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Chapter 3: Problem 29

Suppose a colony of bacteria has doubled in five hours. What is the approximate continuous growth rate of this colony of bacteria?

Short Answer

Expert verified
The approximate continuous growth rate of the colony of bacteria is \(k \approx 0.1386\) per hour.

Step by step solution

01

Identify the given information

The problem states that the colony of bacteria has doubled in five hours. So, we know: Initial amount (A鈧�) = 1 (assuming the initial amount as 1) Final amount (A) = 2 (doubled in size) Time (t) = 5 hours
02

Use the formula for continuous compound growth

The formula for continuous compound growth is: A = A鈧� * e^(kt) Where A is the final amount, A鈧� is the initial amount, k is the continuous growth rate, and t is the time elapsed in hours. We are trying to find the value of k.
03

Plug in the given values and solve for k

Substituting the given values into the formula, we get: 2 = 1 * e^(5k) Now isolate the variable k by following these steps: 1. Divide by the initial amount 2 = e^(5k) 2. Apply the natural logarithm (ln) to both sides of the equation ln(2) = ln(e^(5k)) 3. Simplify the right side of the equation using the logarithmic property ln(a^b) = b * ln(a) ln(2) = 5k * ln(e) Since we know ln(e) = 1, we can simplify further: ln(2) = 5k 4. Finally, divide by 5 to find the value of k k = (ln(2))/5
04

Calculate the approximate value of k

Using a calculator, we get the approximate value of k: k 鈮� 0.1386 Therefore, the approximate continuous growth rate of the colony of bacteria is 0.1386 per hour.

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