Calculus Question

[JohnCU]

A bacteria culture grows with constant relative growth rate. After 2 hours, there are 600 bacteria and after 8 hours the count is 75,000.

Find the initial population.

Here's what I know:

y(t) = y(0)e^(kt), where y(0) is the population at time t.
y(2) = 600
y(8) = 75000

So, I did this...

Letting y(0) = x...

600 = x*e^(2k)

ln600 = lnxe^(2k)
ln600 = (2k)lnxe
ln600 = (2k)lnx + (2k)lne
ln600 = (2k)lnx + 2k
ln600 - 2k = (2k)lnx

(ln600 - 2k) / 2k = lnx

e^((ln600 - 2k) / 2k) = x, and now since everything is in terms of k, now I can figure out y(0), right?

600 = e^((ln600 - 2k) / 2k) * e^(2k)

600 = e^((ln600 - 2k + 4k^2) / 2k)

That seems awfully messed up, though...am I doing this wrong or what?

 

[poopygood]

Who is John F. Kennedy.

 

[sash1]

I don't even know anymore!

definitely looking forward to the AP test soon!

`K

 

[ClueLis]

You need to use the value for y(8) to find k.

Edit: I would also have done it the other way around, finding k first, then solving for y(0).

 

[ngvepforever2]

k=ln(750/6)/6

how?

like Cluelis said used Y(8) to find y(0)

600=y(0)e^2k

600/(e^(2(ln(750/6)/6))=y(0)

120=y(0)

 

[RossGr]

ln600 = lnxe^(2k)
ln600 = (2k)lnxe
ln600 = (2k)lnx + (2k)lne

Should read

ln 600 = lnx + 2k ln e
ln 600 = lnx + 2k

ln (600/x) = 2k

k = ln (600/x) /2

 

[dude8604]

Another way...

y(t) = y(0)e^(kt)
75000 = 600e^(6k)
(75000/600) = e^(6k)
125 = e^(6k)
ln(125) = ln(e^(6k))
4.828 = 6k
k = 0.805

75000 = y(0)e^(.805 * 8)
75000 = 626.407 * y(0)

y(0) = 119.730

 

[JohnCU]

Thanks, I see my error.