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Mathematics 17 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Start 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\[N(4)=12 e^{4k}=8 \rightarrow e^{4k}=\frac{8}{12}=\frac{2}{3}\rightarrow 4k= \ln \frac{2}{3}\rightarrow k = \frac{1}{4}\ln \frac{2}{3}\]

OpenStudy (anonymous):

Eight years from now, t will be 4+8 = 12, so we'll have\[N(12)=12e^{\frac{1}{4}\ln \frac{2}{3}\times 12} =12 e ^{3\ln (\frac{2}{3})}=12 e ^{\ln (\frac{2}{3})^3}=12(\frac{2}{3})^{3}=12 . \frac{2^3}{3^3}=\frac{96}{27}=\frac{32}{9}grams\]

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

You're welcome ;)

OpenStudy (anonymous):

The 4 years ago threw me off!

OpenStudy (anonymous):

Yeah, these things are typically derived from a starting point of t=0...so go back to the farthest point and start counting from there.

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