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mlddmlnog
Group Title
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
 one year ago
mlddmlnog Group Title
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
 one year ago

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mlddmlnog Group TitleBest 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.. (
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
dw:1358381396966:dw
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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:
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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,\[\large C=8\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
thank you so much :) but i think my teacher said that the answer was 18.... o.o
 one year ago

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

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
okay ^^
 one year ago

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

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
ahh.. i';; need some time to understand this. haha but thank you alwayssss :) *
 one year ago
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