Question

The population of a Midwestern city follows the exponential law. If N is the population of the city and t is the time in years, express N as a function of t. If the population doubled in size over an 18-month period and the current population is 10,000, what will the population be 2 years from now?

Answer 1

The population of a Midwestern city follows the exponential law. If $N$ is the population of the city and $t$ is the time in years, express $N$ as a function of $t$. If the population doubled in size over an 18-month period and the current population is 10,000, what will the population be 2 years from now?

Solution

Let $N_0$ – current population, C – constant, then $N = N_0 C^t$.

The population doubled in size after 18-month (1,5 years) period, than:

$$2N_0 = N_0 C^{1.5}, \quad C^{\frac{3}{2}} = 2, \quad C = \sqrt[3]{4}.$$ $$N = N_0 \left(\sqrt[3]{4}\right)^t$$

After 2 years: $N = 10000 \cdot \left(\sqrt[3]{4}\right)^2 = 10000 \cdot \sqrt[3]{16} \approx 25200$.

Answer: $N_0 \left(\sqrt[3]{4}\right)^t, 25200$.