# Maths log problem solving

Question is:
A certain population of bacteria doubles in number every three hours.
If there are initially 1000 bacteria how many will there be after 5 hours? Give your answer to 3 significant figures.

I thought so you have 1000 on first hour,
2000 on third and 4000 on sixth,
so fifth hour is between 2000-4000,
and that each hour no. increases by x1/3
So increase would be 2000x 2/3 = 1333, and total no. on fifth hour would be 2000 + 1333 = 3330 (3s.f.)
But this is incorrect and I'm stuck. Thanks in advance.
Original post by Bookworm524
Question is:
A certain population of bacteria doubles in number every three hours.
If there are initially 1000 bacteria how many will there be after 5 hours? Give your answer to 3 significant figures.

I thought so you have 1000 on first hour,
2000 on third and 4000 on sixth,
so fifth hour is between 2000-4000,
and that each hour no. increases by x1/3
So increase would be 2000x 2/3 = 1333, and total no. on fifth hour would be 2000 + 1333 = 3330 (3s.f.)
But this is incorrect and I'm stuck. Thanks in advance.

Population = 1000 x 2^(5/3) = 3174 = 3200 2.s.f

Population = 1000 x 2^(n/3)
Where n is the number of hours from initial time

Alternatively, since it is 2000 at 3 hours
Population = 2000 x 2^(2/3) = 3174, this is what you seemed to do
But you forgot to raise 2 to the power of 2/3
(edited 8 months ago)
Original post by BankaiGintoki
Population = 1000 x 2^(5/3) = 3174 = 3200 2.s.f

Population = 1000 x 2^(n/3)
Where n is the number of hours from initial time

Alternatively, since it is 2000 at 3 hours
Population = 2000 x 2^(2/3) = 3174, this is what you seemed to do
But you forgot to raise 2 to the power of 2/3

oic, thank you!