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
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!!
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.