Suppose a city with population 200,000 has been growing at a rate of $8 \%$ per year. If this rate continues, find the population of this city in 18 years. The population in 18 years will be approximately (Round to the nearest whole number as needed.)

Step 1 ：Translate the problem into a mathematical formula. The problem is asking for the future value of the population given an initial population, a growth rate, and a time period. This is a compound interest problem, where the population is the principal, the growth rate is the interest rate, and the time is the number of periods. The formula for compound interest is: \[ A = P(1 + r/n)^{nt} \] where: A is the amount of money earned after n years, including interest. P is the principal amount (the initial amount of money). r is the annual interest rate (in decimal form). n is the number of times that interest is compounded per year. t is the time the money is invested for, in years.

Step 2 ：In this case, the population is compounded annually (n=1), so the formula simplifies to: \[ A = P(1 + r)^t \]

Step 3 ：Plug in the given values into this formula to find the future population. P = 200000, r = 0.08, t = 18

Step 4 ：Calculate the future population: A = 200000*(1 + 0.08)^18

Step 5 ：Round the result to the nearest whole number: A = 799204

Step 6 ：Final Answer: The population of the city in 18 years will be approximately \(\boxed{799204}\)