Bacteria Populations

Task

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.

1. Explain, in terms of the structure of the expressions defining $p_1(t)$ and $p_2(t)$, why these two populations never share the same value at any time during the experiment.
2. Explain, in terms of the structure of the expressions defining $p_1(t)$ and $p_3(t)$, why these two populations will be equal at exactly one time during the experiment. Determine this time.

IM 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.

Solutions

Solution: 1

 

1. The first bacteria population has a value of $p_1(0) = 1000$ at the beginning of the experiment while the second population has a value of $p_2(0) = 1500$. The growth rate for the first population is determined by the constant $\frac{1}{3}$ in the exponent of $p_1(t)$. Because $\frac{1}{3}$ is positive, the function $p_1(t)$ grows as time passes. The growth rate for the second population, $\frac{3}{8}$, is larger than than the growth rate for the first population. Since the second population begins at a higher value than the first and it grows at a faster rate, the second population will always be larger than the first.
2. The third bacterial population has an initial value of $p_3(0) = 5000$, five times larger than the first bacteria population. Because the rate of growth for the first population, $\frac{1}{3}$, is larger than the growth rate of the third population, $\frac{1}{4}$, this means that eventually the first population will be larger than the third provided these growth rates continue for a sufficiently long period. Because its rate of growth continues to be faster than that of the third, the first population will be larger than the third population once passing the instant when they are equal.

    

   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 $$ e^{t/12} = 5. $$ Rewriting the equation using the definition of natural log 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.

 

Solution: 2

 

1. It is interesting to see what goes wrong algebraically if we try to set the first population equal to the second population and solve for $t$: $$ 1000e^{t/3} = 1500e^{3t/8}. $$ Dividing both sides by $1000$ and by $e^{3t/8}$ gives $$ e^{-t/24} = \frac{3}{2}. $$ Since only positive values of $t$ make sense within the context of the problem, $e^{-t/24}$ is always less than one and so can never be equal to $\frac{3}{2}$.

    

   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.