im stuck on ii(i) find k3) A culture of bacteria is growing at a rate proportional to the number,

$\displaystyle N$, of bacteria present, so that: $\displaystyle \frac{dN}{dt} = kN$

where $\displaystyle t$ is the time elapsed in hours and $\displaystyle k$ is a constant.

Initially, the culture contains 3000 bacteria,

and after three hours, it contains 9000 bacteria.

(i) Find k.

Thereafter (that is, for t > 3), the culture is washed with a solution

which is harmful to the bacteria, and the bacteria are killed off

at a constant rate of 4500 per hour.

(ii) Find the number of bacteria in the culture for any time t > 3.

(iii) After how much washing time (hours/minutes) do the bacteria die out?

I got $\displaystyle k \:=\: \frac{\ln(3)}{3}$

Which gives:$\displaystyle N(t) \;=\;3000e^{(\frac{ln(3)}{3})t}$...............................$\displaystyle (=\;3000\cdot3^{\frac{t}{3}})$

ii) find the population after 3 hrs (i.e. t>3)

Ok. so first I thought it was $\displaystyle N(t) = 3000e^{\frac{ln3}{3}t} - 4500t$

then i thought it this: $\displaystyle N(t) = (9000-4500)e^{\frac{ln3}{3}t}$

Now im thinking $\displaystyle N(t) = 9000e^{\frac{ln3}{3}t} - 4500t$

yeh so any tips...is it the last one?