Exponential Growth Problem Involving bacteria population??! PLEASE HELP!!! - Precalculus

Question

I am really stuck on this question, particularly questions 3 and 4. Could you please show how you get the answers to those questions and explain? I have been stuck on those for a while and would really like to learn how to achieve the correct answer! I am fairly certain I can figure out problems 1 and 2.

 

 

The count in a bateria culture was 400 after 10 minutes and 1300 after 35 minutes.

What was the initial size of the culture?
Round your answer to the nearest bacteria.

Find the doubling period. Round your answer to one decimal place.

Find the population after 75 minutes. Round your answer to the nearest bacteria.

When will the population reach 14000. Round your answer to one decimal place.

 

 

 

Thank you for the help!!

Answer 1

The exponential growth equation can be written as follows.

 

C=C0*e^(rt)                 (1)

 

or 

 

Ln(C)=Ln(C0)+rt           (2)

 

Here, C0 is the count at t=0, C is the count at time t, and r is the growth rate.

 

You are given

 

C(10)=400

 

and

 

C(35)=1300

 

These two equations allow you to find C0 and r.  The easiest way is to use Equation (2) to get two simultaneous equations in the two unknowns C0 and r.

 

Ln(400)=Ln(C0)+10r              (3)

 

Ln(1300)=Ln(C0)+35r            (4)

 

subtract Equation (3) from Equation (4) to get:

 

15r=Ln(1300)-Ln(400)=Ln(13/4)=1.178655

 

r=1.178655/25=0.0471462

 

Substitute this value for r in Equation (3) to find C0.

 

Ln(C0)=Ln(400)-10r=5.9915-0.471462=5.52

 

C0=e^5.52=249.64, i.e., 250.

 

The doubling period of an exponential growth process satisfies the condition that C=2*C0, or C/C0=2.  Thus, it can be found as follows.

 

C/C0=e^rT

 

Ln(2)=rT

 

T=Ln(2)/r=0.69315/0.0471462=14.7 minutes.

 

Find the count at t=75 as follows.

 

C(75)=250*e^(0.0471462*75)=8582

 

Find when the count will be 14,000 as follows.

 

C(t)=14000=250*e^(0.0471462*t)

 

Ln(14000/250)=0.0471462*t

 

t=Ln(1400/25)/0.0471462=85.4 minutes