1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Exponential question, x = [ 1 / 1 - e^-1.768] x e^-0.884
  1. Ln(1+e^x)=2 Problem (Replies: 4)