91Ó°ÊÓ

We have the given information: - The population in year 1995 (t = 1995) is 800, so: \(Q = ab^{1995-t_0} = 800\). - The population grows by \(2.1\%\) per year, so in one year (t = 1996), Q will be \(800 * 1.021\). Thus: \(Q = ab^{1996-t_0} = 800 * 1.021\). Now, we have two equations with three unknowns (\(a\), \(b\), and \(t_0\)). Let's move to the next step.

First, let's divide the second equation by the first equation to eliminate the constant a: \(\frac{800 * 1.021}{800} = \frac{ab^{1996-t_0}}{ab^{1995-t_0}}\) Now, since both terms have the same base \(b\), we can finish solving for \(b\): \(1.021 = \frac{b^{1996-t_0}}{b^{1995-t_0}} = b^{(1996-t_0)-(1995-t_0)} = b^1\) It is clear that \(b = 1.021\) since the exponent is 1. Now, we can plug the value of \(b\) back into the first equation to solve for \(a\): \(800 = a(1.021)^{1995-t_0}\) At this point, we need a reference point to find \(t_0\) and \(a\). We can say that \(t_0\) denotes the initial year of observation. Therefore, we set \(t_0 = 1994\). Now we can solve for \(a\): \(800 = a(1.021)^{1995-1994} = a(1.021)^1\) So, \(a = \frac{800}{1.021} \approx 783.34\) Now, we have the values of a, b, and \(t_0\): \(a \approx 783.34\), \(b = 1.021\), and \(t_0 = 1994\). Thus, the function representing the population growth is: \(Q \approx 783.34(1.021)^{t-1994}\)