Problem 357 - Precalculus

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 \(P=1650 e^{0.5 t}\) 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

To find out in how many years the population will reach 20,000, set the given equation equal to 20,000. You have: \[ 1650 e^{0.5t} = 20000 \]

Step 2: Solve for the Exponential Term

Divide both sides of the equation by 1650 to isolate the exponential term:\[ e^{0.5t} = \frac{20000}{1650} \approx 12.1212 \]

Step 3: Apply the Natural Logarithm

Take the natural logarithm on both sides to solve for \(t\):\[ \ln(e^{0.5t}) = \ln(12.1212) \]By properties of logarithms, this simplifies to:\[ 0.5t = \ln(12.1212) \]

Step 4: Solve for t

Solve for \( t \) by dividing both sides by 0.5:\[ t = \frac{\ln(12.1212)}{0.5} \]Using a calculator, \( \ln(12.1212) \approx 2.495265 \). Therefore:\[ t \approx \frac{2.495265}{0.5} \approx 4.99053 \]

Step 5: Interpret the Result

Since \( t \approx 4.99 \), it will take approximately 5 years for the population to reach 20,000.

Step 6: Graph the Situation

To graph the situation, plot the function \( P = 1650 e^{0.5t} \) on a coordinate plane, with \(t\) on the x-axis and \(P\) on the y-axis. Mark the point where the population reaches 20,000, approximately at \( t = 5 \) years.

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special logarithm with the base \( e \).The constant \( e \) is approximately equal to 2.71828, and it is an irrational number like \( \pi \). Natural logarithms are incredibly useful when working with exponential equations, especially when the base is \( e \).

Natural logarithms help to "undo" the exponential function, which means they are the inverse of the exponential function. For instance, if we have \( e^{x} = y \), we can easily solve for \( x \) using natural logarithms: \( x = \ln(y) \).

In solving exponential growth problems, natural logarithms allow us to isolate the variable inside the exponent, like we did in the original exercise. By applying \( \ln \) to both sides of the equation, we used the logarithmic identity \( \ln(e^{x}) = x \) to bring down the exponent and solve for the time variable.

Population Modeling

Population modeling is a technique used to predict how populations change over time, utilizing mathematical equations to represent the growth or decline.

One common model is the exponential growth model, which assumes populations increase at a rate proportional to their current size. This scenario describes biological populations under ideal conditions without constraints.

In the equation \( P = 1650 e^{0.5t} \), each part tells us something specific:

• \( 1650 \) is the initial population size.
• \( 0.5 \) is the growth rate (in this context, years), indicating how fast the population grows.
• \( t \) represents time, allowing us to predict future population size.

By understanding these components, we can use the model to determine when the population will reach a certain size, such as 20,000. This is achieved by setting the equation equal to the desired population size and solving for the time \( t \). In our exercise, it took approximately 5 years to reach this goal.

Solving Exponential Equations

Solving exponential equations involves isolating the exponential part and then applying logarithms to eliminate the exponent and solve for the variable.

This process usually involves a few fundamental steps:

• Set Up the Equation: Align the real-world problem with its mathematical representation. In our example, we set the population growth model equal to 20,000.
• Isolate the Exponential Term: Simplify the equation to have the exponential term by itself. This often requires dividing both sides by a constant.
• Apply Natural Logarithms: Use \( \ln \) to both sides of the equation. This step is crucial because it helps you manipulate the equation to bring down the exponent.
• Solve for the Variable: Complete the algebraic steps to solve for the unknown variable. In the case of our population model, we solved for \( t \), the time required for a specific population size.

These steps are core to handling real-life problems involving exponential growth and decay, and understanding them empowers solving similar equations accurately.