Question

sections 4.5 - 4.6
score: 8/15 answered: 8/15
question 9
a bacteria culture initially contains 1500 bacteria and doubles every half hour.
find the size of the bacterial population after 60 minutes.
find the size of the bacterial population after 5 hours.

Explanation:

Step1: Analyze the growth pattern

The bacteria doubles every half - hour. The initial population \(P_0 = 1500\). The general formula for exponential growth when the quantity doubles every time period \(t_0\) is \(P(t)=P_0\times2^{\frac{t}{t_0}}\), where \(t\) is the total time elapsed and \(t_0\) is the time it takes for the quantity to double. Here, \(t_0 = 0.5\) hours (or 30 minutes).

Step2: Find the population after 60 minutes (1 hour)

First, convert 60 minutes to hours, \(t = 1\) hour. Since \(t_0=0.5\) hours, we substitute into the formula \(P(t)=P_0\times2^{\frac{t}{t_0}}\).
\(P(1)=1500\times2^{\frac{1}{0.5}}\)
\(=1500\times2^{2}\)
\(=1500\times4\)
\( = 6000\)

Step3: Find the population after 5 hours

For \(t = 5\) hours, we use the same formula \(P(t)=P_0\times2^{\frac{t}{t_0}}\)
\(P(5)=1500\times2^{\frac{5}{0.5}}\)
\(=1500\times2^{10}\)
We know that \(2^{10}=1024\), so \(P(5)=1500\times1024 = 1536000\)

Answer:

The size of the bacterial population after 60 minutes is \(\boldsymbol{6000}\), and after 5 hours is \(\boldsymbol{1536000}\)