Sure! Let's carefully formulate a mathematical model for population growth given an annual growth rate of 4.2%.
In general, the population growth can be modeled using the exponential growth formula:
[tex]\[ y = P \times (1 + r)^x \][/tex]
Where:
- [tex]\( y \)[/tex] is the future population after [tex]\( x \)[/tex] years.
- [tex]\( P \)[/tex] is the initial population.
- [tex]\( r \)[/tex] is the annual growth rate (expressed as a decimal).
- [tex]\( x \)[/tex] is the number of years.
Given:
- Initial population [tex]\( P = 100,000 \)[/tex]
- Annual growth rate [tex]\( r = 4.2\% = 0.042 \)[/tex]
We substitute these values into our formula to get:
[tex]\[ y = 100,000 \times (1 + 0.042)^x \][/tex]
Simplifying the term inside the parentheses:
[tex]\[ y = 100,000 \times 1.042^x \][/tex]
This is the equation that models the population [tex]\( y \)[/tex] after [tex]\( x \)[/tex] years, given an initial population of 100,000 and an annual growth rate of 4.2%.
### Example Calculation
Let's calculate the population after 5 years as an illustration:
1. Substitute [tex]\( x = 5 \)[/tex] into the equation:
[tex]\[ y = 100,000 \times 1.042^5 \][/tex]
2. Calculate [tex]\( 1.042^5 \)[/tex]:
[tex]\[ 1.042^5 \approx 1.228 \][/tex]
3. Multiply this result by the initial population:
[tex]\[ y \approx 100,000 \times 1.228 = 122,800 \][/tex]
Therefore, the population after 5 years would be approximately 122,800.
In summary:
The equation to model the population [tex]\( y \)[/tex] after [tex]\( x \)[/tex] years is:
[tex]\[ y = 100,000 \times 1.042^x \][/tex]
