SOLUTION: There are initially 3 E-coli bacteria present in a hamburger. If the doubling time for E-coli is 5 minutes, how many E-coli bacteria will be present after 20 minutes?

Click here to see ALL problems on Exponents

Question 121172: There are initially 3 E-coli bacteria present in a hamburger. If the doubling time for E-coli is 5 minutes, how many E-coli bacteria will be present after 20 minutes?
Found 2 solutions by checkley71, bucky:
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
3*2^4=3*48 BACTERIA WILL BE PRESENT AFTER 20 MINUTES.

Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Think of it this way ... each 5 minutes the number of E-coli doubles from what you had at the
start of the 5 minute period.
.
You start with 3 bacteria and after 5 minutes you have double that ... or 6 bacteria.
.
In the next 5 minute period you start with 6 and you double that, so you end that 5 minute period
with 12 bacteria.
.
In the third 5 minute period you start with 12 and you double that so you finish that period
with 24 bacteria.
.
And in the last 5 minute period you start with 24 and end up the period with 48 bacteria.
.
So in 20 minutes (four 5-minute periods) you go from 3 bacteria to 48 bacteria.
.
Can you find a formula that will cover that process so that you won't have to do the same
sort of process for longer periods of time? Sure we can. All we have to do is to find the
number of 5-minute periods involved and then multiply the starting value by 2 for each
5-minute period. So if we call the number we start with E, and we have six 5-minute periods,
we just multiply E by 2 six times:
.
E*2*2*2*2*2*2
.
but you know that 2*2*2*2*2*2 equals 2^6 so you can shorten the formula to:
.
Number of bacteria = E*(2^6)
.
and you can make it more general by recognizing that the exponent is the total time in minutes
divided by 5 minutes or T/5. This makes the equation:
.
Number of bacteria = E*(2^(T/5))
.
Let's try the original problem by letting E = 3 bacteria and T = 20 minutes. Our equation becomes:
.
Number of bacteria = 3*(2^(20/5)) = 3*2^4 = 3*16 = 48 bacteria.
.
This is the answer we got so our formula works OK.
.
Now let's change the problem. Start again with E = 3 bacteria and find out how many bacteria
there are in an hour (60 minutes):
.
Number of bacteria = 3*(2^(60/5)) = 3*2^12 = 3*4096 = 12,288 bacteria.
.
So if you start with 3 bacteria and double every 5 minutes for an hour, you end up with a
colony of 12,288 bacteria.
.
Hope this helps you to understand the problem a little better.
.