If the initial population at $t=0$ is 800 and it keeps doubling, you should start with something like $B = 800 \cdot 2^{\text{(something with t)}}$. Focus on what must go in the exponent.

You want the exponent on 2 to go up by 1 every time 6 hours pass. Which expression, $6t$ or $t/6$, increases by 1 when $t$ increases by 6?

Use the information from the problem: at $t=0$ the population is 800, and at $t=6$ it should be 1600. Plug $t=0$ and $t=6$ into each choice and see which one gives those values.

In Desmos, enter four functions, one for each option. For example, you can type:

• A(t) = 800*2^(6*t)
• B(t) = 800*2^(t/6)
• C(t) = 1600*2^t
• D(t) = 1600*2^(t/6) If Desmos prefers x instead of t, replace t with x in all functions.

Use Desmos to evaluate each function at $t=0$ and $t=6$ (for example, by typing A(0), A(6), B(0), B(6), etc., or by adding a table for each function with $t$-values 0 and 6). You want the function that gives 800 at time 0 and 1600 at time 6.

Whichever function in Desmos has output 800 when $t=0$ and 1600 when $t=6$, and continues to double every additional 6 hours, corresponds to the equation you should choose.

"Doubles every 6 hours" means the population is multiplied by 2 repeatedly after equal time intervals. That is an exponential situation, so the model should look like

where $a$ is the initial amount (when $t=0$).

If a population doubles every 6 hours, then each time $t$ increases by 6, the exponent on 2 should increase by 1.

To make the exponent increase by 1 when $t$ increases by 6, we use $t/6$ in the exponent:

• When $t=0$, exponent $=0/6=0$.
• When $t=6$, exponent $=6/6=1$.
• When $t=12$, exponent $=12/6=2$.

So the exponent must be $t/6$ rather than $6t$. The model now has the form

We are told that at time $t=0$ hours, the population is 800 bacteria. Plug $t=0$ into the model and set $B=800$:

So $a=800$. The full model is

Looking at the answer choices, this matches choice B, so $B = 800(2)^{t/6}$ is the correct equation.