The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. after 1 day, it is observed that 200 bacteria are present. after 3 days, 2000 bacteria are present. what was the initial number of bacteria?

The initial number of bacteria is 63.

Step-by-step explanation:

The population of the bacteria can be modeled by this following differential equation:

dP/dt = Pr

where r is the growth rate.

Solving this equation by the method of variable separation, we end up with:

dP/P = rdt

Integrating both sides, we have equation 1)

1) ln P = rt + P0

Where P0 is the initial number of bacteria

We need to isolate P in equation 1), so we do this

e^(ln P) = e^(rt + P0)

P(t) = P0(e^(rt))

Given that P(1) = 200

P0e^(r) = 200

The same for P(3) = 2000

P0e^(3r) = 2000

Now we solve the following system of two equations, where we want the find two values(P0 and r)

1) P0e^(r) = 200

2) P0e^(3r) = 2000

Isolating P0 in 1) in 1 and replacing it in 2), we have

P0 = 200e^(-r)

200e^(-r)e^(3r) = 2000

We know that e(a)e(b) = e(a+b), so e^(-r)e^(3r) = e(2r), so

e(2r) = 10

we need to find r, so we put ln in both sides of the equation

ln(e(2r)) = ln(10)

2r = 2.30

r = 1.15

Now, from 1), we know that

P0 = 200e^(-r) = 200e^(-1.15) = 63

So the initial number of bacteria is 63.

lol we need to see what the question is asking about

step-by-step explanation:

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol