The population of a particular type of bacteria doubles in number every 12 hours.
a)if there are 8 bcteria initially, write an equation that models the number of bacteria present after x 12 hours.
b)How many bacteria will be present after 5 days?
by using this formula N(t)=N0(1+r)2
The 0 is a small o on the bottom. i could be wrong with the formula.
Since this is a "doubling" process, I would use
N(t) = 8(2)^(t/12) where t is in hours
b) so after 5 days or 120 hours,
N = 8(2)^120/12
= 8192
To solve this problem, we need to use the exponential growth formula. The formula you mentioned, N(t) = N0(1 + r)^(t/d), is a common formula for exponential growth, where N(t) is the final number of bacteria at time t, N0 is the initial number of bacteria, r is the growth rate (expressed as a decimal), t is the total time, and d is the doubling time.
a) The population of bacteria in our case doubles every 12 hours. So, the growth rate would be 1, as it is doubling. The initial number of bacteria is given as 8.
Using the exponential growth formula, the equation that models the number of bacteria present after x 12 hours can be written as:
N(x) = 8(2)^(x/12)
b) To find out how many bacteria will be present after 5 days, we need to convert 5 days into the number of 12-hour intervals. Since each day has 24 hours and each interval is 12 hours, 5 days would be 5 * 24 / 12 = 10 intervals.
Now, substituting x = 10 into the equation, we can find the number of bacteria after 5 days:
N(10) = 8(2)^(10/12)
Simplify it further:
N(10) = 8(2)^(5/6)
N(10) = 8 * (2^(5/6))
Using a calculator or by simplifying further through approximations, you can find the approximate value for N(10) after 5 days.
Keep in mind that as the growth is exponential, the final number of bacteria will be significantly higher than the initial number.