After you set $1.5\times10^6$ equal to the fraction, multiply both sides by the denominator $1+29e^{-0.3t}$, then simplify the numbers $3\times10^6$ and $1.5\times10^6$.

Get the term with $e^{-0.3t}$ alone on one side by moving constants to the other side, then divide to get an equation of the form $e^{(\text{something})} = \text{number}$.

Once you have $e^{(\text{something})} =$ a constant, take the natural log (ln) of both sides, use $\ln(e^x)=x$, and then solve the resulting linear equation for $t$; finally, approximate with your calculator and compare to the choices.

In Desmos, type B(t) = 3*10^6/(1+29*e^(-0.3*t)). Desmos will graph this as a function of $t$ (you can use x instead of t if you prefer: y = 3*10^6/(1+29*e^(-0.3*x))).

Add a new line and enter y = 1.5*10^6 to create a horizontal line representing $1.5\times10^6$ bacteria.

Zoom out or adjust the window so you can see where the logistic curve and the horizontal line intersect. Tap or click on the intersection point; Desmos will display its coordinates. The $x$- (or $t$-) coordinate of this intersection is the time in hours when the population reaches $1.5\times10^6$.

We are told that the population reaches $1.5\times10^6$ bacteria at some time $t$, and the model is

So set $B(t)$ equal to $1.5\times10^6$ and write:

Multiply both sides by the denominator to get rid of the fraction:

Now divide both sides by $1.5\times10^{6}$:

Then subtract 1 from both sides:

Finally, divide both sides by 29:

To undo an exponential with base $e$, take the natural logarithm (ln) of both sides:

Using the property $\ln(e^x) = x$, the left side simplifies to $-0.3t$:

Also, $\ln\left(\frac{1}{29}\right) = -\ln(29)$, so

Divide both sides by $-0.3$ to isolate $t$:

Use a calculator (as allowed on this part of the SAT) to evaluate $\ln(29)$, then divide by $0.3$.

• $\ln(29) \approx 3.37$.
• $$$$

\frac{3.37}{0.3} \approx 11.2.