## mlddmlnog 2 years ago The rate of decay of a radioactive substance is proportional to the amunt of substance present. Four years ago, there 12 grams of substance. Now there ae 8 grams. How many will there be 8 years from now?

1. mlddmlnog

@zepdrix i think you explained this to me already, but could you please do soagain? i adm still not clear on how to do it.. (

2. zepdrix

Hmm I don't remember these :3 gimme a sec to work through it.

3. zepdrix

|dw:1358381396966:dw|

4. zepdrix

Ok ok ok so we're dealing with this type of equation right? with the e thingy? ok ok ummm

5. zepdrix

Hmm this one is a little confusing since they're referring back in time and forward... hmmm $\large y(t)=Ce^{kt}$ So let's call the current time t=0.$\large y(-4)=12=Ce^{-4k},\qquad y(0)=8=Ce^0$ I'm not sure if this will work, we'll try it though D:

6. zepdrix

Let's divide these, $$\huge \frac{y(-4)}{y(0)}$$, Which will give usssssss,$\large \frac{12}{8}=\frac{Ce^{-4k}}{Ce^0}$Which will simplify to,$\large \frac{3}{2}=e^{-4k}$Taking the natural log of both sides gives us,$\large \ln \left(\frac{3}{2}\right)=-4k$ $\large k=-\frac{1}{4}\ln \left(\frac{3}{2}\right)$

7. zepdrix

So we solved for $$k$$. Let's plug it back into our equation and solve for $$C$$.$\huge 12=Ce^{-4\left(-\frac{1}{4}\ln \left(\frac{3}{2}\right)\right)}$Which simplifies to,$\large 12=Ce^{\ln\left(\frac{3}{2}\right)}$And then,$\large 12=C\left(\frac{3}{2}\right)$And thennnnnn,$\large C=8$

8. zepdrix

Hmm I don't think I'm doing this correctly :c I think we have to let t=0 be the starting point, not the current time... I'm just not sure if it's going to work out doing it this way.

9. zepdrix

Do you happen to know the answer before I go too far along this path? XD

10. zepdrix

So I can determine if this is the right way or not.

11. mlddmlnog

thank you so much :) but i think my teacher said that the answer was 18.... o.o

12. zepdrix

Ok lemme punch it in and see if we're on the right track.

13. mlddmlnog

okay ^-^

14. zepdrix

Hmmm No I got like 3.6. Darn I did something wrong :C oh well..

15. mlddmlnog

oh wait. 3.6 might be the answer. i'm not exactly sure if it's 18 or 3.6. i think i heard someone saying that it's 3.6.. i'm not sure... gosh, can i see the work you did to get 3.6 as the answer???? :O

16. zepdrix

lol :O

17. zepdrix

So we solved for C, we solved for k, giving us the following equation,$\huge y(t)=8e^{\left(-\frac{1}{4}\ln\left(\frac{3}{2}\right)\right)t}$ If you plug in t=8, you'll get 3.6

18. mlddmlnog

ahh.. i';; need some time to understand this. haha but thank you alwayssss :) *