SOLUTION: Hi, I'd like some explanation this question please: 3. A population of rats grows at a rate which is proportional to the number of rats, y, existing at the end of month t. Init

Question 953768: Hi, I'd like some explanation this question please:
3. A population of rats grows at a rate which is proportional to the number of rats, y, existing at the end of month t. Initially there are 50 rats, and after 2 months there are 100 rats.
(i) Find a relationship between y and t.
(given answer --> y = 50(sqrt(2))^t)
(ii) Find the number of rats present after 1 year. (given answer --> 3200)
My answer to (i) is different but still gives the correct values for 2 months and 12 months:
y= exp ( (ln2)/2 * t + ln50)
Obviously mine is much less elegant, but I can't see how the given answer is arrived at.
My equation is derived from:
dy/dt = ky
Thanks,
Robert
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We can solve this using the constant of proportionality, k
we are given,
y = 50*k^t and y = 50 for t = 0, hence t is the exponent of k
100 = 50*k^2 and y = 100 for t = 2, then
k^2 = 2
k = square root (2)
therefore
y = 50 *(square root(2)^t)

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