Question

The population \(P\) of a small town is growing according to the function \(P(t)=100 e^{t / 50},\) where \(t\) measures the number of years after \(2010 .\) How long does it take the population to double?

Short answer

Answer: Approximately 34.66 years.

Step-by-step solution

Set up the equation

Set up the equation to find the time when the population doubles, i.e., \(P(2t) = 2 \times P(t)\). Substitute the population function \(P(t)=100 e^{t / 50}\). $$100 e^{\frac{2t}{50}} = 2 \times 100 e^{\frac{t}{50}}$$

Simplify and solve for t

Divide both sides of the equation by \(100e^{\frac{t}{50}}\) $$\frac{100 e^{\frac{2t}{50}}}{100 e^{\frac{t}{50}}} = 2$$ Simplify the equation: $$e^{\frac{t}{50}} = 2$$ Now, we want to solve for t by using the natural logarithm (\(\ln\)) function. $$\frac{t}{50} = ln(2)$$

Determine the time t

Multiply both sides of the equation by 50: $$t = 50\cdot \ln(2)$$ Calculate the value of t: $$t \approx 34.66 \thinspace years$$ Thus, it will take approximately 34.66 years for the population to double.

Key concepts

Population Growth

Population growth refers to the change in the number of individuals in a population over time. It's an important concept in biology, economics, and environmental sciences. In many scenarios, populations grow exponentially, which means the population size increases at a constant rate over time.
The problem referenced uses an exponential growth model, where the population of a small town is described by the equation \(P(t) = 100 e^{t/50}\). This equation showcases how populations can grow exponentially.

• In the equation, \(P(t)\) represents the population at time \(t\).
• The constant \(100\) is the initial population size when \(t = 0\).
• The expression \(e^{t/50}\) signifies the exponential growth factor, where \(e\) is the base of the natural logarithm, approximately equal to 2.718.

Exponential growth models are commonly used because they realistically describe the growth patterns of populations when there are abundant resources and minimal constraints, leading to rapid increases. Understanding this helps to predict future population sizes and prepare for potential impacts of growth.

Doubling Time

Doubling time is the period it takes for a population to grow to twice its size at a constant growth rate. In exponential growth, it is a simple yet powerful metric for understanding the pace at which a population is increasing. The exercise illustrates how to calculate the doubling time using a mathematical formula. To find the doubling time in the given problem, we set the equation \(P(2t) = 2 \times P(t)\). After substituting the given \(P(t)\), the equation becomes \(100 e^{2t/50} = 2 \times 100 e^{t/50}\). Simplifying both sides, the equation reduces to \(e^{t/50} = 2\).

• Doubling time involves solving for \(t\) in the simplified equation.
• By utilizing the properties of the natural logarithm, we find \(\ln(2)\).
• The final calculation shows that \(t = 50 \cdot \ln(2)\), approximately 34.66 years.

This numeric solution gives a clear picture of how long it will take for the population to double in size, helping planners make informed decisions based on growth rates.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function that is the inverse of the exponential function. It plays a critical role when working with exponential growth equations, allowing for the solution of unknown variables like time. In the given problem, we use the natural logarithm to solve for the doubling time of the population.After simplifying the growth equation to \(e^{t/50} = 2\), we employ the natural logarithm to both sides of the equation. This transforms our equation so we can solve for \(t\):

• Taking the natural log on both sides: \(\ln(e^{t/50}) = \ln(2)\).
• Using the property of logarithms \(\ln(e^x) = x\), the equation becomes \(t/50 = \ln(2)\).
• Finally, multiply both sides by 50 to solve for \(t\), leading to \( t = 50 \cdot \ln(2) \).

Understanding the application of the natural logarithm is fundamental for dealing with exponential equations, offering a straightforward method to handle growth-related scenarios. This is especially handy in biological, financial, and environmental modeling, where population or investment growth is analyzed. By knowing how to apply \(\ln\), one can efficiently solve for time or growth rate in exponential models.