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anonymous
 3 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?
anonymous
 3 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.. (

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm I don't remember these :3 gimme a sec to work through it.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Ok ok ok so we're dealing with this type of equation right? with the e thingy? ok ok ummm

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm 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:

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Let'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,\[\large \ln \left(\frac{3}{2}\right)=4k\] \[\large k=\frac{1}{4}\ln \left(\frac{3}{2}\right)\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0So 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,\[\large C=8\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm 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.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Do you happen to know the answer before I go too far along this path? XD

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0So I can determine if this is the right way or not.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you so much :) but i think my teacher said that the answer was 18.... o.o

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Ok lemme punch it in and see if we're on the right track.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm No I got like 3.6. Darn I did something wrong :C oh well..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahh.. i';; need some time to understand this. haha but thank you alwayssss :) *
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