Problem 46 - A Graphical Approach to College Algebra

Question

The growth of bacteria in food products makes it necessary to date some products (such as milk) so that they will be sold and consumed before the bacterial count becomes too high. Suppose that, under certain storage conditions, the number of bacteria present in a product is $$f(t)=500 e^{0.1 t}$$ where \(t\) is time in days after packing of the product and the value of \(f(t)\) is in millions. (a) If the product cannot be safely eaten after the bacterial count reaches \(3,000,000,000,\) how long will this take? (b) If \(t=0\) corresponds to January \(1,\) what date should be placed on the product?

Step-by-Step Solution

Answer

The bacteria count reaches 3 billion in about 87 days, which is March 28.

Step 1: Understand the Function

The function given is \( f(t) = 500 e^{0.1t} \). This represents the bacterial count in millions after \( t \) days. We want to find \( t \) when the bacteria count reaches \( 3,000,000,000 \), which is \( 3,000,000 \) million.

Step 2: Set Up the Equation

We set \( f(t) = 3,000,000 \) because the bacterial count in millions should be \( 3,000,000 \). This gives us the equation \( 500 e^{0.1t} = 3,000,000 \).

Step 3: Solve for e^{0.1t}

Divide both sides of the equation \( 500 e^{0.1t} = 3,000,000 \) by 500 to isolate \( e^{0.1t} \):\[e^{0.1t} = \frac{3,000,000}{500}\]This simplifies to \( e^{0.1t} = 6000 \).

Step 4: Take the Natural Logarithm

To solve for \( t \), take the natural logarithm of both sides:\[\ln(e^{0.1t}) = \ln(6000)\]Simplify using the property \( \ln(e^x) = x \):\[0.1t = \ln(6000)\].

Step 5: Solve for t

To solve for \( t \), divide both sides by 0.1:\[t = \frac{\ln(6000)}{0.1}\]Calculate \( \ln(6000) \) and then divide by 0.1 to find \( t \). \( \ln(6000) \approx 8.69951 \), so\[t = \frac{8.69951}{0.1} = 86.9951 \approx 87\] days.

Step 6: Determine the Date

Since \( t = 0 \) corresponds to January 1, adding 87 days gives us March 28 as the critical date. The product should be consumed by this date.

Key Concepts

Bacterial Growth

Bacterial growth in food products can be a significant concern, requiring attention to safety and consumption timelines. The amount of bacteria in a product can increase exponentially over time. In many cases, this growth is modeled using mathematical functions. This exponential model is especially useful because bacteria populations typically double at regular intervals when conditions are ideal. The rapid increase in numbers can pose health risks, making it necessary to monitor and manage these counts in consumable goods.

For example, in our exercise, the bacterial count is represented by the function \(f(t) = 500 e^{0.1t}\). Here, \(t\) represents the number of days since the product was packed. Such a function helps predict when the bacterial count might reach unsafe levels. Understanding this growth pattern allows manufacturers and consumers to determine an appropriate 'sell by' or 'use by' date to ensure safety. Monitoring bacterial growth is essential because once the count surpasses a certain threshold, the product might not be safe for consumption.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), plays a crucial role when dealing with exponential equations, especially in bacterial growth scenarios.

In mathematical terms, the natural logarithm is the inverse function of the exponential function. It's primarily used to solve equations where the unknown variable is an exponent. If you have an equation of the form \(e^x = y\), taking the natural logarithm of both sides allows us to solve for \(x\).

In our exercise, we applied the natural logarithm to isolate \(t\) in the exponential equation \(e^{0.1t} = 6000\). By applying \(\ln\) on both sides, the exponential \(e\) is effectively 'cancelled,' simplifying the equation to \(0.1t = \ln(6000)\). This transformation makes it possible to solve for \(t\), illustrating the power of logarithms in tackling exponential growth problems.

Solving Exponential Equations

Exponential equations, like the one given in our exercise, are equations where a variable appears in the exponent. These equations are commonly encountered when dealing with growth processes like bacterial growth.

To solve exponential equations, it often involves a series of steps to isolate the variable. Firstly, you aim to express the equation in a simpler form, usually by dividing or factoring. Once simplified to a form like \(e^x = y\), use the natural logarithm to deal with the exponent.

• First, simplify the equation by isolating the exponential expression. For instance, divide both sides by constants to get \(e^{0.1t} = 6000\).
• Next, apply \(\ln\) to both sides. This transforms \(e^{0.1t} = 6000\) into \(0.1t = \ln(6000)\).
• Finally, solve for the variable of interest by performing algebraic operations, such as division. Here, divide by 0.1 to find \(t\).

Through these steps, one can efficiently solve exponential equations, paving the way for practical applications like determining safe consumption dates for perishable products.