Answer
$ t\approx 19.8$ years
Work Step by Step
Graphing, the solution is at $ t\approx 19.8$ years.
(see below)
Algebraically, $ 1800=2600(1-0.51e^{-0.075t})^{3}\qquad $ ... $/\div 2600$
$\displaystyle \frac{9}{13}=(1-0.51e^{-0.075t})^{3}\qquad $ ... $/(...)^{1/3}$
$ 0.8846396=1-0.51e^{-0.075t}\qquad $ ... $/-1$
$-0.11536=-0.51e^{-0.075t}\qquad $ ... $/\div(-0.51)$
$ 0.2261968=e^{-0.075t}\qquad $ ... $/\ln(...)$
$-1.48635=-0.075t\qquad $ ... $/\div(-0.51)$
$ t\approx 19.8$
