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Chapter 4: Problem 14

Suppose a colony of bacteria has a continuous growth rate of \(40 \%\) per hour. If the colony contains 7500 cells now, how many did it contain three hours ago?

Short Answer

Expert verified
Three hours ago, the colony contained approximately 4083 cells.

Step by step solution

01

Write the exponential growth formula

The general formula for exponential growth is: \(N(t) = N_0 * (1 + r)^t\), where: - \(N(t)\) is the number of cells at time t, - \(N_0\) is the initial number of cells, - r is the growth rate, - t is the time in hours.
02

Substitute the given values

In this problem, we're given \(N(t) = 7500\) cells, the growth rate r = 0.4 (40% expressed as a decimal), and t = -3 hours because we want to find the number of bacteria three hours ago. Now, substitute these values into the exponential growth formula: \(7500 = N_0 * (1 + 0.4)^{-3}\).
03

Solve for N_0

Now we need to solve the equation for \(N_0\): \(7500 = N_0 * (1.4)^{-3}\). First, calculate the value of \((1.4)^{-3}\): \((1.4)^{-3} = 0.18367\). Now, the equation is: \(7500 = N_0 * 0.18367\). Now, divide both sides of the equation by 0.18367 to isolate \(N_0\): \(N_0 = \frac{7500}{0.18367} = 4082.7\).
04

Round the result and give the answer

Since the number of bacteria must be a whole number, we will round the result to the nearest whole number: \(N_0 \approx 4083\). So, three hours ago, the colony contained approximately 4083 cells.

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