Plug $t=0$ into the general logistic form $P(t)=\dfrac{12000}{1+Ae^{-kt}}$. The population is 3,000 at that time. What equation do you get for $A$, and what does that tell you about the constant in the denominator?

Once you know the constant in the denominator, plug $t=2$ and $P(2)=4800$ into that form to solve for the growth rate $k$, or plug $t=2$ directly into the remaining answer choices to see which one gives 4,800.

In a logistic growth model, as time increases, the population approaches the carrying capacity from below. Should $e^{kt}$ or $e^{-kt}$ appear in the denominator so that the denominator approaches 1 and the population approaches 12,000?

In Desmos, use $x$ in place of $t$. Type each answer choice as a separate function, for example: $y=12000/(1+3e^{0.35x})$, $y=12000/(1+1.5e^{-0.70x})$, $y=3000(1.6)^x$, and $y=12000/(1+3e^{-0.35x})$.

For each graph, tap on the point where $x=0$ (or add a table for each function and include $x=0$). Note the $y$-values; only the model(s) that give $y\approx3000$ at $x=0$ are consistent with the initial population.

For the remaining candidate(s), tap on $x=2$ (or include $x=2$ in the table) and read off the $y$-values. The correct model will give a value close to 4,800 at $x=2$.

Zoom out horizontally to see what happens as $x$ gets large. The correct logistic model should level off near $y=12000$, while an incorrect model may either grow without bound or approach a different limiting value.

A standard logistic growth model with carrying capacity $L$ has the form

Here the carrying capacity is 12,000 cells, so $L = 12000$:

At the initial observation time $t=0$, the population is $P(0) = 3000$.

Substitute $t=0$ and $P(0)=3000$:

Solve for $A$:

So the model must look like

Any correct choice must have numerator 12000 and denominator $1+3e^{-k t}$ for some positive $k$. This already rules out any option with a different constant (like $1.5$) in the denominator.

We also know that after 2 hours, there are 4,800 cells: $P(2) = 4800$.

Using the partially determined model

substitute $t=2$ and $P(2) = 4800$:

Solve for $k$:

Take the natural log of both sides:

So the exponent in the logistic model is approximately $-0.35t$.

Now substitute $A=3$ and $k\approx 0.35$ back into the logistic form:

This function starts at 3,000 cells when $t=0$, reaches about 4,800 cells at $t=2$, and approaches a maximum of 12,000 cells as $t$ becomes large. Among the answer choices, this expression matches option D, so the correct answer is