/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 The population \(P\) of a small ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified
Answer: Approximately 34.66 years.

Step by step solution

01

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 \(P(t)=100 e^{t / 50}\). $$100 e^{\frac{2t}{50}} = 2 \times 100 e^{\frac{t}{50}}$$
02

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. $$\frac{t}{50} = ln(2)$$
03

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Identify the amplitude and period of the following functions. $$p(t)=2.5 \sin \left(\frac{1}{2}(t-3)\right)$$

The shortest day of the year occurs on the winter solstice (near December 21) and the longest day of the year occurs on the summer solstice (near June 21 ). However, the latest sunrise and the earliest sunset do not occur on the winter solstice, and the earliest sunrise and the latest sunset do not occur on the summer solstice. At latitude \(40^{\circ}\) north, the latest sunrise occurs on January 4 at 7: 25 a.m. ( 14 days after the solstice), and the earliest sunset occurs on December 7 at 4: 37 p.m. ( 14 days before the solstice). Similarly, the earliest sunrise occurs on July 2 at 4: 30 a.m. (14 days after the solstice) and the latest sunset occurs on June 7 at 7: 32 p.m. ( 14 days before the solstice). Using sine functions, devise a function \(s(t)\) that gives the time of sunrise \(t\) days after January 1 and a function \(S(t)\) that gives the time of sunset \(t\) days after January \(1 .\) Assume that \(s\) and \(S\) are measured in minutes and \(s=0\) and \(S=0\) correspond to 4: 00 a.m. Graph the functions. Then graph the length of the day function \(D(t)=S(t)-s(t)\) and show that the longest and shortest days occur on the solstices.

Find a formula for a function describing the given situation. Graph the function and give a domain that makes sense for the problem. Recall that with constant speed. distance \(=\) speed \(\cdot\) time elapsed or \(d=v t\) A function \(y=f(x)\) such that if you ride a bike for \(50 \mathrm{mi}\) at \(x\) miles per hour, you arrive at your destination in \(y\) hours

Make a sketch of the given pairs of functions. Be sure to draw the graphs accurately relative to each other. $$y=x^{4} \text { and } y=x^{6}$$

A capacitor is a device that stores electrical charge. The charge on a capacitor accumulates according to the function \(Q(t)=a\left(1-e^{-t / c}\right),\) where \(t\) is measured in seconds, and \(a\) and \(c > 0\) are physical constants. The steady-state charge is the value that \(Q(t)\) approaches as \(t\) becomes large. a. Graph the charge function for \(t \geq 0\) using \(a=1\) and \(c=10\) Find a graphing window that shows the full range of the function. b. Vary the value of \(a\) holding \(c\) fixed. Describe the effect on the curve. How does the steady-state charge vary with \(a ?\) c. Vary the value of \(c\) holding \(a\) fixed. Describe the effect on the curve. How does the steady-state charge vary with \(c ?\) d. Find a formula that gives the steady-state charge in terms of \(a\) and \(c\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.