One-stray Problem Solving
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A population of bacteria is growing at an hourly rate of r so that at the end of each hour the population of the sample is multiplied by the growth factor of
z = 1+r. The beginning population is 800 bacteria, an additional 300 are present after 1 hour, and an additional 400 are present after the following
hour.
Write an expression for the population at the end of three hours in terms of the growth factor.

Given a bacteria population that is multiplied by z every hour, and the population after 2 hours is 1500, you want the population after 3 hours in terms of z.

Application

The problem statement tells you the population is multiplied by z every hour. After 2 hours, the population is 800 +300 +400 = 1500. After 3 hours, that value will be multiplied by z:

1500z . . . . . population after 3 hours

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Additional comment

The given values of population tell us z = 1100/800 = 1.375 for the first hour, and z = 1500/1100 ≈ 1.364 for the second hour. The average value of z for the two-hour period is √(15/8) ≈ 1.369. The desired expression could be any of ...

800z³ . . . . . . . population 3 hours after the beginning
1100z² . . . . . . . population 2 hours after the first hour
1500z . . . . . . . population 1 hour after the second hour

Presumably z would have different values in each case, and we hesitate to predict what those values might be. Perhaps the population is being rounded to the nearest 10 or 100.

An exponential regression calculation models the population as ...

p = 801.945×1.36812^t . . . . . . . z ≈ 1.36812

Using this formula, we get the given population numbers by rounding to the nearest 10.