Word problem (Calculus)?

Question

Not sure how to approach this, any ideas?

The purchasing power of the 1983 dollar is modeled by P(t) = 3.818e^-0.0506t where t is the number of years after 1960. In what year will the purchasing power of the 1983 dollar be $0.25?

Answer 1

.25 = 3.818e^(-.0506t), and .25/3.818 = .0655 = e^-.0506t, so ln(.0655) = -.0506t, and t = -ln(.0655)/.506 = 53.87. So the purchasing power decreases to $.025 during 2114.

Answer 2

Att=0 (1960 ) the purchasing power was 3.818
0.25 = 3.818e^-0.0506t
so t= -1/0.0506 * ln 0.25/3.818 =approx 54 years after 1960
so 2014