Problem 67 - College Algebra

Question

Solve for the indicated value, and graph the situation showing the solution point. The population of a small town is modeled by the equation \(\mathrm{P}=1650 e^{0.5 t}\) where \(t\) is measured in years. In approximately how many years will the town's population reach \(20,000 ?\)

Step-by-Step Solution

Answer

The town's population will reach 20,000 in approximately 5 years.

Step 1: Set up the equation

We start from the equation given for the population: \( P = 1650 e^{0.5 t} \). We want to find the value of \( t \) when \( P = 20000 \). Thus, we set up the equation: \( 1650 e^{0.5 t} = 20000 \).

Step 2: Isolate the exponential term

To isolate the exponential term \( e^{0.5 t} \), divide both sides of the equation by 1650. This gives us: \( e^{0.5 t} = \frac{20000}{1650} \).

Step 3: Simplify the fraction

Now simplify the fraction \( \frac{20000}{1650} \). This simplifies to approximately \( 12.1212 \). So, \( e^{0.5 t} = 12.1212 \).

Step 4: Solve for the exponent

Take the natural logarithm (ln) on both sides of the equation to solve for \( t \). This gives us: \( \ln(e^{0.5 t}) = \ln(12.1212) \). Since \( \ln(e^x) = x \), we have \( 0.5 t = \ln(12.1212) \).

Step 5: Calculate \( t \)

Compute the natural logarithm \( \ln(12.1212) \) which is approximately \( 2.497 \). Now solve for \( t \) by dividing both sides by 0.5 to get \( t = \frac{2.497}{0.5} \), which is approximately \( 4.994 \).

Step 6: Graph the equation

Plot the graph of \( P = 1650 e^{0.5 t} \) on a set of axes with \( t \) on the horizontal axis and \( P \) on the vertical axis. Mark the point where \( P = 20000 \) occurs, which is approximately at \( t = 4.994 \).

Key Concepts

Exponential Function

An exponential function is a mathematical expression in which a variable appears in the exponent or power. In the population growth model, exponential functions often describe how quantities grow over time. The general form of an exponential function is \( f(t) = a \cdot e^{bt} \), where:

• \( a \) is the initial quantity.
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
• \( b \) is the growth rate.
• \( t \) represents time.

In our problem, the population \( P \) is expressed as an exponential function of time, \( P = 1650e^{0.5t} \). Here, 1650 is the initial population of the town, and 0.5 is the growth rate. As time progresses, the population increases exponentially. This means it doesn't grow by the same amount each year, but increasingly faster as it multiplies by a certain factor. Understanding this function helps in predicting how rapidly the population can grow.

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is the inverse operation of exponentiation with base \( e \). It's essential for solving equations involving exponential functions, especially when you need to "undo" the exponential operation and solve for a variable in the exponent.

In our exercise, we used the natural logarithm to solve for \( t \), the time in years. Once you isolate the exponential term, like \( e^{0.5t} = 12.1212 \), you employ the natural logarithm:

• Apply \( \ln \) to both sides: \( \ln(e^{0.5t}) = \ln(12.1212) \).
• Since \( \ln(e^x) = x \), this simplifies to \( 0.5t = \ln(12.1212) \).

The result provides the necessary step to express \( t \, \), showing its fundamental role in "undoing" the exponential function and allowing us to calculate the desired time in years.

Solving Equations

Solving equations is a vital skill in mathematics, allowing us to find unknown values that make an equation true. In our problem, the task was to find \( t \), when the population reaches a specific number using the equation \( 1650e^{0.5t} = 20000 \).

Here are the essential steps:

• Set up the equation by placing the known final population on one side: \( 1650e^{0.5t} = 20000 \).
• Isolate the exponential term by dividing by the initial population: \( e^{0.5t} = \frac{20000}{1650} \).
• Simplify the equation to make calculations easier: \( e^{0.5t} = 12.1212 \).
• Apply natural logarithm to both sides to isolate the exponent: \( \ln(e^{0.5t}) = \ln(12.1212) \).
• Solve for \( t \) by dividing: \( 0.5t = 2.497 \), leading to \( t = \frac{2.497}{0.5} \), approximately equal to 4.994 years.

This systematic approach illuminates the process and ensures clarity when dealing with exponential equations.

Graphing Exponential Growth

Visual representation of exponential growth through graphing helps to easily understand the nature of growth over time. Graphs provide a clear picture of how quickly or slowly the quantity is increasing as time progresses.

To graph our equation \( P = 1650e^{0.5t} \):

• Use a horizontal axis representing time \( t \) and a vertical axis for population \( P \).
• Plot initial values and points where \( t \) is measurable, such as \( t = 0 \) with \( P = 1650 \).
• Mark the solution point \( t = 4.994 \) for \( P = 20000 \), showing the time when the population reaches 20,000.
• Draw the curve to represent the continuous exponential increase.

This graph helps illustrate the rapid growth phenomenon where the rate of increase accelerates over time, offering an intuitive grasp of exponential dynamics.