In 2006, the population of a country was 50 million and growing at the rate of 1.8% per year. Assuming the percentage growth rate remains constant, express the population, P (in millions), as a function of t, the number of years after 2006.

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- Sep 17th 2008, 02:14 PMasweet1Functions
In 2006, the population of a country was 50 million and growing at the rate of 1.8% per year. Assuming the percentage growth rate remains constant, express the population, P (in millions), as a function of t, the number of years after 2006.

- Sep 17th 2008, 02:24 PMSoroban
Hello, asweet1!

Quote:

In 2006, the population of a country was 50 million and growing at the rate of 1.8% per year.

Assuming the percentage growth rate remains constant, express the population, $\displaystyle P$ (in millions),

as a function of $\displaystyle t$, the number of years after 2006.

If the population increases at a rate of 1.8% per year,

. . each year's population is 1.018 times the previous year's population.

This is identical to a compound interest problem.

The function is: .$\displaystyle P(t) \;=\;50(1.018)^t$