I have the initial equation $$\frac{dP}{dt}=kp$$ where P is the population, t is time, and k is some positive constant. The rest of what I'm given is that P(0) = A, what is the time for the population to double its value?
Now, I've tried doing this in many possible ways but none of them get me to one of the answer choices. $$ A: {\frac{k}{ln(2)}}$$ $$B: \frac{ln(k)}{2}$$ $$C: \frac{ln(2)}{k}$$ $$D: 2A$$ And this is really starting to get on my nerves.
Hint: Assuming $P=p$. Solution to the DE is $$ p(t) = A\exp(kt)$$
And you are looking for $\tau$ so that $p(\tau)=2A=A\exp(k\tau)$.
Edit after question of OP:
$$2A=A\exp(k\tau)$$ $$2=\exp(k\tau)$$ $$\ln(2)=k\tau$$ $$\tau=\frac{\ln(2)}{k}$$