How do u solve this problem: 125 bacteria that double every hour after 6 hours? Book says answer is 8000. How?

Question

This problem deals with Exponential Growth and Decay, it says to predict the population of bacteria for each situation and time period in this case its : 125 bacteria that double every hour after 6 hours. The book says it's 80000 but how did they come to that answer i want to know how to work it out.....This is Algebra 2.....Help please=)

Answer 1

At time = 0, you have 125 bacteria
At time = 1, you have 250
At time = 2, you have 500
At time = 3, you have 1000
At time = 4, you have 2000
At time = 5, you have 4000
At time = 6, you have 8000 (not 80,000!)

At time = T, you have 125 * (2^T)

Answer 2

The bacteria doubles every hour.

If you start with quantity A then

in 1 hour you will have 2A
in 2 hours you will have 2*2A=2^2*A=4A
in 3 hours you will have 2*2*2A=2^3*A=8A

So in 6 hours you will have 2^6*A

A=125 and 2^6=64

64*125=8000.

Answer 3

125 x2=250
250 x2=500
500 x 2=1000
1000x 2=2000
2000 x 2=4000
4000 x 2=8000

8000 bacteria after 6 hours

125 bacteria x 2^6=8000
125 x64=8000

Answer 4

To find the number of bacteria, starting with 125 at t = 0, after t = 6 hours lapsed, start with the logic for each hour that passes:

Thus, when t = 0, we have N(0) = n0 = n0(2)^t = n0(2^0) = 125
t = 1; N(1) = 2 n0 = n0(2)^t = n0(2^1)
t = 2; N(2) = 4 n0 = n0(2)^t = n0(2^2) and so on.

From this logical series, we see in general N(t) = n0 (2^t); for t = 0......T Thus, at the end of the sixth hour (t = 6), we have N(6) = n0(2^6) = 8000

A good way to work problems like this one is to do an equation for each increment (each hour in this case). See what the pattern is for each increment and then generalize on that pattern, e.g., the N(t) = n0 (2^t) is the generalized equation for each of the steps done earlier.

Answer 5

125*2*2*2*2*2*2 (one 2 for each hour) = 125*64 = 8000

Answer 6

In order to have an answer there must be a question!