# Thread: Exponential Growth and Decay

1. ## Exponential Growth and Decay

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?

2. Originally Posted by turtle

Find the amount of time required for a \$2000 investment to double if the annual interest rate r is compounded a) monthly, and b) quarterly

What is r?

Anyway, i'll give you the general solutions and you can plug in r later.

In general, if interest is compounded n times per year, we use

A(t) = A0*(1 + r/n)^nt
where A(t) is the amount of money at any given time, A0 is the initial amount of money, r is the rate of interest, t is time, n is the number of times per year the money is compounded. Here goes.

We want A(t) = 2*A0 (since they said they wanted it to double), so we want

2*A0 = A0*(1+ r/n)^nt
=> 2 = (1 + r/n)^nt .........................divided both sides by A0
=> ln2 = ln(1 + r/n)^nt
=> ln2 = nt*ln(1 + r/n)
=> t = ln2/(n*ln(1 + r/n)), where t is the time in years.

a) Compounded monthly, => n=12
so t = ln2/(12ln(1 + r/12)) years

b) Compounded quarterly, => n=4
so t = ln2/(4ln(1 + r/4)) years

c) annually, => n=1
so t = ln2/ln(1 + r) years

3. Originally Posted by turtle

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?

Ah, you changed the question! I was wondering about that, why did he call it "exponential growth and decay"

4. Originally Posted by turtle

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10000 bacteria. at the end of 5 h there are 40000 bacteria. How many bacteria were present initially?
I hope you dont change the question again while i'm typing this one up.

For exponential decay, we use the formula P(t) = Po*e^rt, where P(t) is the amount after time t, Po is the initial amount, r is the rate of growth, t is the time.

We know P(3) = 10000, P(5) = 40000. So lets take a ratio of the two.

P(5)/P(3) = Poe^5r/Poe^3r = 40000/10000
=> e^2r = 4.................since the Po's cancel, and e^3r into e^5r leaves e^2r
so lne^2r = ln4
=> 2rlne = ln4
=> 2r = ln4
so r = ln4/2 = 0.693 approx.

Now we can find Po by plugging in what we know in either the expression for P(3) or P(5).

Recall: P(t) = Po*e^rt
=> Po = P(t)/e^rt

Say you wanted to use P(5):

Po = 40000/e^(5*0.693) = 1250.9 = 1251 bacteria

Say you wanted to use P(3):

Po = 10000/e^(3*0.693) = 1250.5 = 1251 bacteria

I'm thinking the answer in the back of the book says something like 1250, but that discrepancy would be caused by my rounding off ln4/2