Question

homework assignment 6.7 exponential and logarithmic mod
due friday by 11:59pm points 10 submitting an external tool
homework assignment 6.7 exponential and logarithmic models
score: 5.67/10 answered: 6/10
question 7
the doubling period of a bacterial population is 20 minutes. at time t = 80 minutes, the bacterial
population was 80000.
what was the initial population at time t = 0?
find the size of the bacterial population after 3 hours.
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Explanation:

Step1: Find the initial population

The general formula for exponential growth with a doubling - period \(T\) is \(P(t)=P_0\cdot2^{\frac{t}{T}}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(T\) is the doubling - period.
Given \(T = 20\) minutes and \(t = 80\) minutes, \(P(80)=80000\).
Substitute into the formula: \(80000=P_0\cdot2^{\frac{80}{20}}\).
Since \(\frac{80}{20}=4\), then \(80000 = P_0\cdot2^{4}\).
And \(2^{4}=16\), so \(P_0=\frac{80000}{16}=5000\).

Step2: Find the population after 3 hours

First, convert 3 hours to minutes. Since 1 hour = 60 minutes, 3 hours \(t = 180\) minutes.
Using the formula \(P(t)=P_0\cdot2^{\frac{t}{T}}\) with \(P_0 = 5000\) and \(T = 20\).
Substitute \(t = 180\) into the formula: \(P(180)=5000\cdot2^{\frac{180}{20}}\).
Since \(\frac{180}{20}=9\), then \(P(180)=5000\cdot2^{9}\).
And \(2^{9}=512\), so \(P(180)=5000\times512 = 2560000\).

Answer:

The initial population at \(t = 0\) is \(5000\).
The size of the bacterial population after 3 hours is \(2560000\).