A hospital is conducting a study to see how different environmental conditions influence the growth of streptococcus pneumonia, one of the bacteria which causes pneumonia. Three different populations are studied giving rise to the following equations: \begin{eqnarray*} p_1(t) &=& 1000e^{t/3}, \\ p_2(t) &=& 1500e^{3t/8} , \\ p_3(t) &=& 5000e^{t/4}. \end{eqnarray*} Here \(t\) represents the number of hours since the beginning of the experiment which lasts for 24 hours and \(p_i(t)\) represents the size of the \(i^{th}\) bacteria population.
To find out when the first and third populations are equal, we solve the equation \(p_1(t) = p_3(t)\): $$ 1000e^{t/3} = 5000e^{t/4}. $$ Dividing both sides of the equation by 1000 and then by \(e^{t/4}\) gives $$ e^{t/12} = 5. $$ Taking the natural logarithm of both sides gives $$ \frac{t}{12} = \ln{5}. $$ So the first and third populations are equal when $$ t = 12\ln{5}. $$ Using a calculator to find \(\ln{5}\), the first and third populations are equal after approximately 19 hours and 19 minutes.
Alternatively, we may observe that $$ \frac{1500e^{3t/8}}{1000e^{t/3}} = \frac{3e^{t/24}}{2}. $$ Since \(\frac{3}{2} > 1\) and \(e^{t/24}\) is always larger than \(1\) for positive \(t\), it follows that $$ \frac{1500e^{3t/8}}{1000e^{t/3}} > \frac{3}{2} $$ for all positive values of \(t\) and so the first and second populations are never equal during this experiment.
Commentary
This task provides a real world context for interpreting and solving exponential equations. There are two solutions provided for part (a). The first solution demonstrates how to deduce the conclusion by thinking in terms of the functions and their rates of change. The second approach illustrates a rigorous algebraic demonstration that the two populations can never be equal.