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

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

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

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

mlddmlnog
 one year 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
 one year 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

mlddmlnog
 one year 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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