Log/Ln question?

Question

The population of bacteria is observed to be F(t) = C*e^(2t) where C is some constant. If after one hour a scientist observes the growth rate to be 552 bacteria/hour, What was the initial bacteria population?

Answer 1

The growth rate is given by

F'(t) = 2Ce^(2t).

Plug in t=1 and F'(t)=552. You have

552 = 2Ce^2

so

C = 552/(2*e^2) = 37 (rounded)

Answer 2

the growth rate = dF(t)/dt = 2C*e^(2t). Assuming that t is measured in hours, then the growth rate (1) = 2C*e^2 = 552
C = 37.35 bacteria, which's the initial population

Answer 3

Differentiate to get the formula for growth rate:
F'(t) = 2C*e^(2t)

The information in the question tells you that
F'(t) = 552 when t = 1
so if you substitute these values above you can solve to find C.

Then go back to the original formula for F(t) and substitute (using the value of C you just found, of course) t = 0 to get F(0). Yes, of course it's equal to C.

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Answer 4

let y= C*e^(2t)
dy/dt=2C*e^(2t)
substituting t=1 and dy/dt=552
552=2C*e^2
C=276/e^2
Initial bacteria population=C*e^(2*0)
=C=276/e^2

Answer 5

Sounds like one of your homework problems. If you just want to check to see if you got the right answer, use a computer.