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Question
the doubling period of a bacterial pop. is 20 mins. at time t=80 mins, the pop. was 50000. find the pop. after 5 hours.
Step1: Identify the formula for exponential growth
The formula for exponential growth with doubling time is \( P(t) = P_0 \times 2^{\frac{t}{T}} \), where \( P_0 \) is the initial population, \( t \) is the time elapsed, and \( T \) is the doubling time.
Step2: Determine the given values
We know that at \( t = 80 \) minutes, \( P(80) = 50000 \), and the doubling time \( T = 20 \) minutes. We need to find the population after \( t = 5 \) hours. First, convert 5 hours to minutes: \( 5 \times 60 = 300 \) minutes. Now, we need to find the initial population \( P_0 \) (at \( t = 0 \)) or use the given \( P(80) \) to find \( P(300) \).
First, let's find the number of doubling periods between \( t = 80 \) and \( t = 300 \). The time elapsed from 80 minutes to 300 minutes is \( 300 - 80 = 220 \) minutes. The number of doubling periods \( n = \frac{220}{20} = 11 \). Wait, alternatively, we can express \( P(t) \) in terms of \( P(80) \). Let \( t' = t - 80 \), so \( P(t) = P(80) \times 2^{\frac{t'}{T}} \).
Step3: Calculate the population at \( t = 300 \) minutes
First, find \( t' = 300 - 80 = 220 \) minutes. Then, the number of doubling periods in \( t' \) is \( \frac{220}{20} = 11 \). So, \( P(300) = 50000 \times 2^{11} \).
Calculate \( 2^{11} = 2048 \). Then, \( 50000 \times 2048 = 50000 \times 2000 + 50000 \times 48 = 100000000 + 2400000 = 102400000 \).
Wait, let's check the time again. Wait, the problem says "the doubling period of a bacterial pop. is 20 mins. At time t=80 mins, the pop. was 50000. Find the pop. after 5 hours."
5 hours is 300 minutes. So the time from t=80 to t=300 is 300 - 80 = 220 minutes. The number of doubling periods is 220 / 20 = 11. So the population grows by a factor of \( 2^{11} \) from t=80 to t=300.
So \( P(300) = 50000 \times 2^{11} \).
\( 2^{11} = 2048 \), so \( 50000 \times 2048 = 102400000 \).
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