Forming equations for exponential growth/decay questions

Question

Problem

Dry cleaners use a cleaning fluid that is purified by evaporation and condensation after each cleaning cycle. Every time the fluid is purified, 2.1% of it is lost. The fluid has to be topped up when half of the original fluid remains.

a) Create a model which represents this situation.
b) After how many cycles will the fluid need to be topped up?

Progress

I am thinking that I will need to use something like $y=ca^x$ where $c$ is the initial amount and a equals the decay factor. However I am not certain if this is correct since an initial amount is not given.

$100$ was the first number to come to mind for the initial amount, I just didn't know if it'd have any influence if the number was different for the initial amount just merely because of part b) asking how many cycles.

Answer 1

After our comment conversation, we see that the equation would be $$\text{amount}=\text{initial}(0.979)^x$$

And to see how many cycles it takes to get to half the initial amount would be $$\frac{c}{2}=c(0.979)^x\\\frac{1}{2}=0.979^x$$

Answer 2

Let the initial volume of the container be $V_0$ and the density be $\rho$.

Let the evaporation and condensation be uniform and that 2.1% of the volume is lost everytime the purifying process is over.

Thus the model is $${\rho\times(\dot V_0 - \dot V_1)} = 0.021*\rho\times\dot V_0$$

Cancelling $\rho$, and converting the volumetric rate to volume,

You get$$ V_1 = V_0 - 0.021V_0 = 0.979V_0$$

After the second cycle, $$V_2 = V_1(1-0.021) = V_0*(0.979)*(0.979) = 0.979^2V_0$$

After n cycles, $$V_n = 0.979^nV_0$$

And if $$V_n = 0.5V_0 => 0.5V_0 = 0.979^nV_0$$

$$log(0.5) = nlog(0.979) => n = \frac{log(0.5)}{log(0.979)}$$

$$n = 32.69 = 33\text{ cycles}$$