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many years ago, the towns of franklin and chester agreed to begin posti…

Question

many years ago, the towns of franklin and chester agreed to begin posting their populations on signs just outside of their towns. they also agreed to update their signs once per year at the beginning of the year. during year 1, franklins sign read \franklin: population of 20,000\, while chesters sign read \chester: population of 25,000\. each year, the populations grew. specifically, franklins population grew by 5% each year, and chesters population grew by 500 people each year. what is the first year in which franklins sign shows a larger number than chesters sign? year

Explanation:

Step1: Define population functions

Let $n$ = number of years after Year 1.
Franklin's population: $F(n) = 20000(1.05)^n$
Chester's population: $C(n) = 25000 + 500n$
We need $F(n) > C(n)$.

Step2: Test n=5

Calculate $F(5)=20000(1.05)^5 \approx 20000 \times 1.27628 = 25525.6$
Calculate $C(5)=25000 + 500(5)=27500$
$25525.6 < 27500$, so not yet.

Step3: Test n=6

Calculate $F(6)=20000(1.05)^6 \approx 20000 \times 1.34010 = 26802$
Calculate $C(6)=25000 + 500(6)=28000$
$26802 < 28000$, so not yet.

Step4: Test n=7

Calculate $F(7)=20000(1.05)^7 \approx 20000 \times 1.40710 = 28142$
Calculate $C(7)=25000 + 500(7)=28500$
$28142 < 28500$, so not yet.

Step5: Test n=8

Calculate $F(8)=20000(1.05)^8 \approx 20000 \times 1.47746 = 29549.2$
Calculate $C(8)=25000 + 500(8)=29000$
$29549.2 > 29000$, this works.

Step6: Find the actual year

Year number = $1 + 8 = 9$

Answer:

9