Answer
$2040$
Work Step by Step
We want the $t$ for which $C(t)=2000$
We solve for $t$:
$2000=715e^{0.00356t}$
$2000/715=e^{0.00356t}\qquad $ /ln( .. )
$\ln(2000/715)=0.00356t$
$ t=\displaystyle \frac{\ln(2000/715)}{0.00356}\approx$288.93817
$ 289$ years after $1750$ is $2039$.
To the nearest decade: $2040$
