# Each year for $10$ years, the population of a city increased by $5\%$ of its value in the previous year

Each year for $10$ years, the population of a city increased by $5%$ of its value in the previous year. If the initial population is $200,000$, what was the population after $10$ years?

My solution: $$(200000)(1.05)^{n-1} (200000)(1.05)^{10-1} (200000)(1.05)^{9}$$ I see on the internet for the same question, people are getting $(200000)(1.05)^{10}$ instead. Why is this so? I drew "$10$ lines", and started counting starting from the first line and got $9$, so therefore the population increases by $5\%$ $9$ times, not $10$. Could anyone clear this up?

Thanks!

• Note that you start in year $0$ (not $1$). Thus, following you example with lines, you have to move $10$ times (not $9$). – mfl Mar 29 '15 at 21:06
• I guess you could think the population at the start of year 1 being $200,000$. And at the end of year 10, it was $x$. Therefore it had 10 years worth of time to increase, so it will be multiplied by $1.05$ ten times. – turkeyhundt Mar 29 '15 at 21:06
• If after $1$ year we got $200,000\times 1.05$, why don't we have $200,000\times (1.05)^{10}$ after $10$ years? – Salomo Mar 29 '15 at 21:06
• Your omission of equals signs makes your answer unreadable. It looks like a product rather than two equations. $$200000(1.05)^{n - 1} = 200000(1.05)^{10 - 1} = 200000(1.05)^9$$ – N. F. Taussig Mar 30 '15 at 14:10