Posted by **Anonymous** on Thursday, October 21, 2010 at 8:52pm.

a populatioin of bacteria has an initial size of 500. the population doubles every 7 hours.

a) how many bacteria will there be after 10 hours

b) when will there be 3000 bacter?

I keep getting answers that just don't make sense. Can someone do these out for me......its practice, not homework. I have a test tomorrow.

- Calculus -
**Reiny**, Thursday, October 21, 2010 at 9:01pm
I would set my equation as

N = 500(2)^(t/7) , where 7 is the number of hours

a)when t=10

N = 500(2)^(10/7) = 1345.9 or 1346 bacteria

b) set 3000 = 500(2)^(t/7)

6 = 2^(t/7)

t/7 = log6/log2

t = 7log6/log2 = 18.1

makes sense ....

after 7 hours 1000

after 14 hours 2000

after 21 hours 4000 , 3000 seems reasonable after 18

- Calculus -
**Anonymous**, Thursday, October 21, 2010 at 9:17pm
sorry I forgot to mention I have to use the equation: p(t)= ce^kt

- Calculus -
**Reiny**, Thursday, October 21, 2010 at 9:25pm
ok, then change it to

N = 500(e)^(kt)

when t = 7

1000 = 500(e)^(7k)

2 = (e)^(7k)

7k = ln2

k = ln2/7 = .09902

N = 500(e)^(.09902t)

now follow the same steps as before

a) let t=10 .... ( I got exactly the same answer as before)

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