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What is the rate of decay of awesome-10 which has a half-life of 650 days?

Mathematics
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I can't figure out how to solve these rate of decay questions, so I just made up my own as a way to help me learn this. How would I solve this?
if it has a half life of 650 time units it means every 650 time units 1/2 of the samle dissapears.
Yep. :) How would I find the rate of decay, though?

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Other answers:

halflife= -ln2/k that is the half life equation
I have a formula \[N=N _{0}e ^{kt}\] How/where would I plug everything in?
I'm not sure I understand @Spectrum
I know spectrum
i gave him all right answers
for his teset he did good to
I think they are asking for k you got N=1/2No t=650 so you plug all numbers in and find k
I mean, I don't understand what he meant. I'm having a lot of trouble with logarithms, so I'm not quick on my feet. :P
\[N=\frac{ 1 }{ 2 }N _{0}^{650k}\] Okay, so what do I do next?
Would I put some test number in for No? Like 100 or something?
no,|dw:1354966748637:dw| you plug 1/2No for Nthen you can simplify for No
|dw:1354967053153:dw|
|dw:1354952485674:dw|
|dw:1354952600983:dw|
|dw:1354952743876:dw|
|dw:1354953110349:dw|
I think now it makes sense. Doesn't it?
KIS means keep it simple.
Derp still need help understanding =/
The easiest way I believe is to derive it from calculus (if you've taken Diff EQ , you should know how to). Doing this you get an equation much like yours only N_0 is denoted c \[A=ce^{kt}\]
now we're given two values to find what c is you need to know that \[N(0)=N_0\] \[N_0=ce^0=c\] \[N=N_0e^{kt}\]
same as your equation now we know that after 650 days , we will have half of the original amount soo \[N(650 days)=.5N_0\] \[.5N_0=N_0e^{650k}\] solve for k, divide by \[N_0\]
\[\frac{.5N_0}{N_0}=e^{650k}\] the initial N's cancel so you get \[.5=e^{650k}\] to get rid of e take the ln of both sides \[ln(.5)=650k\] \[\frac{ln(.5)}{650}=k\]
your rate of decay is \[\frac{ln(.5)}{650}=k\] your equation will be \[N=N_0e^{\frac{ln(.5)t}{650}}\]
@Outkast3r09 : same thing I did. That is why I was confused with @mahmit2012 had done
\[\frac{dP}{dt}=kP\] \[\frac{dP}{dt}-kP=0\] \[e^{\int{kdt}}=e^{kt+c}=ce^{kt}\]
you integrate the left side you'll just get that and the right side will be P

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