Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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?

  • This Question is Closed
  1. mlddmlnog
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.. (

  2. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1358381396966:dw|

  4. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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:

  6. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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,\[\large \ln \left(\frac{3}{2}\right)=-4k\] \[\large k=-\frac{1}{4}\ln \left(\frac{3}{2}\right)\]

  7. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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,\[\large C=8\]

  8. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  9. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. mlddmlnog
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. mlddmlnog
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay ^-^

  14. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. mlddmlnog
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  16. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol :O

  17. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  18. mlddmlnog
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.