How many grams of salt are in the tank after 100 minutes have elapsed? This is a question from a GRE mock exam. A similar question has been asked GRE Math subject test #61 Differential Equations, but this does not address my question. The general method of solving these questions are rather well-known, for instance First Order differential equations mixing problem.

A tank initially contains a salt solution of 3 grams of salt dissolved in 100 liters of water. A salt solution containing $0.02$ grams of salt per liter of water is sprayed into the tank at a rate of 4 liters per minute. The sprayed solution is continually mixed with the salt solution in the tank, and the mixture flows out of the tank at a rate of 4 liters per minute. If the mixing is instantaneous, how many grams of salt are in the tank after 100 minutes have elapsed?

Here was my solution. We let $x(t)$ be the grams of salt that are in the solution. We have
$$ x + dx = x + \left( 0.02\times 4 - 4 \times \frac{x}{100}\right)dt .$$
From this, we get $$ \frac{dx}{dt} =  0.02\times 4 - 4 \times \frac{x}{100} = 0.08-0.04x.$$
Upon rearranging, $$\frac{1}{0.08-0.04x} dx = dt \implies \frac{\log |0.08-0.04x|}{-0.04 }= t - C ,$$
and noting that $x = 3$ at $t = 0$ gives $C = \log (0.04)/0.04$. Therefore,
$$\frac{\log |0.08-0.04|}{-0.04} = 100 -  \log (0.04)/0.04 \implies |2-x| = e^{-4}.$$
So which is the answer, from $x = 2 \pm e^{-4}$? The answer sheet indicates that there is only one answer. (Also, did I go wrong anywhere within my solution?)
 A: Please note that initially there is $0.03$ gram of salt per liter and the solution being sprayed contains $0.02$ gram of salt per liter. The mixture flows out at the same rate and so contains more salt than the solution being sprayed in. In summary, the amount of salt in the tank is coming down with time (and goes down to $0.02$ gram per liter at infinity). So from your working,
$\frac{dx}{dt} =  0.02\times 4 - 4 \times \frac{x}{100} = - 0.04 (x - 2)$ and note that $x - 2 \gt 0$
So, $ \ \ln (x-2) =  C - 0.04 t$
At $t = 0, x = 3$ so $C = 0$
At $t = 100,  \ \ln (x-2) =  - 4 \implies x = 2 + e^{-4}$
A: You are loosing this information because when calculating $C$ from the starting values you still have the absolute value, so you’re loosing information.
You have
$$\log|0.08-0.04x| = -0.04(t-C) = -0.04t + C' $$
Then
$$ 0.08-0.04x = \pm \mathrm e^{-0.04t+C'} = \pm \mathrm e^{-0.04t}C'' = \mathrm e^{-0.04t}C''' $$
so
$$ 2-x=\mathrm e^{-0.04t}25C''' = \mathrm e^{-0.04t}C''''$$
With the starting values we get
$$ -1 = C'''' $$
Thus we get
$$ x = 2 + \mathrm e^{-0.04t}$$
