Posted by ronnieday on Monday, March 19, 2012 at 12:43pm.
let the equation be
A = 100 e^(kt)
a) if amount = 95
95 = 100 e^(1k)
.95 = e^k
k = ln .95
so A = 100 e^(ln.95 t)
when t = 50
A = 100 e^(50ln.95) = 7.69 g are left
for half-life time, only 50 g of the original 100g would remain
50 = 100 e^(ln.95 t)
.5 = e^(ln.95 t)
ln.95t = ln.5
t = ln.5/ln.95 = appr13.5 years
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