SOLUTION: The doubling period of a baterial population is 15 minutes. At time t = 90 minutes, the baterial population was 50000. Round your answers to at least 1 decimal place. What was

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The doubling period of a baterial population is 15 minutes. At time t = 90 minutes, the baterial population was 50000. Round your answers to at least 1 decimal place. What was       Log On


   



Question 885241: The doubling period of a baterial population is 15 minutes. At time t = 90 minutes, the baterial population was 50000. Round your answers to at least 1 decimal place.
What was the initial population at time t = 0 ?
Find the size of the baterial population after 4 hours

Found 2 solutions by ankor@dixie-net.com, KMST:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
WITHOUT FORMULAS OR CALCULATOR (but with understanding):
The population of that kind of bacteria doubles every 15 minutes, and after 90 minutes there's 50,000 bacteria.

At time t=90 minutes, 90%2F15=6 doublings have happened since t=0
The population at t=90 minutes is 2%5E6=64 times the population at t=0 ,
so the population at t=0 must have been
50000%2F2%5E6=1541.25
I do not need to round that number, because it is an exact calculation.
Saying that at time t=0 there were 1541%261%2F4 bacteria is a little silly.
I expect the number of bacteria to be an integer.
However, the problem asks to round to at least 1 decimal place, so let's be silly.
How did I calculate that result? With pencil and paper.


At t=4 hours, 4%2A%2860min%2F%2215+min%22%29=4%2A4=16 doubling period have passed since t=0.
That is 16-6=10 doubling periods after t=90 minutes,
so the number of bacteria should be 2%5E10 times what it was at 90 minutes.
That is 50000%2A2%5E10=%2251%2C200%2C000%22 and that is an exact number too.
Should I say it is 51,200,000.00 to have at least one decimal place expressly stated.
How did I calculate that result? With pencil and paper.


WITH FORMULAS AND CALCULATOR:
We are having a case of exponential growth.
P%28t%29= population at t=t minutes
P%28t%29=P%280%29%2A2%5E%28%22t+%2F+15%22%29<-->P%28t%29%2FP%280%29=2%5E%28%22t+%2F+15%22%29
Since calculators allow us to calculate natural logarithms, and base 10 logarithms,
we can also write that as
ln%28P%28t%29%2FP%280%29%29=%28t%2F15%29%2Aln%282%29 or +log%28P%28t%29%2FP%280%29%29=%28t%2F15%29%2Alog%282%29
Going from the natural logarithms version to exponentials on base e we get
P%28t%29%2FP%280%29=e%5E%28%22ln%282%29t+%2F+15%22%29<--->P%28t%29=P%280%29%2Ae%5E%28%22ln%282%29t+%2F+15%22%29
Those are very popular forms. The approximation ln%282%29=0.693 is often used.

At t=90 ,
P%2890%29%2FP%280%29=e%5E%28%22ln%282%2990%2F+15%22%29
50000%2FP%280%29=e%5E%286ln%282%29%29-->P%280%29%2F50000=e%5E%28-6ln%282%29%29-->P%280%29=50000%2Ae%5E%28-+6ln%282%29%29-->P%280%29=50000%2Ae%5E%28-+6ln%282%29%29-->P%280%29=781.25
}}} (using the calculator's value for ln%282%29 )
(Using 0.693 will give you a different decimal part).
At t=4hours=240minutes ,
P%28240%29%2FP%280%29=e%5E%28%22ln%282%29240%2F+15%22%29
P%28240%29%2F781.25=e%5E%2816%2Aln%282%29%29-->P%28240%29=781.25%2Ae%5E%2816%2Aln%282%29%29-->P%28240%29=%2251%2C200%2C000%22