Question

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

Answer 1

Answer:

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

Step-by-step 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:

$N_A=64\,N_0\,(\frac{1}{2} )^t$

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

$N_A=64\,N_0\,(\frac{1}{2} )^4\\N_A=64\,N_0\,(\frac{1}{16} )\\N_A=4\,N_0$

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:

$\frac{N_B}{N_A} =\frac{1024\/N_0}{4\,N_0} =256$

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