So here is the question i received:
The rate of growth of a tumor is directly proportional to the size of the tumor. If the tumor is $5mm$ across at the time $t=0$ and is $8mm$ across $3$ months later, what is the doubling time?
I think I solved it but I need to make sure I did it correctly. Here is how I got my solution. $$y(t)=t+5$$
So to find the doubling time I set that equation equal to $10$ $$10=t+5\implies t=5$$
Would that be correct on how I got to solve this problem?
If it's exponential growth it'd be $$N(t)=N(0)e^{rt}$$ So we can solve for $r$: $$r={\ln({N(t)-N(0)})\over t}$$ So we know $N(0)=5$ and $N(3)=8$ so now we have $$r={{\ln(8)-\ln(5)}\over3}$$ Now we can say that $$2=e^{{{\ln(8)-\ln(5)}\over3}t}$$We can take the natural log of each side and simplify a little to get:$$\ln2={\ln(3)\over 3}t$$ solving for $t$ give us that:$$t={3\ln(2)\over {{\ln(8)-\ln(5)}}}=4.42430954207$$