aquarium polluted water differential equation An aquarium contains 10 gallons of polluted water.A filter is attached to this aquarium which drains off the polluted water at the rate of 5 gallons per hour.and replaces it at the same rate by pure water.How long does it take to reduce the pollution to half of its initial level?
ans is 2log2.
I did not understand 2nd line.If i take x gallon polluted water at any time t  then dx/dt= -5
which is not correct as answer contains log term means some proportionality with x is there.
Please help me solve it.
 A: I think they mean
$$\frac{dy}{dt} = -\frac{y}{2}$$
That is, the rate at which the filter drains and replaces water is $1/2$ the amount of water in the aquarium.  Then
$$y(t) = y(0) \, e^{-t/2}$$
To find the time it takes to drain and replace $1/2$ the water in the aquarium, take logs of both sides:
$$1/2 = e^{-t/2} \implies t = 2 \log{2}$$
A: The differential equation is not $x'(t)=-5$, because the rate at which the pollution leaves is proportional to the pollution in the tank. If you assume the water is perfectly mixed, taking $5$ gallons out will take half of the remaining pollution out (since the tank is $10$ gallons). For example, if there are $3$ gallons of pollution left, and the water is perfectly mixed, then taking out $5$ gallons removes $1.5$ gallons of pollution and $3.5$ gallons of pure water. 
You can carry this observation over to the infinitesimal level: the rate at which the pollution leaves the tank is proportional to half the pollution in the tank. In symbols:
$$x'(t)=-\frac{1}{2}x(t).$$
Solving this, we get $$x(t)=ce^{-t/2}$$ 
for some constant $c$, which we determine by using the information that $x(0)=10$.
Solving, we find $c=10$, so $$x(t)=10e^{-t/2}.$$
We want half of the pollution to remain, or $5$ gallons. We solve for $t$ in
$$5=10e^{-t/2}\Rightarrow 1/2=e^{-t/2}.$$
Taking logarithms, we find $t=2\log 2$.
A: The misleading part is the statement that you are removing "polluted water".  The rate of removing this water is clearly constant. But your goal is to remove the pollutant, the stuff in the water, and its concentration goes down from the first moment.
Assume that at some time $t$ the tank contains $y$ pollutant. It could be grams of salt, ounces of ammonia or whatever. As you pump out a little of this polluted water in a short time $dt$, the amount of $y$ changes. Since the $y$ of pollutant is spread out over the whole 10 gallons, the concentration is $\frac{y}{10}$.  You are removing 5 gallons of this polluted water per hour, so in a time $dt$ you remove a volume $5\times dt$.  So the change in the amount of y, $dy$, is given by$$dy=-\frac{y}{10}\times 5 \times dt $$and the solution to this equations proceeds as above.
