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Exponential question, x = [ 1 / 1 - e^-1.768] x e^-0.884

  1. Aug 27, 2007 #1
    Im trying to setup this equation so I can figure it out, I have tried a couple of ways but I have lost the fundamental rules years ago - can anyone help me set it up for me ? Thanks so much.

    Equation: x = [ 1 / 1 - e^-1.768]e^-0.884
    ; Solve for x

    Answer: 0.498
    Last edited: Aug 28, 2007
  2. jcsd
  3. Aug 27, 2007 #2
    Is that a typo with the second x? What is that second x doing?

    If there is no typo, then x=ax so that x(1-a)=0 so that x=0.
  4. Aug 28, 2007 #3
    latex!!! i'm kind of confused

    however i got the same answer, ~.498
  5. Aug 28, 2007 #4
    Sorry for the confusion

    that x is a multiply sign, I have edited the original post.

    I got it now. Thanks.
    Last edited: Aug 28, 2007
  6. Aug 28, 2007 #5
    i did nothing different than plug in what you had typed out into my calculator


    if you're having trouble typing it into your calculator, this is what i do (for almost everything! and it's so useful for chemistry)

    in your calculator (if you don't know already), store the following

    type in the expression then press STO[tex]\rhd[/tex] then ALPHA then A then ENTER

    let [tex]A=(1-\exp^{-1.768})[/tex]

    let [tex]B=\exp^{-0.884}[/tex]

    so for your answer, you would type in

    Last edited: Aug 28, 2007
  7. Aug 28, 2007 #6

    Gib Z

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    Homework Helper

    I like doing that when I want to store a computation that I do over a large period of time. For my storage memory A, I have the sum of the harmonic series up to some number. Right now Its up to the 432th term, I add to it whenever I feel bored. I like to see how slow it grows :) And also to see it slowly converging to its asymptotic expansion:

    [tex]\sum_{n=1}^k \frac{1}{n}[/tex]~[tex]\log_e k + \gamma[/tex]

    where [itex]\gamma[/itex] is the Euler-Mascheroni Constant.
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