A culture initially has \( P_{0} \) number of Acinetobacter baumannii bacteria. At \( t=1 \) hour, the number of bacteria is measured to be \( \frac{4}{3} P_{0} \). If the rate of growth is proportional to the number of bacteria \( P(t) \) present at time \( t \), determine the time neccessary for the number of bacteria to quadruple.

To solve this problem, we start by noting that the growth of the bacteria is described by the exponential growth model. This is expressed with the following differential equation: \[ \frac{dP}{dt} = kP \] where \( k \) is the growth constant. The solution to this equation gives us: \[ P(t) = P_{0} e^{kt} \] At \( t=1 \) hour, we know the number of bacteria is \( \frac{4}{3} P_{0} \). Plugging this into our equation: \[ \frac{4}{3} P_{0} = P_{0} e^{k \cdot 1} \] Dividing both sides by \( P_{0} \) (assuming \( P_{0} > 0 \)) gives us: \[ \frac{4}{3} = e^{k} \] Taking the natural logarithm of both sides yields: \[ k = \ln\left(\frac{4}{3}\right) \] Next, we want to determine the time required for the population to quadruple; this means we want to find \( t \) such that: \[ P(t) = 4P_{0} \] Substituting into our growth equation: \[ 4P_{0} = P_{0} e^{kt} \] Dividing by \( P_{0} \): \[ 4 = e^{kt} \] Taking the natural logarithm of both sides: \[ \ln(4) = kt \] Substituting in our expression for \( k \): \[ \ln(4) = \ln\left(\frac{4}{3}\right) t \] Now, isolating \( t \) gives us: \[ t = \frac{\ln(4)}{\ln\left(\frac{4}{3}\right)} \] Using a calculator, we can estimate the values: \[ \ln(4) \approx 1.3863 \] \[ \ln\left(\frac{4}{3}\right) \approx 0.2877 \] Thus, \[ t \approx \frac{1.3863}{0.2877} \approx 4.83 \text{ hours} \] Therefore, the time necessary for the number of bacteria to quadruple is approximately **4.83 hours**.