Since the model has form \(P(t) = Ae^{kt}\text{,}\) we know that \(P(0) = A\text{.}\) Because we are given that \(P(0) = 2.475\text{,}\) this shows that \(A = 2.475\text{.}\) To find \(k\text{,}\) we use the fact that \(P(10) = 4.298\text{.}\) Applying this information, \(A = 2.475\text{,}\) and the form of the model, \(P(t) = Ae^{kt}\text{,}\) we see that
\begin{equation*}
4.298 = 2.475 e^{k \cdot 10}\text{.}
\end{equation*}
To solve for \(k\text{,}\) we first isolate \(e^{10k}\) by dividing both sides by \(2.475\) to get
\begin{equation*}
e^{10k} = \frac{4.298}{2.475} \text{.}
\end{equation*}
Taking the natural logarithm of each side, we find
\begin{equation*}
10k = \ln \left( \frac{4.298}{2.475} \right)\text{,}
\end{equation*}
and thus \(k = \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right) \approx 0.05519\text{.}\)
To determine how long it takes for the population to reach \(1\) billion cells, we need to solve the equation \(P(t) = 1000\text{.}\) Using our preceding work to find \(A\) and \(k\text{,}\) we know that we need to solve the equation
\begin{equation*}
1000 = 2.475 e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t}\text{.}
\end{equation*}
We divide both sides by \(2.475\) to get \(e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t} = \frac{1000}{2.475}\text{,}\) and after taking the natural logarithm of each side, we see
\begin{equation*}
\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t = \ln \left( \frac{1000}{2.475} \right)\text{,}
\end{equation*}
so that
\begin{equation*}
t = \frac{10 \ln \left( \frac{1000}{2.475} \right)}{\ln \left( \frac{4.298}{2.475} \right)} \approx 108.741\text{.}
\end{equation*}