From $t=0$ to $t=2$, how does the concentration change? Compute the ratio $\dfrac{c(2)}{c(0)}$ and think about how that ratio appears in an exponential function as the base or as a repeated factor.

Once you know the factor that takes you from $t=0$ to $t=2$, ask: how many of those time intervals are in $t$ hours? How should that number show up in the exponent?

Type these into Desmos as separate functions:

• f(t) = 48(1.5)^t
• g(t) = 48(0.75)^{t/2}
• h(t) = 36(0.75)^t
• k(t) = 48(0.75)^{2t} Adjust the viewing window so you can see $t$ from 0 to about 3 and $c$ from 0 to about 60.

In Desmos, either use the table feature for each function (click the gear icon and choose "Add table") or tap on the graphs at $t=0$ and $t=2$ to see the corresponding $c$-values. Identify which function gives $c=48$ when $t=0$ and $c=36$ when $t=2$; that function represents the correct model.

For an exponential model, a common form is

When $t=0$, any exponential term $b^{\text{(something)}}$ becomes $b^0 = 1$, so $c = a$.

From the table, when $t=0$, $c=48$. That means the starting amount $a$ must be 48, so any correct equation must give $c=48$ when $t=0$.

Check each option at $t=0$:

• A) $c = 48(1.5)^0 = 48(1) = 48$ ✔
• B) $c = 48(0.75)^{0} = 48(1) = 48$ ✔
• C) $c = 36(0.75)^0 = 36(1) = 36$ ✘ (does not match 48)
• D) $c = 48(0.75)^{0} = 48(1) = 48$ ✔

So choice C is impossible. Only A, B, and D still match $c=48$ at $t=0$.

From the table:

• At $t=0$, $c=48$.
• At $t=2$, $c=36$.

The factor that $c$ is multiplied by over 2 hours is

So in each 2-hour interval, the concentration is multiplied by $0.75$ (a decay factor less than 1).

We know:

• Start value (when $t=0$): $48$.
• In each 2-hour period, $c$ is multiplied by $0.75$.

The exponent must count how many 2-hour intervals are in $t$ hours. The number of 2-hour intervals in $t$ hours is $t/2$, so the model should be

This matches answer choice B.