Question 920739
i believe the exponential growth model equation is f = p * e^(rt)


f is the future value
p is the present value
e is the scientific constant of 2.71828...
r is the growth rate per time period.
t is the number of time periods.


in your problem, the bacterial grow from 400 to 900 in 3 hours.


the population is 400 after 2 hours and the population is 900 after 5 hours.


that's 3 hours in between.


your equation becomes:


900 = 400 * e^(3r)


f = 900
p = 400
t = 3 hours
r = what you want to solve for.


divide both sides of this equation by 400 and you get:


900/400 = e^(3r)


simplify to get 9/4 = e^(3r)


take the natural log of both sides of this equation to get:


ln(9/4) = ln(e^(3r))


since ln(e^(3r)) equals 3r * ln(e) and ln(e) = 1, you get:


ln(9/4) = 3r


divide both sides of the equation by 3 to get:


ln(9/4) / 3 = r


solve for r to get r = ln(9/4) / 3 = .2703100721


confirm by replacing r with .2703100721 in the original equation to get:


900 = 400 * e^(.2703100721*3) which becomes:


900 = 900.


this confirms the solution is good.


you are asked to find how many bacteria after 10 hours.


the formula for that is the same general formula of f = p * e^(rt)


you can choose p to be 400 or 900
if you choose 400, then t = 10-2 = 8 hours
if you choose 900, then t = 10 - 5 = 5 hours
you should get the same answer.
i'll do both to show you.


when p = 400
t = 10-2 = 8
r = .2703100721
f = what you want to find.


you get:


f = 400 * e^(.2703100721 * 8) = 3477.069512


when p = 900
t = 10-5 = 5
r = .2703100721
f = what you want to find.


you get:


f = 900 * e^(.2703100721 * 5 = 3477.069512


you're good either way.