given y=Ce^kt where C is the initial value

The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

What is the initial size of the culture?

Thanx in advanced

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- Jun 12th 2009, 10:20 AMjustin016need help growth problem
given y=Ce^kt where C is the initial value

The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

What is the initial size of the culture?

Thanx in advanced - Jun 12th 2009, 11:03 AMapcalculus
Use the two conditions to solve for C and k

Is time measured in hours or minutes?

5000 = C e^(k15)

40000 = C e^(k60)

Divide the second by the first, side by side:

8 = e^(60k-15k)

8 = e^(45k)

ln 8 = 45k

k =ln8 /45

Plug into one of the conditions above to find C. C is the original population, because it is the y-value at t=0.

Good luck!! - Jun 12th 2009, 11:56 AMjustin016
i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

The answer i am looking for is 2500. What did i do wrong? - Jun 12th 2009, 01:10 PMapcalculus
- Jun 12th 2009, 02:58 PMjustin016
- Jun 12th 2009, 08:50 PMmr fantastic
If you don't post your working, how can you expect anyone to answer this? And how can you expect to learn from your mistake(s).

A start:

$\displaystyle 5000 = C e^{\left( \frac{\ln 8}{45}\right) 15} = C e^{\frac{\ln 8}{3}} = C e^{\ln 2}$

where you get that last expression by applying a well known log rule. - Jun 13th 2009, 04:44 PMjustin016
Ce^ln8/3=5000

Ce^ln2=5000

C2=5000

C=5000/2

C=2500

I got confuse where you changed from ln8/3=ln2. How did you do that? - Jun 13th 2009, 05:23 PMmr fantastic
- Jun 13th 2009, 05:26 PMjustin016