Question

Suppose the population of a town is 100,000 in 1999. The population increases at a rate of 4.5% every year. What will be the population of the town in 2005? Round your answer to the nearest whole number.

Answer 1

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.

Answer 2

100,000 * (1.045)^years
2005 Population = 100,000 * (1.045)^6
= 100,000 * 1.3022601248
= 130,226