Question 1022389
Well, theoretically it never falls to no value, but if we take "no value" to mean a value under one penny, we need to find t such that
12000(.85)^t < .01
.85^t < 8.33 x 10^(-7)
Now take the log of both sides and solve for t...
t ln .85 < ln (8.33 x 10^(-7))
t > ln (8.33 x 10^(-7)) / ln .85
(We have to switch the inequality because ln .85  is negative)...
t > 86.16 years