Problem 41 - Intermediate Algebra

Question

The number of bacteria present in a certain culture after \(t\) hours is given by the equation \(Q=Q_{0} e^{0.3 t}\), where \(Q_{0}\) represents the initial number of bacteria. If 6640 bacteria are present after 4 hours, how many bacteria were present initially?

Step-by-Step Solution

Answer

The initial number of bacteria was approximately 2000.

Step 1: Understand the Given Information

We are provided with the formula \(Q = Q_{0} e^{0.3t}\), where \(Q\) represents the number of bacteria at time \(t\), \(Q_{0}\) is the initial number of bacteria, and \(e\) is the base of the natural logarithm. We need to find \(Q_{0}\) given \(Q = 6640\) at \(t = 4\) hours.

Step 2: Substitute Known Values into the Formula

Substitute the known values \(Q = 6640\) and \(t = 4\) into the equation: \[ 6640 = Q_{0} e^{0.3 imes 4} \]

Step 3: Simplify the Exponential Term

Calculate the exponent: \[ 0.3 \times 4 = 1.2 \]Thus, the equation becomes:\[ 6640 = Q_{0} e^{1.2} \]

Step 4: Solve for \(Q_{0}\)

Divide both sides of the equation by \(e^{1.2}\) to solve for \(Q_{0}\):\[ Q_{0} = \frac{6640}{e^{1.2}} \]Calculate \(e^{1.2}\) using a calculator. \(e^{1.2} \approx 3.3201\).Thus, \(Q_{0} \approx \frac{6640}{3.3201} \approx 2000 \).

Step 5: Conclusion

The initial number of bacteria, \(Q_{0}\), was approximately 2000.

Key Concepts

Bacteria Growth

Bacteria, like many other microorganisms, often grow at rates that can be best described by exponential functions. Imagine these tiny creatures multiplying rapidly, doubling their numbers in a matter of hours. This phenomenon is known as exponential growth. In the context of our exercise, the number of bacteria increases according to the formula: \[ Q=Q_{0} e^{0.3 t} \]Here's how it works:

• **\(Q\)** represents the number of bacteria present after a certain time \(t\).
• **\(Q_{0}\)** is the initial amount of bacteria we started with.
• The term **\(e\)** is the base of natural logarithms, approximately equal to 2.718.

The exponential term \(e^{0.3t}\) captures the rapid growth rate. In simple terms, bacteria grow fast because each bacterium can divide to become two, then four, then eight, and so on. As a result, the number of bacteria increases rapidly over time. Understanding this helps us appreciate why controlling bacteria effectively is crucial in fields like healthcare and food safety.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies given initial conditions. In simpler terms, it's about discovering the conditions at the starting point of a system based on how the system behaves at a later time.In our exercise, the equation is:\[ Q = Q_{0} e^{0.3t} \]Here, the challenge is to find the starting number of bacteria, \(Q_{0}\) (which we call the initial condition), given:

• The number of bacteria at a later time, \(Q = 6640\) after \(t = 4\) hours.

By substituting these values into the formula and solving for \(Q_{0}\), we reverse-engineer the growth process. This problem-solving approach is like peeling back the layers of time to find the initial scenario, which is essential in many scientific and engineering applications.

Natural Logarithms

To thoroughly understand the exponential growth equation, it's key to grasp the concept of natural logarithms. The natural logarithm is a logarithm to the base \(e\), where \(e \approx 2.718\).Natural logarithms (\