/*! 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 64 The population \(N(t)\) (in mill... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 64

The population \(N(t)\) (in millions) of India \(t\) years after 1985 may be approximated by the formula \(N(t)=766 e^{0.0182t}\) When will the population reach 1.5 billion?

Short Answer

Expert verified
The population will reach 1.5 billion around the year 2023.

Step by step solution

01

Understand the Formula

The given population model is formulated as \( N(t) = 766 e^{0.0182t} \), where \( N(t) \) represents the population in millions and \( t \) is the number of years after 1985. We need to find \( t \) when \( N(t) = 1500 \) million (since 1.5 billion is equal to 1500 million).
02

Set the Equation

Set the equation using the population value: \( 766 e^{0.0182t} = 1500 \). This equation will allow us to solve for \( t \).
03

Isolate the Exponential

Divide both sides of the equation by 766 to get the exponential term by itself: \( e^{0.0182t} = \frac{1500}{766} \).
04

Take the Natural Logarithm

Take the natural logarithm on both sides to eliminate the exponential: \( \ln(e^{0.0182t}) = \ln\left(\frac{1500}{766}\right) \). This simplifies to \( 0.0182t = \ln\left(\frac{1500}{766}\right) \).
05

Solve for t

Solve for \( t \) by dividing both sides by 0.0182: \( t = \frac{\ln\left(\frac{1500}{766}\right)}{0.0182} \). Calculating this gives \( t \approx 38.14 \).
06

Calculate the Year

Since \( t \) is the number of years after 1985, add \( t \) to 1985 to find the year: 1985 + 38.14 鈮� 2023.14. Therefore, the population will reach 1.5 billion around the year 2023.

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 91影视!

Key Concepts

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

Population Modeling
Population modeling is a mathematical representation of how populations change over time. It's essential in understanding the growth patterns of living organisms, like humans in a specific country. In population modeling, we often use equations that can predict future population numbers based on current statistics and growth rates.

For the exercise, we have a given model: \[ N(t) = 766 e^{0.0182t} \]where:
  • \( N(t) \) is the population in millions.
  • \( t \) is the number of years after 1985.
  • \( 766 \) is the initial population in millions at year 1985.
  • \( e \) is the base of the natural logarithm.
  • \( 0.0182 \) is the annual growth rate.
Population models help us identify trends and make predictions about future population sizes. In this case, the model lets us determine when the population will reach 1.5 billion people.
Natural Logarithm
The natural logarithm is a logarithm to the base of the mathematical constant \( e \). It's a crucial concept in solving exponential equations like the one in the exercise. When dealing with equations where the unknown variable is in the exponent, we can use natural logarithms to simplify them.

For instance, in the exercise, the equation was:\[ e^{0.0182t} = \frac{1500}{766} \]Taking the natural logarithm of both sides helps us get rid of the exponent:\[ \ln(e^{0.0182t}) = \ln\left(\frac{1500}{766}\right) \]
This step is vital because the natural logarithm of \( e^x \) is simply \( x \). This property makes it easier to isolate the variables and solve for them. Thus, we convert:\[ 0.0182t = \ln\left(\frac{1500}{766}\right) \]
Year Calculation
After determining the value of \( t \), which represents the number of years since 1985, it's straightforward to calculate the actual year. This is the final step in using a population model prediction for a real-world year.

Given the equation:\[ t = \frac{\ln\left(\frac{1500}{766}\right)}{0.0182} \]we solved it to get \( t \approx 38.14 \). This means that 38.14 years will have passed from 1985.

To find the specific year, simply add this duration to 1985:
  • 1985 + 38.14 = 2023.14
Rounding this result indicates that the population will approximately reach 1.5 billion sometime in the year 2023. This type of calculation is invaluable to transform abstract mathematical solutions into meaningful time frames.

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

Electron energy The energy \(E(x)\) of an electron after passing through material of thickness \(x\) is given by the equation \(E(x)=E_{0} e^{-x k_{0} y},\) where \(E_{0}\) is the initial energy and \(x_{0}\) is the radiation length.Express, in terms of \(E_{0}\), the energy of an electron after it passes through material of thickness \(x_{0}\) (b) Express, in terms of \(x_{0}\), the thickness at which the electron loses \(99 \%\) of its initial energy.

Graph \(f\) on the interval \((0,200] .\) Find an approximate equation for the horizontal asymptote. $$f(x)=\left(1+\frac{2}{x}\right)^{x}$$

Reaction to a stimulus Let \(R\) denote the reaction of a subject to a stimulus of strength \(x\). There are many possibilities for \(R\) and \(x .\) If the stimulus \(x\) is saltiness (in grams of salt per liter), \(R\) may be the subject's estimate of how salty the soIution tasted, based on a scale from 0 to \(10 .\) One relationship between \(R\) and \(x\) is given by the Weber-Fechner formula, \(R(x)=a \log \left(x / x_{0}\right),\) where \(a\) is a positive constant and \(x_{0}\) is called the threshold stimulus. Find \(R\left(x_{0}\right)\) Find a relationship between \(R(x)\) and \(R(2 x)\)

Sketch the graph of \(f\). $$f(x)=\log (x+10)$$

According to Newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. The face of a household iron cools from \(125^{\circ}\) to \(100^{\circ}\) in 30 minutes in a room that remains at a constant temperature of \(75^{\circ} .\) From calculus, the temperature \(f(t)\) of the face after \(t\) hours of cooling is given by \(f(t)=50(2)^{-2 t}+75\) (a) Assuming \(t=0\) corresponds to 1: 00 PM., approximate to the nearest tenth of a degree the temperature of the face at 2: 00 P.M., 3: 30 P.M., and 4: 00 P.M. (b) Sketch the graph of \(f\) for \(0 \leq t \leq 4\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation
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.