bacteria initial population

Question

An initial population of bacteria grows continuously at a rate of 2.8% per day. What is the population after 48 hours if the initial bacertial population is 180?

Answer 1

Hi Hailey,

 

I would use P = Ae^rt, where growth rate is 0.028 bacteria per day, t = 2days, A = 180

P = 180e^(0.056)

= 190 bacteria (rounded to the nearest whole number)

Answer 2

A growth rate of 2.8% per day means that at the end of a day, the population is 1.028 times what it was at the start of the day.  Doing this for 2 days in a row (48 hours) with an initial population of 180,

 

P(2 days) = 180(1.028)² ≅ 190

Answer 3

We can use an exponential equation to solve this.  We should know that in this scenario A=I(1+r)ⁿ  With our information, the intial population (I) is 180, the growth rate (r) is .028, and the number of time intervals (n) is 2 days.  If we plug this all in we know that A=180(1.028)² which is equal to 190.22112.

 

Side note: The wording in this problem is a bit ambiguous as it describes continuous growth and also a daily growth rate.  It is possible that you need to convert the daily growth rate into a continuous rate and solve using A=Pe^rt

Answer 4

A = a₀e^rt

A = 180e^(1.028)(2)

 

Can you calculate?