# Bacteria Problem

• March 30th 2006, 08:17 AM
mazdatuner
Bacteria Problem
This poblem Involves bacteria...
there are initially 4000 bacteria per mililiter in a sample. after one hour their concentration increases to 6000 bacteria per mililiter.
a) what is the general exponetial equasion?
b) How many bacteria will there be after 2.5 hours
c) how many hours does it take to reach 8000 bacteria per mililiter.

so far i have isolated that the equasion N(t)=N e^kt is used to solve it
with N(t) being the final number
N being the initial number (4000)
k being a positive constant
and t being the number of hours

from here I have 6000= 4000 e^k*1

any help would be greatly apreciated thank you
• March 30th 2006, 09:12 AM
topsquark
Quote:

Originally Posted by mazdatuner
This poblem Involves bacteria...
there are initially 4000 bacteria per mililiter in a sample. after one hour their concentration increases to 6000 bacteria per mililiter.
a) what is the general exponetial equasion?
b) How many bacteria will there be after 2.5 hours
c) how many hours does it take to reach 8000 bacteria per mililiter.

so far i have isolated that the equasion N(t)=N e^kt is used to solve it
with N(t) being the final number
N being the initial number (4000)
k being a positive constant
and t being the number of hours

from here I have 6000= 4000 e^k*1

any help would be greatly apreciated thank you

a) To get k:
$6000=4000e^k$

$6000/4000=3/2=e^k$

$ln(3/2)=ln(e^k)=k$.
I don't have my calculator on me (and I'm too lazy to dig out the Window's version.) You don't HAVE to have a decimal form for this, but that generally seems to be the standard way to express it.

b) $N(2.5)=4000e^{k*2.5}$. Just calculate it using the k value from a)

c) N(t) = 8000
$8000=4000e^{kt}$

$2 = e^{kt}$

$ln2 = ln(e^{kt})=kt$

$t = ln2/k$.

-Dan