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A radioactive substance decays according to :

$$x' = -ax$$

where $a>0$ is a constant. After $2$ days there are $1,000$ grams and after $7$ days there are $300$ grams. How many grams were there initially?

I'm unsure how I'd go on with doing this, any help would be greatly appreciated...

Cheers.

EDIT: The step I got upto was separation of variables/integration:

$$x = e^{-at}$$

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    $\begingroup$ Hint: your equation says that $dx/dt = -ax$ so that $dx/x = -adt$. $\endgroup$
    – Eric Thoma
    Commented Dec 11, 2013 at 22:58
  • $\begingroup$ Just what I was thinking! Separation of variables followed by integration :) please let me know if that is correct so far, thank you Eric! $\endgroup$
    – user101985
    Commented Dec 11, 2013 at 23:00
  • $\begingroup$ Exactly. If you get stuck the answer is another very good hint. $\endgroup$
    – Eric Thoma
    Commented Dec 11, 2013 at 23:02
  • $\begingroup$ Okay, so doing so, I get: x = e^(-at). How would I make the 2 equations though now since I have the data for 2 days = 1000 grams and 7 days = 300 grams. Thanks again! $\endgroup$
    – user101985
    Commented Dec 11, 2013 at 23:14
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    $\begingroup$ To get multicharacter exponents, put them in braces. e^{-at} gives $e^{-at}$ This works for subscripts and items in fractions, as well. $\endgroup$ Commented Dec 11, 2013 at 23:56

1 Answer 1

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Can you solve the differential equation? You should have a solution with two constants-the initial amount and $a$. The two pieces of data give you two equations in two unknowns to find these constants.

Added: your solution needs a constant of integration. The solution should be $x=c\cdot e^{-at}$. Now plug in the data you are given: $$1000=c\cdot e^{-2a}\\300=c \cdot e^{-7a}$$ Now solve those for $a,c$ and $x(0)$ is just $c$

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  • $\begingroup$ Would I do a separation of variables and then integrate both sides to find a general equation ? Hope you can answer, thanks for your reply mate. $\endgroup$
    – user101985
    Commented Dec 11, 2013 at 22:57
  • $\begingroup$ Yes, that is a fine approach. $\endgroup$ Commented Dec 11, 2013 at 23:03
  • $\begingroup$ Great edit, how would you move onto solving? Would I just divide the two first? Hope you can respond again, thanks for your quality response. $\endgroup$
    – user101985
    Commented Dec 12, 2013 at 0:43
  • $\begingroup$ Yes, if you divide them, $c$ goes away and you can solve for $a$. Put that $a$ into one and you can solve for $c$. $\endgroup$ Commented Dec 12, 2013 at 2:00
  • $\begingroup$ Hey Ross, great response. I got the value of a but I'm unsure if it would make a difference if I inputted that value into either of the equations? Because if I input into one or the other, they both have different values. Hope you can answer, that should be my last inquiry. Thanks! $\endgroup$
    – user101985
    Commented Dec 12, 2013 at 6:31

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