Answer:
1) 13,434
2) During 2038 (by 2039)
Step-by-step explanation:
We can model this problem using an exponential equation.
General form of an exponential equation: [tex]y=ab^x[/tex]
where:
- a is the initial value
- b is the base (or growth factor) in decimal form
- x is the independent variable
- y is the dependent variable
If b > 1 then it is an increasing function
If 0 < b < 1 then it is a decreasing function
As the population increases by 3% each year, each year it will be 103% of the previous year. Therefore, the growth factor in decimal form is 1.03
Given:
- a = 8,888
- b = 1.03
- x = time (in years)
- y = population
Substituting the given values into the equation:
[tex]y=8888(1.03)^x[/tex]
To find the population in 2025:
2025 - 2011 = 14
Therefore, set x = 14 and solve for y:
[tex]\begin{aligned}\implies y &=8888(1.03)^{14}\\& =13433.89747\\& = 13,434\end{aligned}[/tex]
To find the year in which the population will reach 20000, set y = 20000 and solve for x:
[tex]\implies 8888(1.03)^x=20000[/tex]
[tex]\implies (1.03)^x=\dfrac{20000}{8888}[/tex]
[tex]\implies \ln (1.03)^x=\ln \left(\dfrac{20000}{8888}\right)[/tex]
[tex]\implies x\ln (1.03)=\ln \left(\dfrac{20000}{8888}\right)[/tex]
[tex]\implies x=\dfrac{\ln \left(\frac{20000}{8888}\right)}{\ln (1.03)}[/tex]
[tex]\implies x=27.43785809[/tex]
2011 + 27.437... = 2038.437...
Therefore, the population will reach 20,000 some time during 2038.
