I have no idea how to do this exercise. The problem: Population growth rate is proportional to the the amount of population. At the time between $t=0$ and $t=1$ the population has increased three times. After how long will the population be 100 times larger than at $t=0$?
My understanding of it: $P'-kP=0$
$Pe^{-kt}=C$ where $C$ - any constant.
$P(t)=Ce^{kt}$ This is general solution to the problem.
Then at $t=0$ $P(0)=C\cdot e^0 = C$ but the value at $t=0$ is not given so there is nothing I can do. I know that logarithm is involved in solution, but I don't know why and what are the steps to resolve this problem. Can someone help?
ِAssume that $P(0)=C=a$, and $P(1)=Ce^{k}=3a$, now divide $\frac{P(1)}{P(0)}$
we get: $e^k=3$, so $\boxed{k=\ln(3)}$
We obtained earlier that $P(t)=Ce^{kt}$, this means that $P(t)=ae^{t\ln(3)}$
Replacing $P(t)$ by $100a$ [this is the required population], we get:
$100a=ae^{t\ln(3)}$, dividing by $a$ we get: $100=e^{t\ln(3)}$, taking the natural logarithm on both sides we get: $\ln(100)=t\ln(3)$, dividing both sides by $\ln(3)$ to get the required time, $t=\frac{\ln(100)}{\ln(3)}\approx 4.1918...$