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Solving for x in an exponential

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data
    My problem is: e^-(xt) <= y, where t = 10 and y = 10^-6

    So: e^-(x10) <= 10^-6

    I have to find a value x that would make the probabilty less than or equal to 10^-6.

    2. Relevant equations



    3. The attempt at a solution
    I am not sure but my attempt in finding x is:

    x*10 =ln(10^-6)
    x=ln(10^-6) / 10

    I am not so sure that is right. I'm having somewhat of difficulty treating the e in the problem. Any help is greatly appreciated. Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 19, 2007 #2

    danago

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    Gold Member

    Thats right in taking natural logs of both sides to start with, but you seem to have missed a negative sign in your first step.
     
  4. Apr 19, 2007 #3
    Ok, so I would get ln(-x*10) <= ln 10^-6
    x = ln(10^-6) / -10)
    If I do that I calculate x to be 13.8. Does this seem right?
    Put x= 13.8 back into the original equation. Does it satisfy the equation?
     
    Last edited by a moderator: Apr 19, 2007
  5. Apr 19, 2007 #4

    hotvette

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    Homework Helper

    In problems like this, you can always confirm your answer by going back to the original equation. Since you've found x and you know t, just calculate e-xt and see if it gives the answer you want.
    .
     
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