SOLUTION: A bacteria culture started with 6500 bacteria and increased by 35% in the first 2 hours. How long does it take the population to triple in size? Use Q(t)=6500e^1.35/2(t)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: A bacteria culture started with 6500 bacteria and increased by 35% in the first 2 hours. How long does it take the population to triple in size? Use Q(t)=6500e^1.35/2(t)      Log On


   

Question 821723: A bacteria culture started with 6500 bacteria and increased by 35% in the first 2 hours.
How long does it take the population to triple in size?
Use Q(t)=6500e^1.35/2(t)
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
---
f(t)= P*e^(tk)
where:
P = initial value
k = growth or decay factor per unit of time
t = time
---
f(0) = P = 6500
f(t)= 6500*e^(tk)
---
f(2) = 6500*1.35 = 8775
8775 = 6500*e^(2k)
8775/6500 = e^(2k)
ln(8775/6500) = ln( e^(2k) )
ln(8775/6500) = 2k
k = ln(8775/6500)/2
k = 0.1500523 growth factor per hour
---
time to triple the population:
f(t)= 3*6500 = 6500*e^(tk)
3 = e^(tk)
ln(3) = ln( e^(tk) )
ln(3) = tk
t = ln(3)/k
t = 7.3215293
---
answer:
time to triple the population = 7.3 hours = 7 hours and 19.3 minutes
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php