how many will there be 5 hours from the initial time?

Question

A certain type of bacteria increases continuously at a rate proportional to the number present. If there are 500 present at a given time and 800 present 2 hours later, how many will there be 5 hours from the initial time?

Answer 1

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