Exponential Growth and Decay; Modeling Data Help......

The doubling period of a baterial population is, 10 minutes. At time, t = 80 minutes, the baterial population was 60000. For some constant A, the formula for the population is, p (t) = Ae^kt where, k = ln (2) / 10 . What was the initial population at time ? ____________

Find the size of the baterial population after 3 hours. ___________

1. Find A by plugging in values you already know and solving for A

60000 = Ae^(8 ln (2) )

a= 60000 / e^(ln 256)

plug 0 into t to get intial population

plug 18 into t to get populatino after 3 hours

p(t) = A e^kt

Doubling time is 10 minutes
A e^10k = 2A
e^10k =2
10k = ln(2)
k=ln(2)/10

k = 0.069315

P(t) = A e^0.069315t

When t=80, P(t) = 60000
60000 = A e^0.069315(80)
60,000 = 256.00577 A
A = 60,000 /256.00577 =234.37
A=235 ----- initial population

P(t) = 235 e^0.069315t

To find the population after 3 hours (180 minutes), substitute t=180
P(180) = 235 e^[0.069315 (180)]
= 61 606 966

p(80) = 60000 = Ae^((ln(2)/10)*80)
e^((ln(2)/10)*80) = e^(8*ln(2)) = e^(ln(2^8)) = 2^8 = 256
A = 60000/256 = 234.375
The initial population is approximately 234 bacteria.

p(3*60) = p(180) = 234.375e^((ln(2)/10)*180) = 234.375*2^18 = 61 million

step 1:
ok so the formula for the populations(After plugging in for k) is p(t)=Ae^(ln(2)/10)t
this can be rewritten(using the property of ln) as p(t)=Ae^(ln(2))(1/10)t
This becomes p(t)=A2^(t/10)
step2:
We know that at time 80 the population is 60000
so p(80)=60000=A2^(80/10)=A2^(8)
this becomes 60000=A(256) divide both sides by 256, 60000/256=256/256A
A=234.375, ok Know since we know what A is we can find the Initial population(YAY!)
Initial Population is when time is zero of course, so plug in 0 for t and all the values we know now.