your value of is correct ...
see new post below ...
A culture of bacteria is growing at a rate proportional to the number,
N, of bacteria present, so that
dN/dt= kN
where t is the time elapsed in hours and k is a constant. Initially, the culture contains 3000 bacteria, and after three hours of uninterrupted growth, it contains 9000 bacteria.
(i) Find k.
Thereafter (that is, for t > 3) the culture is carefully washed with a solution
which is harmful to the bacteria, and the bacteria are killed o by the solution
at a constant rate of 75 each minute. [Thus, the rate of change in the number
of bacteria is as described above, but diminished by 4500 per hour.]
(ii) Find the number of bacteria in the culture for any time t > 3.
(iii) After how much washing time (in hours and minutes) do the bacteria die out?
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For (i)
dN/dt=kN
1/N*dN/dt=k
int(1/N)*(dN/dt)*dt=int k dt
int(1/N)*dN=kt+C
ln ㅣNㅣ=kt+C
ㅣNㅣ=e^(kt+c)
ㅣNㅣ=e^c*e^kt
and N=ħe^c*e^kt
=Ae^kt, where A=ħe^c
∴N=Ae^kt
when N(0)=3000,
3000=A
so, N(t)=3000e^kt,
when N(3)=9000
9000=3000e^kt
3=e^3k
3k ln e= ln 3
3k= ln 3
k=(1/3) ln (3), which is 0.3662 in 4 decimal places....
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First, can u plz check whether this answer is right or wrong??
and..i have no idea how to approach to question (ii) and (iii)
Thank you for your kindness