Anyone want to help me with my homework?

[Tab]

A 50mg sample of a radioactive substances decays over the course of three days to 35mg. Assuming it decays exponentially, determine the rate of decay...

Okay...

50^(3X)=35

Right? How the hell do I work with that exactly?

Anyway, I got another formula... Just look at the problem above to see where everything goes...

A=P(1-r)^3

35=50(1-r)^3

35=50-50r^3

-15=-50r^3

-15/-50=r^3

.3=r^3

Thats how far I got, now how do I manipulate that futher? I think the answer is something like .0003

A 50mg sample of a radioactive substance decays to 35mg over the course of three days. Assume the remaining sample decayed in an exponential manner, determine the daily rate of decay. Give the results as a percentage.

 

[PurdueRy]

50^(3X)=35

that is not a decay equation....

also, where the heck did you get the number 35??

 

[Tab]

Originally posted by: PurdueRy
50^(3X)=35

that is not a decay equation....

also, where the heck did you get the number 35??

Thats what I orginally came up with... for a formula to work it out... Oh, seems I didn't complete the question..

 

[PurdueRy]

Originally posted by: Tab

Originally posted by: PurdueRy
50^(3X)=35

that is not a decay equation....

also, where the heck did you get the number 35??

Thats what I orginally came up with... for a formula to work it out...

well you could solve that equation, but it doesn't make sense for the problem. Might wanna just post the full problem.

 

[PurdueRy]

EDITED See work below, after OP posted full problem

 

[Tab]

Originally posted by: PurdueRy
A=P(1-r)^3

35=50(1-r)^3

35/50=(1-r)^3

(35/50)^(1/3)=(1-r)

-(35/50)^(1/3)+1=r

r=.112

If you did the beginning right at least...

Whoa, hmm... You sure thats right anyway to check it?...

 

[Syringer]

Originally posted by: Tab
A 50mg sample of a radioactive substances decays over the course of three days to 35mg. Assuming it decays exponentially, determine the rate of decay...

Okay...

50^(3X)=35

Right? How the hell do I work with that exactly?

Anyway, I got another formula... Just look at the problem above to see where everything goes...

A=P(1-r)^3

35=50(1-r)^3

35=50-50r^3

That's not how it works..but neither is the way you did the problem either..not that I know how to do it

 

[PurdueRy]

Now that I have the real problem:

N(t)=N0e^(-rt)

35 = 50e^(-r(3))

35/50 = e^(-3r)
ln(35/50) = -3r
r = ln(35/50)/-3
r = .119

r = 11.9% daily

This is the correct way to do it

 

[Ophir]

Originally posted by: PurdueRy
Now that I have the real problem:

N(t)=N0e^(-rt)

35 = 50e^(-r(3))

35/50 = e^(-3r)
ln(35/50) = -3r
r = ln(35/50)/-3
r = .119

r = 11.9% daily

That's much better.

 

[Tab]

If anyone is bored as I am at 1:49AM

A newly formed lake was stocked with blue gills four years ago. After two years it was estimated that the lake contained 5000 bluegills, and after four years, it was estimated that the lake contained 8000 bluegills. If the bluegill population has been growing exp. how many bluegills were initially stocked.

I assume I could just use A=(1+r)^t

A=final amount
B=orginal amount
r=rate
t=time

Now, it's just werid because I don't have the time exactly or the intial amount.

 

[Tab]

Originally posted by: Ophir

Originally posted by: PurdueRy
Now that I have the real problem:

N(t)=N0e^(-rt)

35 = 50e^(-r(3))

35/50 = e^(-3r)
ln(35/50) = -3r
r = ln(35/50)/-3
r = .119

r = 11.9% daily

That's much better.

Thats confusing as fsck.

 

[PurdueRy]

Originally posted by: Tab
If anyone is bored as I am at 1:49AM

A newly formed lake was stocked with blue gills four years ago. After two years it was estimated that the lake contained 5000 bluegills, and after four years, it was estimated that the lake contained 8000 bluegills. If the bluegill population has been growing exp. how many bluegills were initially stocked.

N(t)=N0e^(-rt)

5000 = xe^(2r)
8000 = xe^(4r)
x = 8000/e^(4r)

5000 = (8000/e^(4r))e^(2r)
5000 = 8000e^(-2r)
5000/8000 = e^(-2r)
ln(5000/8000) = -2r
r = (ln(5000/8000)/-2)
r=.235

5000 = xe^(2*.235)

x = 3125


That's the last of your hw problems I am doing for you

 

[PurdueRy]

Originally posted by: Tab

Originally posted by: Ophir

Originally posted by: PurdueRy
Now that I have the real problem:

N(t)=N0e^(-rt)

35 = 50e^(-r(3))

35/50 = e^(-3r)
ln(35/50) = -3r
r = ln(35/50)/-3
r = .119

r = 11.9% daily

That's much better.

Thats confusing as fsck.

It's solved step by step, you can't get much clearer

 

[Tab]

Thanks PurdueRy

I'd give you a beer but it's late so here is a


BTW - I don't think I should have to use logs for these... We haven't even gone over them in class, yet.

 

[PurdueRy]

Originally posted by: Tab
Thanks PurdueRy

I'd give you a beer but it's late so here is a


BTW - I don't think I should have to use logs for these... We haven't even gone over them in class, yet.

Note that I changed the last answer I gave a bit...didn't go far enough...