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Help finding half life

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the half life (in hours) of a radioactive substance that is reduced by 5% in 65 hours.


    2. Relevant equations



    3. The attempt at a solution
    So looking around I find that A=A0e^-kt

    but Im not really sure how to solve this without having an initial ammount.
     
  2. jcsd
  3. Aug 31, 2009 #2

    zcd

    User Avatar

    You don't need an initial amount. You're given the rate at which it decays, so mathematically:
    [tex]A=A_{0}e^{0k}[/tex]
    [tex]0.95A=A_{0}e^{65k}[/tex]
     
    Last edited: Aug 31, 2009
  4. Aug 31, 2009 #3
    never mind figured it out. but if anyone xould help me wih this one it would be greatly appreciated.

    Solve for x: 4x=ex + 1

    how do you solve for x when its on both sides. How ever I did it the x would cancel.
     
  5. Aug 31, 2009 #4
    Didn't see your post but thanks for the help. I used Q=Q0(1/2)(t/T) and it worked out for all my half-life problems.
     
  6. Aug 31, 2009 #5

    Dick

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    Science Advisor
    Homework Helper

    Take the log of both sides.
     
  7. Aug 31, 2009 #6
    ok this is what im doing

    log of both sides, i get

    (X) log(4) = (x + 1) log(e)

    .60X = (X + 1) .43

    I dont know how to get both Xs to one side. I have done this with natural log aswell and get

    X ln (4) = X+1

    but how do I move the X

    I even tried doing

    4^X = exe1

    If someone could walk me through this one please, I have been trying so much crap my algebra is getting all messed up.
     
  8. Aug 31, 2009 #7
    How would you solve

    [tex]ax = x + 1\; ?[/tex]

    Once you've figured that out, just let a = ln(4).

    --Elucidus
     
  9. Aug 31, 2009 #8
    Start from xln(4)=x+1. Are you familiar with subtraction :D? What could you subtract from the right that would bring all x's to the left?
     
  10. Aug 31, 2009 #9
    ok so X is 2.588 right?

    thats what I was doing the whole time aswell but for some reason I would see x-x instead of 1.38X-X lol thank you. I know I was missing something. I hate when im doing things this late and my mind is just not working right. on the last few I would not be able to do anything and then the next morning when its already to late (due at 5 AM online) the answers would come to me.
     
  11. Sep 1, 2009 #10
    Ok now I cant get Solve for X : 4e6x=6e7x

    then get 6x ln(4e) = 7x ln(6e)

    6x (2.38) = 7x (2.79)

    Now how do I get Xs on the same side without canceling them? if I divide then the Xs cancel.
     
  12. Sep 1, 2009 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is now the same as Ax= Bx. How would you solve that? Seriously, if you can't solve equations like that, you need to go back and review basic algebra before you try problems involving exponentials and logarithms.
     
  13. Sep 1, 2009 #12
    Be careful! You are abusing the power rule of logarithms here.

    The natural log of 4e6x is not 6x (ln 4e). It is ln(4) + 6x.

    --Elucidus
     
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