A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
The rate of decay of a radioactive substance is proportional to the amount of substance present. Four years ago there were 12 grams of substance. Now there are 8 grams. How many grams will there be 8 years from now?
anonymous
 5 years ago
The rate of decay of a radioactive substance is proportional to the amount of substance present. Four years ago there were 12 grams of substance. Now there are 8 grams. How many grams will there be 8 years from now?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Start time from 4 years ago. I think I've helped you with deriving these types of formulas before, so to save time, I won't go into deriving the expression for exponential growth or decay. Here, it is,\[N(t)=N(0)e^{kt}\]Now, 4 years ago is out time t=0, and then, we had 12 g of substance: \[N(t=0)=N(0)=12g\]Four years from time 0 (i.e. now), we have 8g, so\[N(4)=12 e^{4k}=8 \rightarrow e^{4k}=\frac{8}{12}=\frac{2}{3}\rightarrow 4k= \ln \frac{2}{3}\rightarrow k = \frac{1}{4}\ln \frac{2}{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Eight years from now, t will be 4+8 = 12, so we'll have\[N(12)=12e^{\frac{1}{4}\ln \frac{2}{3}\times 12} =12 e ^{3\ln (\frac{2}{3})}=12 e ^{\ln (\frac{2}{3})^3}=12(\frac{2}{3})^{3}=12 . \frac{2^3}{3^3}=\frac{96}{27}=\frac{32}{9}grams\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 4 years ago threw me off!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, these things are typically derived from a starting point of t=0...so go back to the farthest point and start counting from there.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.