Answer
$2030$
Work Step by Step
We want the $t$ for which $C(t)=390$.
Thus, we plug in the $C$ value and solve for $t$:
$390=280e^{0.00119t}$
$390/280=e^{0.00119t}\qquad $ /ln( .. )
$\ln(39/28)=0.00119t$
$ t=\displaystyle \frac{\ln(39/28)}{0.00119}\approx$278.45137
$ 278$ years after $1750$ is $2028$.
To the nearest decade: $2030$
