After you find $A$, plug $t=2$ and $C=9$ into your new equation. Subtract the baseline $3$ from both sides so that only the exponential term is on the right.

Once you reach an equation of the form $\text{(number)} = e^{\text{(expression in }k)}$, take the natural log (ln) of both sides. Use the fact that $\ln(1/2) = -\ln 2$ to solve for $k$.

Your work should give you a specific value for $A$ and for $k$. Look at the answer choices and see which one uses the same coefficient in front of $e$ and the same factor multiplying $t$ in the exponent.

In Desmos, add a table or individual points for the measured concentrations: $(0,15)$ and $(2,9)$. These represent the two times and their corresponding $C$ values.

Type each option into Desmos one at a time, for example:

• y = 12*exp(-ln(2)*x) + 3
• y = 9*exp(-(ln(2)/2)*x) + 3
• y = 12*exp(-(ln(2)/2)*x) + 3
• y = 9*exp(-ln(2)*x) + 3 (Use ln(2) for $\ln 2$ and exp(...) for $e^{(...)}$.)

For each equation you graph, see whether its curve passes exactly through both plotted points $(0,15)$ and $(2,9)$. The equation whose graph goes through both points corresponds to the correct model.

The model is

At time $t=0$, the concentration is $15$ mg/L, so plug in $t=0$ and $C=15$:

Since $e^0 = 1$, this becomes

so

Now the equation is

This tells you the coefficient in front of the exponential must be $12$, not $9$.

At time $t=2$, the concentration is $9$ mg/L. Plug $t=2$ and $C=9$ into

Subtract $3$ from both sides to isolate the exponential term:

Now divide both sides by $12$:

This equation will let you solve for $k$.

From the previous step, you have

Take the natural logarithm (ln) of both sides:

Recall that $\ln\left(\frac{1}{2}\right) = -\ln 2$, so

Divide both sides by $-2$ to solve for $k$:

So the exponent in the model is $-\dfrac{\ln 2}{2}t$.

Substitute $A = 12$ and $k = \dfrac{\ln 2}{2}$ back into the model

This gives

Comparing with the answer choices, this equation matches choice C.