Problem 37 - Intermediate Algebra

Question

For a certain strain of bacteria, the number present after \(t\) hours is given by the equation \(Q=Q_{0} e^{0.34 t}\), where \(Q_{0}\) represents the initial number of bacteria. How long will it take 400 bacteria to increase to 4000 bacteria?

Step-by-Step Solution

Answer

It takes approximately 6.8 hours for 400 bacteria to increase to 4000.

Step 1: Define the Variables

Identify the known values and the unknown we need to solve for. We know that the initial number of bacteria is \(Q_0 = 400\) and the final amount we want is \(Q = 4000\). We are solving for time \(t\).

Step 2: Set Up the Equation

The equation to model the number of bacteria after \(t\) hours is given by \(Q = Q_0 e^{0.34t}\). Substitute the known values: \(4000 = 400e^{0.34t}\).

Step 3: Simplify the Equation

Divide both sides by 400 to isolate the exponential expression: \(10 = e^{0.34t}\).

Step 4: Solve for Time Using Logarithms

Apply the natural logarithm \(\ln\) to both sides of the equation: \(\ln(10) = \ln(e^{0.34t})\). This simplifies to \(\ln(10) = 0.34t\) because \(\ln(e^x) = x\).

Step 5: Calculate the Value of \(t\)

Solve for \(t\) by dividing both sides by 0.34: \(t = \frac{\ln(10)}{0.34}\). Calculate this using a calculator to get \(t \approx 6.795\).

Key Concepts

Bacterial Growth

Bacterial growth in the context of the given problem can be understood as an exponential growth process. This means that the bacteria population increases at a rate proportional to its current size. In the context of biology, when bacteria reproduce, they often divide at regular intervals, leading to rapid increases in population. This can be mathematically modeled using the formula:

• \(Q = Q_0 e^{kt}\)

where \(Q_0\) is the initial quantity of bacteria, \(Q\) is the quantity after time \(t\), \(e\) is the base of the natural logarithm (approximately 2.71828), and \(k\) is the growth rate.
In this case, our specific problem gives us a growth rate \(k\) of 0.34. Understanding this helps us see how quickly bacteria can multiply under optimal conditions. Exponential growth is seen in many natural processes, illustrating how powerful and fast nature can be.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\). The number \(e\) (approximately 2.71828) is an important constant in mathematics known as Euler's number. Logarithms are mathematical tools used to solve equations involving exponents by converting multiplicative relationships into additive ones.
In our exercise, we apply the natural logarithm to solve for the time \(t\) when the equation involves an exponent on \(e\). Specifically, we use the property:

• \(\ln(e^x) = x\)

which allows us to turn the equation \(10 = e^{0.34t}\) into \(\ln(10) = 0.34t\). This transformation simplifies solving for \(t\) because it isolates \(t\) as a straightforward arithmetic problem, making it a crucial step in the process of solving exponential equations.

Initial Value Problem

An initial value problem in mathematics seeks to find a function that satisfies a given relationship involving derivatives, accompanied by specified values, called initial conditions, at a starting point. In this problem, we start with an initial number of bacteria, which provides the necessary starting point for calculating the subsequent growth of the bacteria over time.

• Initial condition: \(Q_0 = 400\)
• Final target: \(Q = 4000\)

With the initial condition known, the problem involves finding the required time \(t\) for these bacteria to multiply to the specified target quantity. Solving initial value problems, especially in real-life applications like biology, allows us to understand how systems evolve from a given state to another over time. They often provide insights into the dynamics of growth and change in various natural and engineered systems.