging equation (Replies: 3)

  2. Rearranging equations (Replies: 3)

  3. Rearranging equations (Replies: 1)

  4. Rearranging equation (Replies: 3)