Radioactive decay problem. Please help me !

I've just been really thrown off by what this problem is asking me.

Given:

The decay of a radioactive material may be modeled by assuming that the amount A(t) of material present (in grams) at time t (minutes) decays at a rate proportional to the amount present, that is dA/dt= -kA for some positive constant k. Every subpart of this question refers to exactly the same radioactive material.

Question:

a. derive an equation for the amount A(t) present at time t in terms of the constant k and the amount A(o) present at time t=0

b. if A(5) = 1/3A(3), find K

c. at what time t will the amount A(t) be 1/4A(0)

My attempt (?):

for part a I wasnt sure if it was referring to just giving the equation A(t)=A(0)e^-(kt)

and I have no idea how to go about b or c. please help me if possible ! I think i'm over thinking the problem. I'm just really thrown off by the lack of numbers. Thank you!

Re: Radioactive decay problem. Please help me !

Maybe integration will help. Integrate both sides and gain an expression for A. Then use the fact when t=5 the answer is 1/3 t=3 to find k.

Re: Radioactive decay problem. Please help me !

For (a) I learned to separate the variables giving $\displaystyle \frac{dA}{A}=-kdt$ and then integrate both sides. A little algebra (and identifying what C, the constant of integration, is) should result in the answer you gave.

For (b), plug in your expression for A(t) with t=3 and t=5. A(0) cancels, and you can solve for k.

For (c), it's pretty much the same thing - plug in your expression for A(t) and your answer for k from part (b), and A(0) cancels so you can solve for t.

- Hollywood

Re: Radioactive decay problem. Please help me !

Thanks for the reply ! It looks like the answer I got for part b is k=(ln 3)/-2 .. does that seem right ? or completely off. haha

Re: Radioactive decay problem. Please help me !

I don't get a minus sign, just k=(ln 3)/2.

- Hollywood

Re: Radioactive decay problem. Please help me !

woops you are right. thanks so much for your help!!!!