If a quantity doubles every 6 hours, what is the growth factor (the base) and what should the exponent be counting: hours, or 6-hour intervals?

Try plugging $t=0$ and $t=6$ into each option. Which function gives 8,000 at $t=0$ and exactly 16,000 at $t=6$?

In Desmos, define four functions using $t$ as the variable:

• $N_1(t) = 8000 \cdot 2^t$
• $N_2(t) = 8000 \cdot 2^{t/6}$
• $N_3(t) = 8000 \cdot 1.06^t$
• $N_4(t) = 8000 \cdot 1.06^{t/2}$.

Use the table feature or tap on each graph at $t=0$ and note the $N$-value. Eliminate any functions that do not give $N = 8000$ when $t=0$.

Now look at each remaining function’s value at $t=6$. The correct model is the one whose $N$-value at $t=6$ is exactly double its value at $t=0$ (that is, it should give 16,000 when it started at 8,000).

We are told:

• At 9:00 a.m. (this is $t=0$), there are 8,000 infected cells.
• The number of infected cells doubles every 6 hours.

An exponential growth model has the form

Here, the initial amount is 8,000, the growth factor is 2 (because it doubles), and the exponent must represent how many 6-hour intervals have passed.

If $t$ is measured in hours, then the number of 6-hour intervals that have passed is

For example:

• After 6 hours, $t=6$, and $t/6 = 1$ interval has passed.
• After 12 hours, $t=12$, and $t/6 = 2$ intervals have passed.

This means the exponent on 2 should be $t/6$, not just $t$.

Using the general form $N = 8{,}000 \cdot 2^{t/6}$, verify it matches the description:

• At $t=0$ (9:00 a.m.):

which matches the given starting amount.

• At $t=6$ (6 hours later):

which is exactly double 8,000.

• At $t=12$ (12 hours later):

which is double 16,000.

So this function correctly starts at 8,000 and doubles every 6 hours.

From the answer choices, the one that uses initial value 8,000, growth factor 2, and exponent $t/6$ (to represent one doubling every 6 hours) is