differential equations

[tiff_tiff]

how would i solve this question???

[Phenomenol]

I'm in a good mood, so here is the solution to the first part of the question:

[Special At Specialist]

Following on from Phenomenol's solution:
Q(t) = Q₀e^{k(t₀ - t)}

If the half-life is 5568 years, then that means there will be twice as much carbon at time zero as there will be after 5568 years:
Q(0) = 2*Q(5568)
Q₀e^kt₀ = 2*Q₀e^{k*(t₀ - 5568)}
e^kt₀ = e^ln(2)e^{k*(t₀ - 5568)}
kt₀ = ln(2) + kt₀ - 5568k
5568k = ln(2)
k = ln(2) / 5568

Now they tell us that Q / Q₀ = 0.3 = 3/10
We want to substitute the value of k into our equation Q(t).
First note that e^k = e^ln(2)/5568 = 2^1/5568
Now Q(t) = Q₀*2^{(t₀ - t)/5568}
But also Q(t) = 3/10 * Q₀
So therefore Q₀*2^{(t₀ - t)/5568} = 3/10 * Q₀
Which means 2^{(t₀ - t)/5568} = 3/10
(t₀ - t) / 5568 = log2(3/10)
t₀ - t = 5568*log2(3/10)
t = t₀ - 5568*log2(3/10)