Math Problem Statement

In

20002000

the population of a country reached

11

​billion, and in

20252025

it is projected to be

1.11.1

billion.

​(a)

Find values for C and a so that

​P(x)equals=Ca Superscript x minus 2000Cax−2000

models the population of a country in year x.year x.

​(b)

Estimate the​ country's population in

​(c)

Use P to determine the year when the​ country's population might reach

1.31.3

billion.

Question content area bottom

Part 1

​(a)

Cequals=enter your response here

​(Type an integer or decimal rounded to five decimal places as​ needed.)

Solution

Given the population model P(x)=Cax2000P(x) = Ca^{x-2000}, we need to determine the values of CC and aa based on the provided population dataGiven the population model P(x)=Cax2000P(x) = Ca^{x-2000}, we need to determine the values of CC and aa based on the provided population

  1. In 2000, the population was 11 billion.
  2. In 2025, the population is projected to be 1.1 billion.

We have two data points: P(2000)=11P(2000) = 11 P(2025)=1.1P(2025) = 1.1

Substitute these values into the model:

  1. For x=2000x = 2000: P(2000)=Ca20002000=Ca0=CP(2000) = C \cdot a^{2000-2000} = C \cdot a^0 = C C=11C = 11

  2. For x=2025x = 2025: P(2025)=11a20252000=1.1P(2025) = 11 \cdot a^{2025-2000} = 1.1 11a25=1.111 \cdot a^{25} = 1.1

Solve for aa: a25=1.111a^{25} = \frac{1.1}{11} a25=0.1a^{25} = 0.1 a=(0.1)1/25a = (0.1)^{1/25}

Calculate aa: a0.39811a \approx 0.39811 (rounded to five decimal places)

Thus, the values are: C=11C = 11 a0.39811a \approx 0.39811

So the model is: P(x)=11×0.39811x2000P(x) = 11 \times 0.39811^{x-2000}

Part (b): Estimate the country's population in 2008

Substitute x=2008x = 2008 into the model: P(2008)=11×0.3981120082000P(2008) = 11 \times 0.39811^{2008-2000} P(2008)=11×0.398118P(2008) = 11 \times 0.39811^{8}

Calculate P(2008)P(2008): P(2008)11×0.01079P(2008) \approx 11 \times 0.01079 P(2008)0.11869P(2008) \approx 0.11869

Therefore, the estimated population in 2008 is approximately 0.118690.11869 billion, or 118.69 million.

Part (c): Determine the year when the population might reach 1.3 billion

Set P(x)=1.3P(x) = 1.3: 1.3=11×0.39811x20001.3 = 11 \times 0.39811^{x-2000} 1.311=0.39811x2000\frac{1.3}{11} = 0.39811^{x-2000} 0.118180.39811x20000.11818 \approx 0.39811^{x-2000}

Take the natural logarithm of both sides: ln(0.11818)=ln(0.39811x2000)\ln(0.11818) = \ln(0.39811^{x-2000}) ln(0.11818)=(x2000)ln(0.39811)\ln(0.11818) = (x-2000) \ln(0.39811)

Solve for xx: x2000=ln(0.11818)ln(0.39811)x - 2000 = \frac{\ln(0.11818)}{\ln(0.39811)} x20002.1320.920x - 2000 \approx \frac{-2.132}{-0.920} x20002.316x - 2000 \approx 2.316 x2002.316x \approx 2002.316

Therefore, the population might reach 1.3 billion approximately in the year 2002 (early in the year).

Would you like more details on any step or have any questions?

Here are some related questions for further practice:

  1. How does the growth rate affect the population projection?
  2. What if the initial population in 2000 was different? How would that change the model?
  3. How can you use logarithms to solve for the growth rate in other contexts?
  4. What is the significance of the base aa in the exponential model?
  5. How would changes in population growth trends affect long-term projections?
  6. How can this model be verified against real-world data?
  7. How does exponential decay differ from exponential growth in population models?
  8. What other factors might influence the accuracy of population projections?

Tip: When working with exponential models, using logarithms can simplify the process of solving for unknown variables.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Population Growth Models
Logarithms

Formulas

Exponential growth model: P(x) = Ca^{x-2000}
Logarithmic equation for solving growth projections

Theorems

Exponential growth properties
Logarithmic properties for solving equations

Suitable Grade Level

Grades 10-12