Inactive Tutor answered • 03/06/22
Hello Bethany!
I'll try and help you out!
Based on their description, the bacteria in question grow over time according to this general ("power-law") formula:
P(n) = P0 * (1 + r)n
where P(n) is the "final" population after n hours, P0 is a constant (that turns out to be equal to the initial population), and r is the "rate" of growth per hour.
In this problem, we want to solve for P(n = 5), the population after 5 hours, but we don't directly know P0 or r. To solve for those two unknown parameters, we can use the data points given to us in the problem!
"...there are 500 present at a given time..." means that...
P(n = 0) = 500
"...and 800 present 2 hours later..." means that...
P(n = 2) = 800
Let's plug that first data point into our general formula...
P(n) = P0 * (1 + r)n
P(n = 0) = 500 = P0 * (1 + r)(0) (anything to the zeroth power is just = 1)
500 = P0 * (1)
P0 = 500
So our first data point allowed us to learn the value of P0, which is actually just equal to the initial population value. Let's apply the 2nd data point now...
P(n) = P0 * (1 + r)n
P(n = 2) = 800 = (500) * (1 + r)(2)
(8 / 5) = (1 + r)2 (take sqrt() of both sides...)
√(8/5) = (1 + r)
r = √(8/5) - 1
r = 0.265
So now we know both unknown parameters in the equation, now we have the power to use the equation to solve for the population at any time...
P(n) = P0 * (1 + r)n
P(n) = 500 * (1 + 0.265)n
P(n) = 500 * (1.265)n
Plug in n = 5, our time of interest for this problem...
P(n=5) = 500 * (1.265)5
P(n=5) = 500 * 3.239
P(n=5) = 1620 bacteria
So the # of bacteria after 5 hours should be 1620 bacteria.
Note that is doesn't make sense to have like, 0.4 of a bacterium, so it is wise to round to the nearest whole number here.
Hope that helps!
--Zach
