Please Help ASAP!!!!

Question

A bacteria has a doubling period of 6 days. If there are 3250 bacteria present now, how many will there be in 33 days?
First we must find the daily growth rate (Round this to four decimal places).
A. The growth rate is?

 

B. Then we use this rate to answer the question.
There will be how many bacteria?

Answer 1

t=0  , then B=3250

t=6  , then B=6500

 

6500 is double 3250.

 

We can use the condition and equation for growth to solve for growth rate:

 

3250e^6k = 6500

 

where k is the growth rate.

 

e^6k = 2

 

6kln(e) = ln(2)

 

6k = 0.6931

 

  k = 0.1155

 

This is the growth rate.

 

Use the formula

 

B(t) = 3250e^0.1155t

 

to find B(33).

Answer 2

N(t) = number of bacteria as function of days elapsed

 

t = number of days elapsed

 

N₀ = initial number of bacteria (at t = 0 days)

 

Since the population doubles every 6 days, divide the number of days elapsed by 6 to get the number of periods over which the population has doubled, and use the figure as the exponent in 2^x.

 

N(t) = N₀ 2^t/6

 

N(33) = 3250 (2^33/6) = 147,078

 

At the end of each day, the population has increased by a factor of 2^1/6 = 1.1225 from what it was at the start of the day.