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    Hi. I've been looking at this since yesterday and I just cannot solve it (My basic algebra has gone out of the window coz I've not been so much out of practise, but I didn't really work on maths last year and now paying for it )

    But anyway could someone help me with this please?

    Radioactive mass, Mg is given my M=25e^-0.0012t, where t = time (secs) from 1st obvservation.

    What is the half life?

    Thanks

    For mathematicians who have never studied chemistry

    A half life is the time it takes for a radioactive mass to decay to half of its mass
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    you know that at t=0, the mass is 25, so if you put M=25/2 you can solve for t to find the half life
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    (Original post by sonofdot)
    you know that at t=0, the mass is 25, so if you put M=25/2 you can solve for t to find the half life
    Its the solving I'm having the problem with

    using 12.5 for M is about as far as I got (plus variations of taking e across as ln)
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    (Original post by Loz17)
    Its the solving I'm having the problem with

    using 12.5 for M is about as far as I got (plus variations of taking e across as ln)
    Well it sounds like you're on the right lines - if you post your working it may be easier to see where you're getting stuck
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    (Original post by sonofdot)
    Well it sounds like you're on the right lines - if you post your working it may be easier to see where you're getting stuck
    12.5=25e^-0.0012t

    12.5ln(-0.0012t) = 25

    I dunno if I did that right

    Its been almost a year since I did this last and I never really practised it then
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    (Original post by Loz17)
    12.5=25e^-0.0012t

    12.5ln(-0.0012t) = 25

    I dunno if I did that right

    Its been almost a year since I did this last and I never really practised it then
    That's not quite right. Remember that ln is the inverse of the exponential function, so \ln(e^{stuff}) = stuff. The easiest way to get this to work is to get e^stuff on its own on one side of the equation, so divide through by 25 to get 0.5 = e^(-0.0012t) and then try taking logs of both sides
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    Ahh yea

    But ln(-something) is impossible isn't it?
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    (Original post by Loz17)
    Ahh yea

    But ln(-something) is impossible isn't it?
    you have got ln here only -something read sonfodot's post
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    (Original post by rbnphlp)
    you have got ln here only -something read sonfodot's post
    I mean

    Once I have ln(0.5) = ln(-0.0012)t

    How do I get ln(-0.0012)? It says math error on calculator

    Sorry, I'm not the best at maths at all
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    (Original post by Loz17)
    I mean

    Once I have ln(0.5) = ln(-0.0012)t

    How do I get ln(-0.0012)? It says math error on calculator

    Sorry, I'm not the best at maths at all
    not quite. you get \ln(0.5) = \ln(e^{-0.0012t}). Since all of the -0.0012t is the power of e, when you log it you just get -0.0012t on that side
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    (Original post by Loz17)
    I mean

    Once I have ln(0.5) = ln(-0.0012)t

    How do I get ln(-0.0012)? It says math error on calculator

    Sorry, I'm not the best at maths at all
    Thats where the error is
    As sonfodots post says when you take ln(e^something) you only get something not ln(something)...

    so here you would have
    ln(0.5)=-0.0012t
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12.5 = 25e^{-0.0012t}

    

\frac{12.5}{25} = e^{-0.0012t}

    

ln(\frac{12.5}{25}) = -0.0012t

    Do you see how you go from e to ln?
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    (Original post by rbnphlp)
    Thats where the error is
    As sonfodots post says when you take ln(e^something) you only get something not ln(something)...

    so here you would have
    ln(0.5)=-0.0012t
    Ahhh ok I get it now

    Thank you both
 
 
 
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