Question

Could someone help me understand exponential growth and decay?

Answer 1

Here is a sample question:

The rate of decay of a radioactive substance is proportional to the amount of substance present. What is the half-life of a radioactive substance if it takes 2 years for one-third of the substance to decay?

Answer 2

I have the most trouble plugging into the formula.

Answer 3

ok.. so the formula is something like

\[P = A_{0}e^{-kt}\]

is that right..?

Answer 4

Looks correct, yeah

Answer 5

Yeah I guess like if we start out with some amount of sample let's call it Ao it's going to eventually disappear. like I guess if we had the rate of formation of A is equivalent to -k[A]0 dt.

\[-k[A_{o}]dt = dA\]

\[\int\limits \frac{ dA }{ [A_{o}]} = \int\limits -k~dt\]

\[\ln(\frac{ [A] }{ [A]_{o} }) = -kt\]

yeah we play around with this and solve for the final concentration of A

\[e^{\ln(\frac{ [A] }{ [A]_{o} })} = e^{-kt}\]

\[\frac{ [A] }{ [A]_{o} } = e^{-kt}\]

we do some simple moving around

\[[A] = [A_{o}]e^{-kt}\]

so like yeah we end up getting an exponential decay function, where A is the final concentration you end up with

Answer 6

you also could say that \[[A] =\frac{ 1 }{ 3 }[A_{o}]\]

Answer 7

ok... so the explanation is that

\[A_{0} = Initial ~amount\]

t = time and -k is the constant of decay..

you are also told that the initial amount reduces to 1/3 after 2 years

so you can use this to find k, if you let the initial quantity be 1

so

\[\frac{1}{3} = 1 \times e^{-k \times 2}\]

so to solve for k you need to take the log of both sides... does that make sense..?

Answer 8

So its about being able to manipulate the formula to get what you need

Answer 9

exactly

Answer 10

k is the constant of decay so you could actually find that and then find the half life easy

Answer 11

because you don't need to know the exact concentrations or amount of sample, all you need is the fact that A = 0.5[A_{0}] and you can solve it for t_{1/2}

Answer 12

that's correct.... the 1st step is normally to find the constant of decay or growth

Answer 13

ok so 1 is arbitrary in this question.
I see what you did here. It's just comfusing, seems like a really dynamic formula

Answer 14

the normal step is to use logs to find the value of k

Answer 15

\[k=-\frac{ \ln \frac{ 1 }{ 3 } }{ 2 }\]

Answer 16

yeah :)

Answer 17

and now I plug back in and solve for t?

Answer 18

Another thing too. you see why we don't necessarily need the starting concentration

\[[\cancel\A_{o}]e^{-kt_{\frac{ 1 }{ 2 }}}= \frac{ 1 }{ 2 }[\cancel\A_{o}]\]

\[[A] = \frac{ 1 }{ 2 }[A_{o}]\]

Answer 19

Does the equation on the right hand side of the equal sign represent what is left over after the half life time?

Answer 20

yeah that's how much sample you have left over

Answer 21

gotcha. So how do I solve for the half life now that I have the constant of decay?

Answer 22

Get t by itself :)

Answer 23

oh okay, lets see

Answer 24

yeah like don't plug anything in yet until after you've re-arranged it

Answer 25

ok it looks kind of crazy though

Answer 26

\[\frac{ 1 }{ 2 } = e ^{-kt}\]

\[\frac{ \ln(\frac{ 1 }{ 2 }) }{ \frac{ \ln(\frac{ 1 }{ 3 }) }{ 2 } }=t\]

Answer 27

\[e^{-kt} = 0.5\]

Answer 28

\[\ln(e^{-kt}) = \ln(\frac{ 1 }{ 2 })\]

Answer 29

Follow you so far

Answer 30

\[-kt = \ln(\frac{ 1 }{ 2 })\]

\[\frac{ \ln(\frac{ 1 }{ 2 }) }{ -k } = t\]


now then and only then do you now plug in the value for k after you're done re-arranging stuff


\[k = \frac{ -\ln(\frac{ 1 }{ 3 } )}{ 2 }\]

Answer 31

so now I plug in k

Answer 32

Keep typing what you're typing, but I got a little lost there. What happened to our t?

Answer 33

\[\frac{ \ln(\frac{ 1 }{ 2 }) }{ \frac{ (\ln(\frac{ 1 }{ 3 } )}{ 2 }} = \frac{ 2*\ln(\frac{ 1 }{ 2 }) }{ \ln(\frac{ 1 }{ 3} )} = k~ ~1.262\]

Answer 34

sorry that should be t

Answer 35

we plug in our value for k and we get a value for t

Answer 36

How did you simplify the natural logs?

Answer 37

Well I didn't actually. I just found their exact value put it into a calculator.

Answer 38

Gotcha. Thanks for taking the time to explain that to me lol. I'll go back over it a few times and practice a few more problems. It's one of the things covered on my test so I'm trying to really get a grasp on it. Thanks again for all of your help!

Answer 39

Make sure you try to derive it yourself

Answer 40

that should help more

Answer 41

I'm not sure if I'm reading the example problem correctly or not...
but it seems like it's talking about the substance decaying `by 1/3` in 2 years,
not decaying `to 1/3` in 2 years.

Just an observation :d