Hey all, need help.
How do you find k once you have intergrated 1/NdN=kdt?
I've got to:
N=Ae^kt
And i'm told:
120=Ae^20k
Thanks
Trev
Hello, Trev!
Did they give us an initial condition?
$\displaystyle \frac{dN}{N} \:=\:k\,dt$ . . . and we are given: .$\displaystyle N(20) \:=\:120$
We have: .$\displaystyle \frac{dN}{N} \;=\;k\,dt$
. . Integrate: .$\displaystyle \ln N \;=\;kt + c$
. . Then: .$\displaystyle N \;=\; e^{kt+c} \;=\;e^{kt}\cdot e^c\;=\; Ce^{kt} $
Assume that, when $\displaystyle t = 0,\;N \:=\:N_o$
. . Then we have: .$\displaystyle N_o \:=\: Ce^0\quad\Rightarrow\quad C \:=\:N_o$
. . The function (so far) is: .$\displaystyle N \;=\;N_oe^{kt}$
We are told that: .$\displaystyle N(20) \:=\:120$
. . Then we have: .$\displaystyle 120 \;=\;N_oe^{20k}\quad\Rightarrow\quad e^{20k} \;=\;\frac{120}{N_o}$
. . Take logs: .$\displaystyle \ln\left(e^{20k}\right) \;=\;\ln\left(\frac{120}{N_o}\right) \quad\Rightarrow\quad 20k(\ln e) \;=\; \ln\left(\frac{120}{N_o}\right)$
. . Hence: .$\displaystyle k \;=\;\frac{1}{20}\ln\left(\frac{120}{N_o}\right) $
Therefore: .$\displaystyle N(t) \;=\;N_oe^{\left[\frac{1}{20}\ln(\frac{120}{N_o})\right]t} $
Edit: Darn, just realized that this identical to kalagota's solution.
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