Answer:
130,226.
Step-by-step explanation:
We have been given that the population of a town is 100,000 in 1999. The population increases at a rate of 4.5% every year.
We will use exponential growth function to solve our given problem. We know that exponential function is in form \(f(x)=a\cdot b^x\), where,
\(f(x)\) = Final value,
a = Initial value,
b = For growth b is in form \(1+r\), where r represents rate in decimal form.
Let us convert 4.5% into decimal form.
\(4.5\%=\frac{4.5}{100}=0.04\)
Upon substituting our given values in exponential growth formula, we will get:
\(f(x)=100,000\cdot (1.045)^x\), where x represents number of years after 1999.
To find population of town in 2005, we will substitute \(x=6\) (2005-1999=6) in our function.
\(f(6)=100,000\cdot (1.045)^6\)
\(f(6)=100,000\cdot 1.3022601248475156\)
\(f(6)=130,226.01248475156\)
\(f(6)\approx 130,226\)
Therefore, the population of the town in 2005 will be 130,226.
