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Using Exponential Functions to Determine the Growth of Bacterial Populations

Exponential functions are useful for predicting the growth of something after a certain amount of time and it has many real-world applications in science and lives. One such example is the ability to accurately predict bacterial growth, that is the asexual reproduction of a bacterium into two daughter cells repeatedly, over time. To approach this problem, the fundamental principles of exponential functions needs to be explained.

In mathematics, exponential functions differ from polynomial functions such that the former has the independent variable raised as an exponent. Therefore, the simplest exponential function is described as f(x)=ab^x where a is the initial amount, b is the rate of growth or the rate of decay if it is closer to zero, and x is the independent variable. The growth or decay formula A=Pe^rt will also be useful to solve related problems, where A represents the final amount, P is the initial amount, r is the growth or decay rate, e is the Euler constant, and t is time. The simplest form of an exponential function, without any transformations applied to it, shall be sufficient to solve the problem since we do not need to draw a graph to answer it. Now let us solve a bacterial growth problem using this knowledge:


Problem: A biologist is researching a newly-discovered species of bacteria. At time = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant k for the bacteria? (Round k to two decimal places.)


For these types of problems, it is important to recognize the unit of time (here hours), so that we keep all of our units the same when answering the question. The question tells us that when the time is zero, A = 100 so that is the initial value, and that at t = 6, the biologist measures 450 so that is the final value. However, the only variable that a value is not known for is k, which is what we are trying to find. To solve this problem, the things known are inputted into the growth formula and k is then solved for:

450 = 100e^6k 4.5 = e^6k ln(4.5)/ = 6k ln(4.5/6 = k k = 0.250679566129

Therefore, the growth constant k or rate in this problem is 0.25/hour. Most exponential problems follow the similar problem model above, and if we read the problem correctly and identify what we need to know versus what we already have, exponential problems should not be a problem to solve. Exponential functions can help us solve a variety of problems, and this is just one of them!

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