Solve for t over the real numbers:
e^(-0.9 t) = 0.73
e^(-0.9 t) = e^(-9 t/10) and 0.73 = 73/100:
e^(-9 t/10) = 73/100
Take reciporicals of both sides:
e^(9 t/10) = 100/73
Take the natural logarithm of both sides:
(9 t)/10 = log(100/73)
Multiply both sides by 10/9:
Answer: t = 10/9 log(100/73) =0.34967860.......this is the "natural log".
