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    see attached,

    t= 22 but i cant show it

    Thanks
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    Where did you get that question from? I can't solve it.

    I get to t= (5 ^30)* e^(-1.5t)

    how to solve that?
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    t is not equal to 22,plug it in to both sides and they will give different answers.
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    t=22 doesn't solve it
    e^((ln22)/30) = 1.108..
    5e^(-0.05*22) = 1.664..
    Plotting it, it looks like t is somewhere around 30 (not exactly 30 though). I can't work out how to do it algebraically though.
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    (Original post by Bezza)
    t=22 doesn't solve it
    e^((ln22)/30) = 1.108..
    5e^(-0.05*22) = 1.664..
    Plotting it, it looks like t is somewhere around 30 (not exactly 30 though). I can't work out how to do it algebraically though.
    i dont think its possible algebraically,i end up with
    t^1/30 + 0.05t=ln5.
    maybe u can use an iterative process?
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    The question is June01 AQA

    Can be found here: http://www.aqa.org.uk/qual/gceasa/mat_assess.html

    Q3 b)ii)

    Q.

    1000e^1/30ln2t = 5000e^-0.05t

    thats the start of the question.

    And the mark scheme reckons it goes to:

    1/30ln2t + 0.05t = ln5 (but i cant get to this stage!!!)

    Then they say t=22 for the 4th mark.
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    Thanks for your efforts guys.

    I've managed to solve it !!

    i forgot that when dividing powers you have to subtract powers!! e.g. e^-3/e^-2 = e^(-3)-(-2) = e^-1 BASIC school boy error!

    Thats why i got it wrong!!

    t=22.01!
 
 
 
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