Bacteria can multiply at an alarming rate when each bacteria splits into two new cells, thus doubling. If we start with only bacteria which can double every hour, how many bacteria will we have by the end of the day?
A number was not specified as the starting amount, so I am assuming the starting number will be 1.
You would use a formula like:
y=a(1+r)^x
a = the initial amount before the growth begins
r = growth rate
x = the number of intervals
y= end result
r=1 since the rate is only doubling.
x=24 since there are 24 hours in a day
y=1(1+1.00)^24
I believe the right way to solve for this should be.
A= y * 2^r (considering the rate is doubling, that is what the 2 is there for)
A= ending value. (total number of bacteria at the end of 24 hrs)
y= starting value (can be 1 bacteria, can be 10, depends on the experiment)
2=doubling
r= rate or interval (24 hours/one day)
To determine the number of bacteria by the end of the day, we need to consider the doubling rate and the total number of hours in a day.
If each bacterium doubles every hour, we can calculate the number of bacteria at each hour throughout the day, assuming we start with one bacterium:
Hour 1: 1 bacterium
Hour 2: 2 bacteria (1 original + 1 new from the first hour)
Hour 3: 4 bacteria (2 original + 2 new from the second hour)
Hour 4: 8 bacteria (4 original + 4 new from the third hour)
...
Hour 24: (2^23) bacteria
To find the total number of bacteria by the end of the day, we can sum the values from each hour.
Using the formula for summing powers of 2:
Total number of bacteria = 2^0 + 2^1 + 2^2 + ... + 2^23
To simplify the calculation, we can use the formula for the sum of a geometric series:
Total number of bacteria = (1 - 2^(n+1)) / (1 - 2), where n = 23
Plugging in the values:
Total number of bacteria = (1 - 2^24) / (1 - 2)
= (1 - 16,777,216) / (-1)
= 16,777,215
Therefore, by the end of the day, there would be approximately 16,777,215 bacteria.