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
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.
3 answers
sorry I forgot to mention I have to use the equation: p(t)= ce^kt
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)
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)
3 answers