Step 1 :\(0.12 = e^{-0.00012t}\)
Step 2 :Take the natural logarithm of both sides: \(\ln(0.12) = -0.00012t\)
Step 3 :Divide both sides by -0.00012 to isolate t: \(t = \frac{\ln(0.12)}{-0.00012}\)
Step 4 :Calculate t: \(t \approx 18407.5\)
Step 5 :Round to the nearest year: \(\boxed{18408}\)