exponential problem

Question

The count in a bacteria culture was 700 after 10 minutes and 1600 after 30 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?     

Find the doubling period.     

Find the population after 65 minutes.     

When will the population reach 14000.  

Answer 1

Exponential growth is modeled by the equation: P(t) = P₀e^kt , where P(t) is population at time t, P₀ is initial population and k is the growth rate.

We know that 700 = P₀e^k(10) and 1600 = P₀e^k(30) so use 700 as P₀ and use 1600 as P(t) where t = 30-10

so t = 20.

Then we have: 1600 = 700e^k(20) Divide both sides by 700 to isolate the exponential: (1600/700) = e^20k

Now, take the natural log of both sides: ln(1600/700) = ln(e^20k) -> ln(1600/700) = 20k (because ln(e)^a=alne and lne = 1)

Divide both sides by 20, and you get: [ln(1600/700)]/20 = k (put this in a calculator to get an approximation for k) (k=0.0413)

Now that you have k, rewrite the exponential equation with k: P(t) = P₀e^0.0413t

We can now use the given time and bacteria count to find P₀. Plug in 700 for P(t) and 10 for t:

700 = P₀e^(0.0413*10) ---> P₀ = 700/(e^0.413) ---> P₀ = 463.1774 (initial size of culture)

For doubling period: You are being asked to find the time taken to double the initial "population" use the formula t = (ln2)/k with the "k" value found above.

For population after 65 minutes: write equation using the k and the P₀ that we found above: P(t) = 463.18e^0.0413t, plug in 65 for t and solve.

For population to reach 14000, use the same equation as above, P(t) = 463.18e^0.0413t, but this time, plug in 14000 where P(t) is (because this is "final population") and solve for t by isolating the exponential (the "e" part) and then taking the natural log (ln) of both sides.

Answer 2

To begin, we need the exponential growth formula: N(t)= N₀e^kt, where N₀ is the initial population, k is the growth constant, and t is time.

From here we are given two populations at two times. We can calculate the growth constant and from there solve for the initial population.

So, are N(t)=1600, N₀=700, and t=30-10= 20. So no we can plug that into the equation:

1600= 700 e^20k

1600/700= e^20k

ln(1600/700)= 20k

0.827=20k

k=0.0413

Now we can solve for the initial N₀. Pick one of the two, and find N₀.

700= N₀e^10*0.0413

700/(e^10*0.0413)= N₀

N₀= 463.1≈463

To find the time in which N₀ doubled, it basically becomes this:

2= e^0.0413*t

ln(2)= 0.0413t

t= 16.8 minutes

For the population after 65 minutes:

N(t)= 463e^0.0413(65)

N(t)= 463*14.7

N(t)= 6783.35≈6783

To find the time when the population reaches 14000, you set that as N(t).

14000=463e^0.0413t

14000/463= e^0.0413t

ln(14000/463)= 0.0413t

3.41/0.0413=t

t= 82.5