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.
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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**.
