Exponential and Logarithmic Models

Question

The doubling period of a bacterial population is  minutes. At time  minutes, the bacterial population was 80000.The initial population at time  787.45Correct  Find the size of the bacterial population after 4 hours.

Answer 1

2 = e^14r where r=growth rate, 14 minutes= doubling time

take natural logs of both sides

ln2 = 14r

r = ln2/14 = 0.0495 = 4.95% growth rate

BUT

80,000=787.45e^100r

ln80000/787.45) =4.621 = 100r

r = 0.04621 which is probably more accurate as 14 minutes as doubling time was probably rounded off to nearest minute

4 hours = 60x4 = 240 minutes

use r=4.621% growth rate

A=787.45e^240(.04621) =

787.45e^11.0904

= 787.45(65539)= 51,608,651.27= about 52 million after 4 hours

doubling time means

2 = e^.04621t where t=doubling time in minutes

ln2 = .04621t

t = ln2/.04621 = 14.99994 = 15 minutes for doubling

Odds are there's a typo and the 14 was supposed to be 15 for the doubling time. 4 is close on the keyboard to 5.