Posted by **Vanessa** on Monday, July 14, 2008 at 10:46am.

A population of 500 E. coli bacteria doubles every 15 minutes. Use this information to find an expression for this population growth. Using this expression, find what the population would be in 87 minutes. Use an exponential model.

so we're supposed to use P(t)= (Po)(e)^kt

so P(o) would be 500,

and then would t be 15 or 0.25 and wowhat would k be? 2?

- Calculus -
**Reiny**, Monday, July 14, 2008 at 11:43am
in P(t) = 500 e^kt

it depends whether you want to to be in minutes or hours.

let's do it for minutes, then

1000 = 500 e^15k

2 = e^15k

15k = ln2

k = (ln2)/15

if you want t to be hours then use .25 for t in

1000 = 500 e^.25K

.

.

K = ln2/.25 or K = 4ln2

- Calculus -
**Vanessa**, Monday, July 14, 2008 at 12:51pm
okay, so then in 87 minutes what would it be?

- Calculus -
**Reiny**, Monday, July 14, 2008 at 1:42pm
let's use our first one, the one in minutes

P(87) = 500 e^(87*ln2/15)

= 500 e^4.020254

= 27857.6

is our answer reasonable??

remember it doubles every 15 min

so after 15 min --> 1000

after 30 min ---> 2000

after 45 min ---> 4000

after 60 min ---> 8000

after 75 min ---> 16000

after 90 min ---> 32000

we could check to see if we get 32000 for 90 minutes

P(90) = 500 e^(90*ln2/15)

= 500(64) = 32000 YES!!!!

- Calculus -
**Reiny**, Monday, July 14, 2008 at 7:18pm
e is one of these very strange transcendental numbers like pi

one defn of e is

1 + 1/1! + 1/2! + 1/3! + ... to infinitity

where something like 5! = 1*2*3*4*5

or

e = Limit (1 + 1/n)^n where n approaches infinitity

The Swiss/German mathematician Euler spent a lot of time with the concept of that number and e is often called Euler's number

- Calculus -
**Anonymous**, Saturday, June 16, 2012 at 6:14am
e is approximately 2.71828

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