Posted by james on Tuesday, January 22, 2013 at 10:15pm.
For half-life question, a good equation to use is
amount = starting value (1/2)^(t/k) where k is the half-life
so :
.35 = 1(1/2)^(t/2000) , t in years
.35 = (.5)^(t/2000
log .35 = log ( (.5)^(t/2000)
log .35 = (t/2000) log .5
t/2000 = log .35/log .5
t = 2000( log .35/log .5) = 3029.14
it would take appr 3029 years
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