Question

Organisms A and B start out with the same population size.

Organism A's population doubles every day. After 6 days, the
population stops growing and a virus cuts it in half every day for 4
days.
Organism B's population grows at the same rate but is not infected
with the virus. After 10 days, how much larger is organism B's
population than organism A's population?

Answer 1

After 10 days, population B has grown to be 256 times the size of population A.

Explanation:

Answer 2

At the end of ten days, the size of population B is 256 times that of population A

Explanation:

We work under the premise that population A and B start both with the same number of individuals. Let's call such initial population
`N_0`

Now, we write the exponential expression that describes population A as a function of days (t) for the first 6 days:

`N_A=N_0\,(2)^t`

which represents the starting point with
`N_0` individuals on day zero, doubling after one day (t= 1), and keeping on doubling the following days for 6 days.

So at the end of 6 days, population A would have the following number of individuals:

`N_A=N_0\,(2)^6\\N_A=N_0\,(64)\\N_A=64\,N_0`

That is 64 times the starting number of individuals.

After this, the population stops growing and starts reducing to one-half each day. This behavior can be represented by:

`N_A=64\,N_0\,((1)/(2) )^t`

therefore after 4 days in this pattern, this culture has the following number of organisms:

`N_A=64\,N_0\,((1)/(2) )^4\\N_A=64\,N_0\,((1)/(16) )\\N_A=4\,N_0`

which is now just four times what the culture started with.

Now, on the other hand, population B grows doubling each day without interruption, so at the end of 10 days its size is given by:

`N_B=N_0\,(2)^t\\N_B=N_0\,(2)^10\\N_B=N_0\1024\\N_B=1024\,N_0`

that is it has 1024 times the initial number of organisms.

So if we compare both populations at day 10:

`(N_B)/(N_A) =(1024\/N_0)/(4\,N_0) =256`

Therefore, at the end of ten days, population B is 256 times the size of population A