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 

20082008.

​(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.)

Given the population model \( P(x) = Ca^{x-2000} \), we need to determine the values of \( C \) and \( a \) based on the provided population dataGiven the population model \( P(x) = Ca^{x-2000} \), we need to determine the values of \( C \) and \( a \) 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:
\[ P(2000) = 11 \]
\[ P(2025) = 1.1 \]

Substitute these values into the model:

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

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

Solve for \( a \):
\[ a^{25} = \frac{1.1}{11} \]
\[ a^{25} = 0.1 \]
\[ a = (0.1)^{1/25} \]

Calculate \( a \):
\[ a \approx 0.39811 \] (rounded to five decimal places)

Thus, the values are:
\[ C = 11 \]
\[ a \approx 0.39811 \]

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

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

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

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

Therefore, the estimated population in 2008 is approximately \( 0.11869 \) billion, or 118.69 million.

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

Set \( P(x) = 1.3 \):
\[ 1.3 = 11 \times 0.39811^{x-2000} \]
\[ \frac{1.3}{11} = 0.39811^{x-2000} \]
\[ 0.11818 \approx 0.39811^{x-2000} \]

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

Solve for \( x \):
\[ x - 2000 = \frac{\ln(0.11818)}{\ln(0.39811)} \]
\[ x - 2000 \approx \frac{-2.132}{-0.920} \]
\[ x - 2000 \approx 2.316 \]
\[ x \approx 2002.316 \]

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 \( a \) 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.

Learn how to model and project population growth using exponential functions. Solve for constants C and a in the equation P(x) = Ca^{x-2000} based on population data from 2000 and projections for 2025. Estimate the population in 2008 and determine the year when the population might reach 1.3 billion.

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