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Chapter 4: Problem 13

Suppose a colony of bacteria has a continuous growth rate of \(30 \%\) per hour. If the colony contains 8000 cells now, how many did it contain five hours ago?

Short Answer

Expert verified
The bacterial colony contained approximately 29,703 cells five hours ago.

Step by step solution

01

Identify given information

We are given the following information: - Growth rate (\(r\)): 30%, which can be converted to a decimal as 0.30 - Current population (\(P_t\)): 8000 cells - Time period (\(t\)): 5 hours ago (we will use a negative value for time since we are going back in time)
02

Convert the growth rate to a decimal

We need to convert the given growth rate from a percentage to a decimal for calculations. To do this, divide the percentage by 100: \[r = \frac{30}{100} = 0.30\]
03

Insert values into exponential growth formula and solve

Using the exponential growth formula, we have \(P_t = P_0 \cdot (1 + r)^t\). Our goal is to find the initial population (\(P_0\)), so we will rearrange the equation to solve for \(P_0\). \[ \begin{aligned} P_t &= P_0 \cdot (1 + r)^t\\ \Rightarrow P_0 &= \frac{P_t}{(1+r)^t} \end{aligned} \] Now, substitute the given values into the equation: \[ \begin{aligned} P_0 &= \frac{8000}{(1 + 0.30)^{-5}}\\ P_0 &= \frac{8000}{(1.30)^{-5}}\\ P_0 &= 8000 \cdot (1.30)^{5}\\ P_0 &= 8000 \times 3.71293 \end{aligned} \]
04

Calculate the initial population

After substituting the given values, now calculate the initial population: \[ \begin{aligned} P_0 &= 8000 \times 3.71293\\ P_0 &= 29,703.44 \end{aligned} \] Since the number of bacterial cells must be a whole number, we can round the result to the nearest whole number: \(P_0 \approx 29,703\)
05

Conclusion

The bacterial colony contained approximately 29,703 cells five hours ago.

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