Question

The number of bacteria in a Petrie dish are modelled using the exponential function `N(t) = Ae^(Bt)` where `t` is in minutes. The initial population is `2500`. After `20` minutes the population is `8400`. How many bacteria after `60` minutes?

Answer 1

`N(t) = 2500 e^(1/20 ln(3.36) t)`

`N(60) ~~ 95000`

Explanation:

We are given:

`N(t) = Ae^(Bt)`

`N(0) = 2500`

`N(20) = 8400`

So:

`8400/2500 = (N(20))/(N(0))`

`color(white)(8400/2500) = (Ae^(B(color(blue)(20))))/(Ae^(B(color(blue)(0))))`

`color(white)(8400/2500) = e^(20B)`

Taking natural logarithms of both ends:

`20B = ln(8400/2500) = ln(3.36)`

So:

`B=1/20 ln(3.36)`

Then:

`2500 = N(0) = Ae^0 = A`

So:

`N(t) = 2500 e^(1/20 ln(3.36) t)`

and:

`N(60) = 2500 e^(3 ln(3.36)) = 2500(3.36^3) = 94832.64`

That exact figure is what the model gives us, but not only can you not have a fractional number of bacteria, but the original counts were only stated to `2` significant figures. Therefore we should probably state the figure for `60` minutes to `2` significant figures too, i.e. `95000`.