Biologists stocked a lake with 300 fish and estimated the carrying capacity (the

Question

Biologists stocked a lake with 300 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 4800. The number of fish doubled in the first year. Assuming that the size of the fish population satisfies the logistic equation ,find K, P(t), and how long it will take to reach 2400 (half of the carrying capacity)

This question is asked many times and hase multiple answers, but theyre too difficult to understand or they are actually done incorrectly.  Can i get a clear step by step on how to solve these types and this exact problem

Explanation / Answer

The formula (logistic equation) you have to use is as follows:

P(t) = Po*K/(Po + (K - Po)e^-kt)
Note: Notice the capitol K and the small k (they represent different things)

K = max carrying capacity
Po = initial population

Given:
Po = 300
P1 = 300 * 3 = 900.....since the population is doubled
K = 6000

Thus, plug into the equation:
so you get,
P(t) = (300*6000)/(300+5700*e^-kt)

Now we know tht P1 is 900 (after the population is tripled)
thus, we can plug in P(t) = 900 and t=1 (t=1, because the population triples after 1 year)

so, we get,

900 = (300*6000)/(300(1+19*e^-k(1)))
thus, 300 cancels out

we are left with:
900 = 6000/(1+19*e^-k)

now solve for k (small k):
1+19*e^-k = 6000/900
19*e^-k = (60/9)-1
e^-k = ((60/9)-1)/19

now take ln of both sides, thus,

-k = ln(0.298)
-k = -1.210
k = 1.210

now, plug this k value into the equation, and you have your answer!

so, the final answer is:
P(t) = 6000/(1+19*e^-1.210t)

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