Originally Posted by

**matheagle** I haven't done one of these in years, so I had to solved it from scratch.

Assuming that dX/dt=kX, we have the separable DE...

$\displaystyle {dX\over dt}=kX$ or $\displaystyle {dX\over X}=kdt$

integrating we have $\displaystyle \ln X=kt+c$.

Solving for X(t) we have $\displaystyle X(t)=ae^{kt}$

where $\displaystyle a=e^c=X(0)$.

From what you said, I believe that $\displaystyle X(0)=1000$

so $\displaystyle X(t)=1000e^{kt}$. Next we solve for k.

But the trick is not to do it completely.

$\displaystyle X(2)=2000=1000e^{2k}$, so $\displaystyle 2=e^{2k}$.

There is no need to try to approximate k.

The base is here 2, not e. Instead, insert this into our function...

$\displaystyle X(t)=1000e^{kt}=1000(e^{2k})^{t/2}=1000(2)^{t/2}$.

Now check to see if this makes sense. Look at $\displaystyle X(0)$ and $\displaystyle X(2)$.

Finally, set $\displaystyle 1000(2)^{t/2}=5000$, which yields

$\displaystyle t=2\ln 5/\ln2$.