Respuesta :
We have been given that Oswego has a population of 7,800 people and is growing at a rate of 6% per year.Rye Brook has a population of 9,400 people and is growing at a rate of 4% per year.
We will use exponential growth function to represent population in both cities after x years.
[tex]y=a\cdot (1+r)^x[/tex], where
y = Final amount,
a = Initial amount,
r = Growth rate in decimal form,
t = Time.
[tex]6\%=\frac{6}{100}=0.06[/tex]
Population of Oswego after x years would be [tex]y=7,800(1+0.06)^x[/tex].
[tex]4\%=\frac{4}{100}=0.04[/tex]
Population of Rye Brook after x years would be [tex]y=9,400(1+0.04)^x[/tex].
Now we will set an inequality such as population of Oswego is greater than Rye Brook.
[tex]7800\cdot(1.06)^t>9400\cdot (1.04)^t[/tex]
Let us take natural log on both sides.
[tex]\text{ln}(7800\cdot(1.06)^t)>\text{ln}(9400\cdot (1.04)^t)[/tex]
[tex]\text{ln}(7800)+\text{ln}((1.06)^t)>\text{ln}(9400)+\text{ln}((1.04)^t)[/tex]
[tex]\text{ln}(7800)-\text{ln}(7800)+\text{ln}((1.06)^t)>\text{ln}(9400)-\text{ln}(7800)+\text{ln}((1.04)^t)[/tex]
[tex]\text{ln}((1.06)^t)>\text{ln}(\frac{9400}{7800})+\text{ln}((1.04)^t)[/tex]
[tex]\text{ln}((1.06)^t)>\ln \left(\frac{47}{39}\right)+\text{ln}((1.04)^t)[/tex]
[tex]\text{ln}((1.06)^t)-\text{ln}((1.04)^t)>\ln \left(\frac{47}{39}\right)+\text{ln}((1.04)^t)-\text{ln}((1.04)^t)[/tex]
[tex]t\cdot \text{ln}(1.06)-t\cdot \text{ln}(1.04)>\ln \left(\frac{47}{39}\right)[/tex]
[tex]t(\text{ln}(1.06)-\text{ln}(1.04))>\ln \left(\frac{47}{39}\right)[/tex]
[tex]\frac{t(\text{ln}(1.06)-\text{ln}(1.04))}{\text{ln}(1.06)-\text{ln}(1.04))}>\frac{\ln \left(\frac{47}{39}\right)}{\text{ln}(1.06)-\text{ln}(1.04))}[/tex]
[tex]t>9.79546[/tex]
Therefore, in 10 years Oswego will have a greater population than Rye Brook.
